# Introduction

This is the final and concluding part of my series on ‘Practical Machine Learning with R and Python’. In this series I included the implementations of the most common Machine Learning algorithms in R and Python. The algorithms implemented were

1. Practical Machine Learning with R and Python – Part 1 In this initial post, I touch upon regression of a continuous target variable. Specifically I touch upon Univariate, Multivariate, Polynomial regression and KNN regression in both R and Python
2. Practical Machine Learning with R and Python – Part 2 In this post, I discuss Logistic Regression, KNN classification and Cross Validation error for both LOOCV and K-Fold in both R and Python
3. Practical Machine Learning with R and Python – Part 3 This 3rd part included feature selection in Machine Learning. Specifically I touch best fit, forward fit, backward fit, ridge(L2 regularization) & lasso (L1 regularization). The post includes equivalent code in R and Python.
4. Practical Machine Learning with R and Python – Part 4 In this part I discussed SVMs, Decision Trees, Validation, Precision-Recall, AUC and ROC curves
5. Practical Machine Learning with R and Python – Part 5  In this penultimate part, I touch upon B-splines, natural splines, smoothing spline, Generalized Additive Models(GAMs), Decision Trees, Random Forests and Gradient Boosted Treess.

In this last part I cover Unsupervised Learning. Specifically I cover the implementations of Principal Component Analysis (PCA). K-Means and Heirarchical Clustering. You can download this R Markdown file from Github at MachineLearning-RandPython-Part6

The content of this post and much more is now available as a compact book  on Amazon in both formats – as Paperback ($9.99) and a Kindle version($6.99/Rs449/). see ‘Practical Machine Learning with R and Python – Machine Learning in stereo

## 1.1a Principal Component Analysis (PCA) – R code

Principal Component Analysis is used to reduce the dimensionality of the input. In the code below 8 x 8 pixel of handwritten digits is reduced into its principal components. Then a scatter plot of the first 2 principal components give a very good visial representation of the data

library(dplyr)
library(ggplot2)
#Note: This example is adapted from an the example in the book Python Datascience handbook by
# Jake VanderPlas (https://jakevdp.github.io/PythonDataScienceHandbook/05.09-principal-component-analysis.html)

# Read the digits data (From sklearn datasets)
# Create a digits classes target variable
digitClasses <- factor(digits$X0.000000000000000000e.00.29) #Invoke the Principal Componsent analysis on columns 1-64 digitsPCA=prcomp(digits[,1:64]) # Create a dataframe of PCA df <- data.frame(digitsPCA$x)
# Bind the digit classes
df1 <- cbind(df,digitClasses)
# Plot only the first 2 Principal components as a scatter plot. This plot uses only the
# first 2 principal components
ggplot(df1,aes(x=PC1,y=PC2,col=digitClasses)) + geom_point() +
ggtitle("Top 2 Principal Components")

## 1.1 b Variance explained vs no principal components – R code

In the code below the variance explained vs the number of principal components is plotted. It can be seen that with 20 Principal components almost 90% of the variance is explained by this reduced dimensional model.

# Read the digits data (from sklearn datasets)
# Digits target
digitClasses <- factor(digits$X0.000000000000000000e.00.29) digitsPCA=prcomp(digits[,1:64]) # Get the Standard Deviation sd=digitsPCA$sdev
# Compute the variance
digitsVar=digitsPCA$sdev^2 #Compute the percent variance explained percentVarExp=digitsVar/sum(digitsVar) # Plot the percent variance exlained as a function of the number of principal components #plot(cumsum(percentVarExp), xlab="Principal Component", # ylab="Cumulative Proportion of Variance Explained", # main="Principal Components vs % Variance explained",ylim=c(0,1),type='l',lwd=2, # col="blue") ## 1.1c Principal Component Analysis (PCA) – Python code import numpy as np from sklearn.decomposition import PCA from sklearn import decomposition from sklearn import datasets import matplotlib.pyplot as plt from sklearn.datasets import load_digits # Load the digits data digits = load_digits() # Select only the first 2 principal components pca = PCA(2) # project from 64 to 2 dimensions #Compute the first 2 PCA projected = pca.fit_transform(digits.data) # Plot a scatter plot of the first 2 principal components plt.scatter(projected[:, 0], projected[:, 1], c=digits.target, edgecolor='none', alpha=0.5, cmap=plt.cm.get_cmap('spectral', 10)) plt.xlabel('PCA 1') plt.ylabel('PCA 2') plt.colorbar(); plt.title("Top 2 Principal Components") plt.savefig('fig1.png', bbox_inches='tight') ## 1.1 b Variance vs no principal components ## – Python code import numpy as np from sklearn.decomposition import PCA from sklearn import decomposition from sklearn import datasets import matplotlib.pyplot as plt from sklearn.datasets import load_digits digits = load_digits() # Select all 64 principal components pca = PCA(64) # project from 64 to 2 dimensions projected = pca.fit_transform(digits.data) # Obtain the explained variance for each principal component varianceExp= pca.explained_variance_ratio_ # Compute the total sum of variance totVarExp=np.cumsum(np.round(pca.explained_variance_ratio_, decimals=4)*100) # Plot the variance explained as a function of the number of principal components plt.plot(totVarExp) plt.xlabel('No of principal components') plt.ylabel('% variance explained') plt.title('No of Principal Components vs Total Variance explained') plt.savefig('fig2.png', bbox_inches='tight') ## 1.2a K-Means – R code In the code first the scatter plot of the first 2 Principal Components of the handwritten digits is plotted as a scatter plot. Over this plot 10 centroids of the 10 different clusters corresponding the 10 diferent digits is plotted over the original scatter plot. library(ggplot2) # Read the digits data digits= read.csv("digits.csv") # Create digit classes target variable digitClasses <- factor(digits$X0.000000000000000000e.00.29)

# Compute the Principal COmponents
digitsPCA=prcomp(digits[,1:64])

# Create a data frame of Principal components and the digit classes
df <- data.frame(digitsPCA$x) df1 <- cbind(df,digitClasses) # Pick only the first 2 principal components a<- df[,1:2] # Compute K Means of 10 clusters and allow for 1000 iterations k<-kmeans(a,10,1000) # Create a dataframe of the centroids of the clusters df2<-data.frame(k$centers)

#Plot the first 2 principal components with the K Means centroids
ggplot(df1,aes(x=PC1,y=PC2,col=digitClasses)) + geom_point() +
geom_point(data=df2,aes(x=PC1,y=PC2),col="black",size = 4) +
ggtitle("Top 2 Principal Components with KMeans clustering") 

## 1.2b K-Means – Python code

The centroids of the 10 different handwritten digits is plotted over the scatter plot of the first 2 principal components.

import numpy as np
from sklearn.decomposition import PCA
from sklearn import decomposition
from sklearn import datasets
import matplotlib.pyplot as plt
from sklearn.datasets import load_digits
from sklearn.cluster import KMeans

# Select only the 1st 2 principal components
pca = PCA(2)  # project from 64 to 2 dimensions
projected = pca.fit_transform(digits.data)

# Create 10 different clusters
kmeans = KMeans(n_clusters=10)

# Compute  the clusters
kmeans.fit(projected)
y_kmeans = kmeans.predict(projected)
# Get the cluster centroids
centers = kmeans.cluster_centers_
centers

#Create a scatter plot of the first 2 principal components
plt.scatter(projected[:, 0], projected[:, 1],
c=digits.target, edgecolor='none', alpha=0.5,
cmap=plt.cm.get_cmap('spectral', 10))
plt.xlabel('PCA 1')
plt.ylabel('PCA 2')
plt.colorbar();
# Overlay the centroids on the scatter plot
plt.scatter(centers[:, 0], centers[:, 1], c='darkblue', s=100)
plt.savefig('fig3.png', bbox_inches='tight')

## 1.3a Heirarchical clusters – R code

Herirachical clusters is another type of unsupervised learning. It successively joins the closest pair of objects (points or clusters) in succession based on some ‘distance’ metric. In this type of clustering we do not have choose the number of centroids. We can cut the created dendrogram mat an appropriate height to get a desired and reasonable number of clusters These are the following ‘distance’ metrics used while combining successive objects

• Ward
• Complete
• Single
• Average
• Centroid
# Read the IRIS dataset
iris <- datasets::iris
iris2 <- iris[,-5]
species <- iris[,5]

#Compute the distance matrix
d_iris <- dist(iris2)

# Use the 'average' method to for the clsuters
hc_iris <- hclust(d_iris, method = "average")

# Plot the clusters
plot(hc_iris)

# Cut tree into 3 groups
sub_grp <- cutree(hc_iris, k = 3)

# Number of members in each cluster
table(sub_grp)
## sub_grp
##  1  2  3
## 50 64 36
# Draw rectangles around the clusters
rect.hclust(hc_iris, k = 3, border = 2:5)

## 1.3a Heirarchical clusters – Python code

from sklearn.datasets import load_iris
import matplotlib.pyplot as plt
from scipy.cluster.hierarchy import dendrogram, linkage
# Load the IRIS data set

# Generate the linkage matrix using the average method
Z = linkage(iris.data, 'average')

#Plot the dendrogram
#dendrogram(Z)
#plt.xlabel('Data')
#plt.ylabel('Distance')
#plt.suptitle('Samples clustering', fontweight='bold', fontsize=14);
#plt.savefig('fig4.png', bbox_inches='tight')

# Conclusion

This is the last and concluding part of my series on Practical Machine Learning with R and Python. These parallel implementations of R and Python can be used as a quick reference while working on a large project. A person who is adept in one of the languages R or Python, can quickly absorb code in the other language.

Hope you find this series useful!

More interesting things to come. Watch this space!

References

1. Statistical Learning, Prof Trevor Hastie & Prof Robert Tibesherani, Online Stanford
2. Applied Machine Learning in Python Prof Kevyn-Collin Thomson, University Of Michigan, Coursera

To see all posts see ‘Index of posts

# Practical Machine Learning with R and Python – Part 5

This is the 5th and probably penultimate part of my series on ‘Practical Machine Learning with R and Python’. The earlier parts of this series included

1. Practical Machine Learning with R and Python – Part 1 In this initial post, I touch upon univariate, multivariate, polynomial regression and KNN regression in R and Python
2.Practical Machine Learning with R and Python – Part 2 In this post, I discuss Logistic Regression, KNN classification and cross validation error for both LOOCV and K-Fold in both R and Python
3.Practical Machine Learning with R and Python – Part 3 This post covered ‘feature selection’ in Machine Learning. Specifically I touch best fit, forward fit, backward fit, ridge(L2 regularization) & lasso (L1 regularization). The post includes equivalent code in R and Python.
4.Practical Machine Learning with R and Python – Part 4 In this part I discussed SVMs, Decision Trees, validation, precision recall, and roc curves

This post ‘Practical Machine Learning with R and Python – Part 5’ discusses regression with B-splines, natural splines, smoothing splines, generalized additive models (GAMS), bagging, random forest and boosting

As with my previous posts in this series, this post is largely based on the following 2 MOOC courses

1. Statistical Learning, Prof Trevor Hastie & Prof Robert Tibesherani, Online Stanford
2. Applied Machine Learning in Python Prof Kevyn-Collin Thomson, University Of Michigan, Coursera

You can download this R Markdown file and associated data files from Github at MachineLearning-RandPython-Part5

The content of this post and much more is now available as a compact book  on Amazon in both formats – as Paperback ($9.99) and a Kindle version($6.99/Rs449/). see ‘Practical Machine Learning with R and Python – Machine Learning in stereo

For this part I have used the data sets from UCI Machine Learning repository(Communities and Crime and Auto MPG)

## 1. Splines

When performing regression (continuous or logistic) between a target variable and a feature (or a set of features), a single polynomial for the entire range of the data set usually does not perform a good fit.Rather we would need to provide we could fit
regression curves for different section of the data set.

There are several techniques which do this for e.g. piecewise-constant functions, piecewise-linear functions, piecewise-quadratic/cubic/4th order polynomial functions etc. One such set of functions are the cubic splines which fit cubic polynomials to successive sections of the dataset. The points where the cubic splines join, are called ‘knots’.

Since each section has a different cubic spline, there could be discontinuities (or breaks) at these knots. To prevent these discontinuities ‘natural splines’ and ‘smoothing splines’ ensure that the seperate cubic functions have 2nd order continuity at these knots with the adjacent splines. 2nd order continuity implies that the value, 1st order derivative and 2nd order derivative at these knots are equal.

A cubic spline with knots $\alpha_{k}$ , k=1,2,3,..K is a piece-wise cubic polynomial with continuous derivative up to order 2 at each knot. We can write $y_{i} = \beta_{0} +\beta_{1}b_{1}(x_{i}) +\beta_{2}b_{2}(x_{i}) + .. + \beta_{K+3}b_{K+3}(x_{i}) + \epsilon_{i}$.
For each ($x{i},y{i}$), $b_{i}$ are called ‘basis’ functions, where  $b_{1}(x_{i})=x_{i}$$b_{2}(x_{i})=x_{i}^2$, $b_{3}(x_{i})=x_{i}^3$, $b_{k+3}(x_{i})=(x_{i} -\alpha_{k})^3$ where k=1,2,3… K The 1st and 2nd derivatives of cubic splines are continuous at the knots. Hence splines provide a smooth continuous fit to the data by fitting different splines to different sections of the data

## 1.1a Fit a 4th degree polynomial – R code

In the code below a non-linear function (a 4th order polynomial) is used to fit the data. Usually when we fit a single polynomial to the entire data set the tails of the fit tend to vary a lot particularly if there are fewer points at the ends. Splines help in reducing this variation at the extremities

library(dplyr)
library(ggplot2)
source('RFunctions-1.R')
# Read the data
df=read.csv("auto_mpg.csv",stringsAsFactors = FALSE) # Data from UCI
df1 <- as.data.frame(sapply(df,as.numeric))
#Select specific columns
df2 <- df1 %>% dplyr::select(cylinder,displacement, horsepower,weight, acceleration, year,mpg)
auto <- df2[complete.cases(df2),]
# Fit a 4th degree polynomial
fit=lm(mpg~poly(horsepower,4),data=auto)
#Display a summary of fit
summary(fit)
##
## Call:
## lm(formula = mpg ~ poly(horsepower, 4), data = auto)
##
## Residuals:
##      Min       1Q   Median       3Q      Max
## -14.8820  -2.5802  -0.1682   2.2100  16.1434
##
## Coefficients:
##                       Estimate Std. Error t value Pr(>|t|)
## (Intercept)            23.4459     0.2209 106.161   <2e-16 ***
## poly(horsepower, 4)1 -120.1377     4.3727 -27.475   <2e-16 ***
## poly(horsepower, 4)2   44.0895     4.3727  10.083   <2e-16 ***
## poly(horsepower, 4)3   -3.9488     4.3727  -0.903    0.367
## poly(horsepower, 4)4   -5.1878     4.3727  -1.186    0.236
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 4.373 on 387 degrees of freedom
## Multiple R-squared:  0.6893, Adjusted R-squared:  0.6861
## F-statistic: 214.7 on 4 and 387 DF,  p-value: < 2.2e-16
#Get the range of horsepower
hp <- range(auto$horsepower) #Create a sequence to be used for plotting hpGrid <- seq(hp[1],hp[2],by=10) #Predict for these values of horsepower. Set Standard error as TRUE pred=predict(fit,newdata=list(horsepower=hpGrid),se=TRUE) #Compute bands on either side that is 2xSE seBands=cbind(pred$fit+2*pred$se.fit,pred$fit-2*pred$se.fit) #Plot the fit with Standard Error bands plot(auto$horsepower,auto$mpg,xlim=hp,cex=.5,col="black",xlab="Horsepower", ylab="MPG", main="Polynomial of degree 4") lines(hpGrid,pred$fit,lwd=2,col="blue")
matlines(hpGrid,seBands,lwd=2,col="blue",lty=3)

## 1.1b Fit a 4th degree polynomial – Python code

import numpy as np
import pandas as pd
import os
import matplotlib.pyplot as plt
from sklearn.model_selection import train_test_split
from sklearn.preprocessing import PolynomialFeatures
from sklearn.linear_model import LinearRegression
#Read the auto data
# Select columns
autoDF1=autoDF[['mpg','cylinder','displacement','horsepower','weight','acceleration','year']]
# Convert all columns to numeric
autoDF2 = autoDF1.apply(pd.to_numeric, errors='coerce')

#Drop NAs
autoDF3=autoDF2.dropna()
autoDF3.shape
X=autoDF3[['horsepower']]
y=autoDF3['mpg']
#Create a polynomial of degree 4
poly = PolynomialFeatures(degree=4)
X_poly = poly.fit_transform(X)

# Fit a polynomial regression line
linreg = LinearRegression().fit(X_poly, y)
# Create a range of values
hpGrid = np.arange(np.min(X),np.max(X),10)
hp=hpGrid.reshape(-1,1)
# Transform to 4th degree
poly = PolynomialFeatures(degree=4)
hp_poly = poly.fit_transform(hp)

#Create a scatter plot
plt.scatter(X,y)
# Fit the prediction
ypred=linreg.predict(hp_poly)
plt.title("Poylnomial of degree 4")
fig2=plt.xlabel("Horsepower")
fig2=plt.ylabel("MPG")
# Draw the regression curve
plt.plot(hp,ypred,c="red")
plt.savefig('fig1.png', bbox_inches='tight')

## 1.1c Fit a B-Spline – R Code

In the code below a B- Spline is fit to data. The B-spline requires the manual selection of knots

#Splines
library(splines)
# Fit a B-spline to the data. Select knots at 60,75,100,150
fit=lm(mpg~bs(horsepower,df=6,knots=c(60,75,100,150)),data=auto)
# Use the fitted regresion to predict
pred=predict(fit,newdata=list(horsepower=hpGrid),se=T)
# Create a scatter plot
plot(auto$horsepower,auto$mpg,xlim=hp,cex=.5,col="black",xlab="Horsepower",
ylab="MPG", main="B-Spline with 4 knots")
#Draw lines with 2 Standard Errors on either side
lines(hpGrid,pred$fit,lwd=2) lines(hpGrid,pred$fit+2*pred$se,lty="dashed") lines(hpGrid,pred$fit-2*pred$se,lty="dashed") abline(v=c(60,75,100,150),lty=2,col="darkgreen") ## 1.1d Fit a Natural Spline – R Code Here a ‘Natural Spline’ is used to fit .The Natural Spline extrapolates beyond the boundary knots and the ends of the function are much more constrained than a regular spline or a global polynomoial where the ends can wag a lot more. Natural splines do not require the explicit selection of knots # There is no need to select the knots here. There is a smoothing parameter which # can be specified by the degrees of freedom 'df' parameter. The natural spline fit2=lm(mpg~ns(horsepower,df=4),data=auto) pred=predict(fit2,newdata=list(horsepower=hpGrid),se=T) plot(auto$horsepower,auto$mpg,xlim=hp,cex=.5,col="black",xlab="Horsepower", ylab="MPG", main="Natural Splines") lines(hpGrid,pred$fit,lwd=2)
lines(hpGrid,pred$fit+2*pred$se,lty="dashed")
lines(hpGrid,pred$fit-2*pred$se,lty="dashed")

## 1.1.e Fit a Smoothing Spline – R code

Here a smoothing spline is used. Smoothing splines also do not require the explicit setting of knots. We can change the ‘degrees of freedom(df)’ paramater to get the best fit

# Smoothing spline has a smoothing parameter, the degrees of freedom
# This is too wiggly
plot(auto$horsepower,auto$mpg,xlim=hp,cex=.5,col="black",xlab="Horsepower",
ylab="MPG", main="Smoothing Splines")

# Here df is set to 16. This has a lot of variance
fit=smooth.spline(auto$horsepower,auto$mpg,df=16)
lines(fit,col="red",lwd=2)

# We can use Cross Validation to allow the spline to pick the value of this smpopothing paramter. We do not need to set the degrees of freedom 'df'
fit=smooth.spline(auto$horsepower,auto$mpg,cv=TRUE)
lines(fit,col="blue",lwd=2)

## 1.1e Splines – Python

There isn’t as much treatment of splines in Python and SKLearn. I did find the LSQUnivariate, UnivariateSpline spline. The LSQUnivariate spline requires the explcit setting of knots

import numpy as np
import pandas as pd
import os
import matplotlib.pyplot as plt
from scipy.interpolate import LSQUnivariateSpline
autoDF.shape
autoDF.columns
autoDF1=autoDF[['mpg','cylinder','displacement','horsepower','weight','acceleration','year']]
autoDF2 = autoDF1.apply(pd.to_numeric, errors='coerce')
auto=autoDF2.dropna()
auto=auto[['horsepower','mpg']].sort_values('horsepower')

# Set the knots manually
knots=[65,75,100,150]
# Create an array for X & y
X=np.array(auto['horsepower'])
y=np.array(auto['mpg'])
# Fit a LSQunivariate spline
s = LSQUnivariateSpline(X,y,knots)

#Plot the spline
xs = np.linspace(40,230,1000)
ys = s(xs)
plt.scatter(X, y)
plt.plot(xs, ys)
plt.savefig('fig2.png', bbox_inches='tight')


## 1.2 Generalized Additiive models (GAMs)

Generalized Additive Models (GAMs) is a really powerful ML tool.

$y_{i} = \beta_{0} + f_{1}(x_{i1}) + f_{2}(x_{i2}) + .. +f_{p}(x_{ip}) + \epsilon_{i}$

In GAMs we use a different functions for each of the variables. GAMs give a much better fit since we can choose any function for the different sections

## 1.2a Generalized Additive Models (GAMs) – R Code

The plot below show the smooth spline that is fit for each of the features horsepower, cylinder, displacement, year and acceleration. We can use any function for example loess, 4rd order polynomial etc.

library(gam)
# Fit a smoothing spline for horsepower, cyliner, displacement and acceleration
gam=gam(mpg~s(horsepower,4)+s(cylinder,5)+s(displacement,4)+s(year,4)+s(acceleration,5),data=auto)
# Display the summary of the fit. This give the significance of each of the paramwetr
# Also an ANOVA is given for each combination of the features
summary(gam)
##
## Call: gam(formula = mpg ~ s(horsepower, 4) + s(cylinder, 5) + s(displacement,
##     4) + s(year, 4) + s(acceleration, 5), data = auto)
## Deviance Residuals:
##     Min      1Q  Median      3Q     Max
## -8.3190 -1.4436 -0.0261  1.2279 12.0873
##
## (Dispersion Parameter for gaussian family taken to be 6.9943)
##
##     Null Deviance: 23818.99 on 391 degrees of freedom
## Residual Deviance: 2587.881 on 370 degrees of freedom
## AIC: 1898.282
##
## Number of Local Scoring Iterations: 3
##
## Anova for Parametric Effects
##                     Df  Sum Sq Mean Sq  F value    Pr(>F)
## s(horsepower, 4)     1 15632.8 15632.8 2235.085 < 2.2e-16 ***
## s(cylinder, 5)       1   508.2   508.2   72.666 3.958e-16 ***
## s(displacement, 4)   1   374.3   374.3   53.514 1.606e-12 ***
## s(year, 4)           1  2263.2  2263.2  323.583 < 2.2e-16 ***
## s(acceleration, 5)   1   372.4   372.4   53.246 1.809e-12 ***
## Residuals          370  2587.9     7.0
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Anova for Nonparametric Effects
##                    Npar Df Npar F     Pr(F)
## (Intercept)
## s(horsepower, 4)         3 13.825 1.453e-08 ***
## s(cylinder, 5)           3 17.668 9.712e-11 ***
## s(displacement, 4)       3 44.573 < 2.2e-16 ***
## s(year, 4)               3 23.364 7.183e-14 ***
## s(acceleration, 5)       4  3.848  0.004453 **
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
par(mfrow=c(2,3))
plot(gam,se=TRUE)

## 1.2b Generalized Additive Models (GAMs) – Python Code

I did not find the equivalent of GAMs in SKlearn in Python. There was an early prototype (2012) in Github. Looks like it is still work in progress or has probably been abandoned.

## 1.3 Tree based Machine Learning Models

Tree based Machine Learning are all based on the ‘bootstrapping’ technique. In bootstrapping given a sample of size N, we create datasets of size N by sampling this original dataset with replacement. Machine Learning models are built on the different bootstrapped samples and then averaged.

Decision Trees as seen above have the tendency to overfit. There are several techniques that help to avoid this namely a) Bagging b) Random Forests c) Boosting

### Bagging, Random Forest and Gradient Boosting

Bagging: Bagging, or Bootstrap Aggregation decreases the variance of predictions, by creating separate Decisiion Tree based ML models on the different samples and then averaging these ML models

Random Forests: Bagging is a greedy algorithm and tries to produce splits based on all variables which try to minimize the error. However the different ML models have a high correlation. Random Forests remove this shortcoming, by using a variable and random set of features to split on. Hence the features chosen and the resulting trees are uncorrelated. When these ML models are averaged the performance is much better.

Boosting: Gradient Boosted Decision Trees also use an ensemble of trees but they don’t build Machine Learning models with random set of features at each step. Rather small and simple trees are built. Successive trees try to minimize the error from the earlier trees.

Out of Bag (OOB) Error: In Random Forest and Gradient Boosting for each bootstrap sample taken from the dataset, there will be samples left out. These are known as Out of Bag samples.Classification accuracy carried out on these OOB samples is known as OOB error

## 1.31a Decision Trees – R Code

The code below creates a Decision tree with the cancer training data. The summary of the fit is output. Based on the ML model, the predict function is used on test data and a confusion matrix is output.

# Read the cancer data
library(tree)
library(caret)
library(e1071)
cancer <- read.csv("cancer.csv",stringsAsFactors = FALSE)
cancer <- cancer[,2:32]
cancer$target <- as.factor(cancer$target)
train_idx <- trainTestSplit(cancer,trainPercent=75,seed=5)
train <- cancer[train_idx, ]
test <- cancer[-train_idx, ]

# Create Decision Tree
cancerStatus=tree(target~.,train)
summary(cancerStatus)
##
## Classification tree:
## tree(formula = target ~ ., data = train)
## Variables actually used in tree construction:
## [1] "worst.perimeter"      "worst.concave.points" "area.error"
## [4] "worst.texture"        "mean.texture"         "mean.concave.points"
## Number of terminal nodes:  9
## Residual mean deviance:  0.1218 = 50.8 / 417
## Misclassification error rate: 0.02347 = 10 / 426
pred <- predict(cancerStatus,newdata=test,type="class")
confusionMatrix(pred,test$target) ## Confusion Matrix and Statistics ## ## Reference ## Prediction 0 1 ## 0 49 7 ## 1 8 78 ## ## Accuracy : 0.8944 ## 95% CI : (0.8318, 0.9397) ## No Information Rate : 0.5986 ## P-Value [Acc > NIR] : 4.641e-15 ## ## Kappa : 0.7795 ## Mcnemar's Test P-Value : 1 ## ## Sensitivity : 0.8596 ## Specificity : 0.9176 ## Pos Pred Value : 0.8750 ## Neg Pred Value : 0.9070 ## Prevalence : 0.4014 ## Detection Rate : 0.3451 ## Detection Prevalence : 0.3944 ## Balanced Accuracy : 0.8886 ## ## 'Positive' Class : 0 ##  # Plot decision tree with labels plot(cancerStatus) text(cancerStatus,pretty=0) ## 1.31b Decision Trees – Cross Validation – R Code We can also perform a Cross Validation on the data to identify the Decision Tree which will give the minimum deviance. library(tree) cancer <- read.csv("cancer.csv",stringsAsFactors = FALSE) cancer <- cancer[,2:32] cancer$target <- as.factor(cancer$target) train_idx <- trainTestSplit(cancer,trainPercent=75,seed=5) train <- cancer[train_idx, ] test <- cancer[-train_idx, ] # Create Decision Tree cancerStatus=tree(target~.,train) # Execute 10 fold cross validation cvCancer=cv.tree(cancerStatus) plot(cvCancer) # Plot the plot(cvCancer$size,cvCancer$dev,type='b') prunedCancer=prune.tree(cancerStatus,best=4) plot(prunedCancer) text(prunedCancer,pretty=0) pred <- predict(prunedCancer,newdata=test,type="class") confusionMatrix(pred,test$target)
## Confusion Matrix and Statistics
##
##           Reference
## Prediction  0  1
##          0 50  7
##          1  7 78
##
##                Accuracy : 0.9014
##                  95% CI : (0.8401, 0.945)
##     No Information Rate : 0.5986
##     P-Value [Acc > NIR] : 7.988e-16
##
##                   Kappa : 0.7948
##  Mcnemar's Test P-Value : 1
##
##             Sensitivity : 0.8772
##             Specificity : 0.9176
##          Pos Pred Value : 0.8772
##          Neg Pred Value : 0.9176
##              Prevalence : 0.4014
##          Detection Rate : 0.3521
##    Detection Prevalence : 0.4014
##       Balanced Accuracy : 0.8974
##
##        'Positive' Class : 0
## 

## 1.31c Decision Trees – Python Code

Below is the Python code for creating Decision Trees. The accuracy, precision, recall and F1 score is computed on the test data set.

import numpy as np
import pandas as pd
import os
import matplotlib.pyplot as plt
from sklearn.metrics import confusion_matrix
from sklearn import tree
from sklearn.datasets import load_breast_cancer
from sklearn.model_selection import train_test_split
from sklearn.tree import DecisionTreeClassifier
from sklearn.datasets import make_classification, make_blobs
from sklearn.metrics import accuracy_score, precision_score, recall_score, f1_score
import graphviz

(X_cancer, y_cancer) = load_breast_cancer(return_X_y = True)

X_train, X_test, y_train, y_test = train_test_split(X_cancer, y_cancer,
random_state = 0)
clf = DecisionTreeClassifier().fit(X_train, y_train)

print('Accuracy of Decision Tree classifier on training set: {:.2f}'
.format(clf.score(X_train, y_train)))
print('Accuracy of Decision Tree classifier on test set: {:.2f}'
.format(clf.score(X_test, y_test)))

y_predicted=clf.predict(X_test)
confusion = confusion_matrix(y_test, y_predicted)
print('Accuracy: {:.2f}'.format(accuracy_score(y_test, y_predicted)))
print('Precision: {:.2f}'.format(precision_score(y_test, y_predicted)))
print('Recall: {:.2f}'.format(recall_score(y_test, y_predicted)))
print('F1: {:.2f}'.format(f1_score(y_test, y_predicted)))

# Plot the Decision Tree
clf = DecisionTreeClassifier(max_depth=2).fit(X_train, y_train)
dot_data = tree.export_graphviz(clf, out_file=None,
feature_names=cancer.feature_names,
class_names=cancer.target_names,
filled=True, rounded=True,
special_characters=True)
graph = graphviz.Source(dot_data)
graph
## Accuracy of Decision Tree classifier on training set: 1.00
## Accuracy of Decision Tree classifier on test set: 0.87
## Accuracy: 0.87
## Precision: 0.97
## Recall: 0.82
## F1: 0.89

## 1.31d Decision Trees – Cross Validation – Python Code

In the code below 5-fold cross validation is performed for different depths of the tree and the accuracy is computed. The accuracy on the test set seems to plateau when the depth is 8. But it is seen to increase again from 10 to 12. More analysis needs to be done here


import numpy as np
import pandas as pd
import os
import matplotlib.pyplot as plt
from sklearn.datasets import load_breast_cancer
from sklearn.tree import DecisionTreeClassifier
(X_cancer, y_cancer) = load_breast_cancer(return_X_y = True)
from sklearn.cross_validation import train_test_split, KFold
def computeCVAccuracy(X,y,folds):
accuracy=[]
foldAcc=[]
depth=[1,2,3,4,5,6,7,8,9,10,11,12]
nK=len(X)/float(folds)
xval_err=0
for i in depth:
kf = KFold(len(X),n_folds=folds)
for train_index, test_index in kf:
X_train, X_test = X.iloc[train_index], X.iloc[test_index]
y_train, y_test = y.iloc[train_index], y.iloc[test_index]
clf = DecisionTreeClassifier(max_depth = i).fit(X_train, y_train)
score=clf.score(X_test, y_test)
accuracy.append(score)

foldAcc.append(np.mean(accuracy))

return(foldAcc)

cvAccuracy=computeCVAccuracy(pd.DataFrame(X_cancer),pd.DataFrame(y_cancer),folds=10)

df1=pd.DataFrame(cvAccuracy)
df1.columns=['cvAccuracy']
df=df1.reindex([1,2,3,4,5,6,7,8,9,10,11,12])
df.plot()
plt.title("Decision Tree - 10-fold Cross Validation Accuracy vs Depth of tree")
plt.xlabel("Depth of tree")
plt.ylabel("Accuracy")
plt.savefig('fig3.png', bbox_inches='tight')

## 1.4a Random Forest – R code

A Random Forest is fit using the Boston data. The summary shows that 4 variables were randomly chosen at each split and the resulting ML model explains 88.72% of the test data. Also the variable importance is plotted. It can be seen that ‘rooms’ and ‘status’ are the most influential features in the model

library(randomForest)
df=read.csv("Boston.csv",stringsAsFactors = FALSE) # Data from MASS - SL

# Select specific columns
Boston <- df %>% dplyr::select("crimeRate","zone","indus","charles","nox","rooms","age",                          "distances","highways","tax","teacherRatio","color",
"status","medianValue")

# Fit a Random Forest on the Boston training data
rfBoston=randomForest(medianValue~.,data=Boston)
# Display the summatu of the fit. It can be seen that the MSE is 10.88
# and the percentage variance explained is 86.14%. About 4 variables were tried at each # #split for a maximum tree of 500.
# The MSE and percent variance is on Out of Bag trees
rfBoston
##
## Call:
##  randomForest(formula = medianValue ~ ., data = Boston)
##                Type of random forest: regression
##                      Number of trees: 500
## No. of variables tried at each split: 4
##
##           Mean of squared residuals: 9.521672
##                     % Var explained: 88.72
#List and plot the variable importances
importance(rfBoston)
##              IncNodePurity
## crimeRate        2602.1550
## zone              258.8057
## indus            2599.6635
## charles           240.2879
## nox              2748.8485
## rooms           12011.6178
## age              1083.3242
## distances        2432.8962
## highways          393.5599
## tax              1348.6987
## teacherRatio     2841.5151
## color             731.4387
## status          12735.4046
varImpPlot(rfBoston)

## 1.4b Random Forest-OOB and Cross Validation Error – R code

The figure below shows the OOB error and the Cross Validation error vs the ‘mtry’. Here mtry indicates the number of random features that are chosen at each split. The lowest test error occurs when mtry = 8

library(randomForest)
df=read.csv("Boston.csv",stringsAsFactors = FALSE) # Data from MASS - SL

# Select specific columns
Boston <- df %>% dplyr::select("crimeRate","zone","indus","charles","nox","rooms","age",                          "distances","highways","tax","teacherRatio","color",
"status","medianValue")
# Split as training and tst sets
train_idx <- trainTestSplit(Boston,trainPercent=75,seed=5)
train <- Boston[train_idx, ]
test <- Boston[-train_idx, ]

#Initialize OOD and testError
oobError <- NULL
testError <- NULL
# In the code below the number of variables to consider at each split is increased
# from 1 - 13(max features) and the OOB error and the MSE is computed
for(i in 1:13){
fitRF=randomForest(medianValue~.,data=train,mtry=i,ntree=400)
oobError[i] <-fitRF$mse[400] pred <- predict(fitRF,newdata=test) testError[i] <- mean((pred-test$medianValue)^2)
}

# We can see the OOB and Test Error. It can be seen that the Random Forest performs
# best with the lowers MSE at mtry=6
matplot(1:13,cbind(testError,oobError),pch=19,col=c("red","blue"),
type="b",xlab="mtry(no of varaibles at each split)", ylab="Mean Squared Error",
main="Random Forest - OOB and Test Error")
legend("topright",legend=c("OOB","Test"),pch=19,col=c("red","blue"))

## 1.4c Random Forest – Python code

The python code for Random Forest Regression is shown below. The training and test score is computed. The variable importance shows that ‘rooms’ and ‘status’ are the most influential of the variables

import numpy as np
import pandas as pd
import os
import matplotlib.pyplot as plt
from sklearn.model_selection import train_test_split
from sklearn.ensemble import RandomForestRegressor
df = pd.read_csv("Boston.csv",encoding = "ISO-8859-1")

X=df[['crimeRate','zone', 'indus','charles','nox','rooms', 'age','distances','highways','tax',
'teacherRatio','color','status']]
y=df['medianValue']

X_train, X_test, y_train, y_test = train_test_split(X, y, random_state = 0)

regr = RandomForestRegressor(max_depth=4, random_state=0)
regr.fit(X_train, y_train)

print('R-squared score (training): {:.3f}'
.format(regr.score(X_train, y_train)))
print('R-squared score (test): {:.3f}'
.format(regr.score(X_test, y_test)))

feature_names=['crimeRate','zone', 'indus','charles','nox','rooms', 'age','distances','highways','tax',
'teacherRatio','color','status']
print(regr.feature_importances_)
plt.figure(figsize=(10,6),dpi=80)
c_features=X_train.shape[1]
plt.barh(np.arange(c_features),regr.feature_importances_)
plt.xlabel("Feature importance")
plt.ylabel("Feature name")

plt.yticks(np.arange(c_features), feature_names)
plt.tight_layout()

plt.savefig('fig4.png', bbox_inches='tight')

## R-squared score (training): 0.917
## R-squared score (test): 0.734
## [ 0.03437382  0.          0.00580335  0.          0.00731004  0.36461548
##   0.00638577  0.03432173  0.0041244   0.01732328  0.01074148  0.0012638
##   0.51373683]

## 1.4d Random Forest – Cross Validation and OOB Error – Python code

As with R the ‘max_features’ determines the random number of features the random forest will use at each split. The plot shows that when max_features=8 the MSE is lowest

import numpy as np
import pandas as pd
import os
import matplotlib.pyplot as plt
from sklearn.model_selection import train_test_split
from sklearn.ensemble import RandomForestRegressor
from sklearn.model_selection import cross_val_score
df = pd.read_csv("Boston.csv",encoding = "ISO-8859-1")

X=df[['crimeRate','zone', 'indus','charles','nox','rooms', 'age','distances','highways','tax',
'teacherRatio','color','status']]
y=df['medianValue']

cvError=[]
oobError=[]
oobMSE=[]
for i in range(1,13):
regr = RandomForestRegressor(max_depth=4, n_estimators=400,max_features=i,oob_score=True,random_state=0)
mse= np.mean(cross_val_score(regr, X, y, cv=5,scoring = 'neg_mean_squared_error'))
# Since this is neg_mean_squared_error I have inverted the sign to get MSE
cvError.append(-mse)
# Fit on all data to compute OOB error
regr.fit(X, y)
# Record the OOB error for each max_features=i setting
oob = 1 - regr.oob_score_
oobError.append(oob)
# Get the Out of Bag prediction
oobPred=regr.oob_prediction_
# Compute the Mean Squared Error between OOB Prediction and target
mseOOB=np.mean(np.square(oobPred-y))
oobMSE.append(mseOOB)

# Plot the CV Error and OOB Error
# Set max_features
maxFeatures=np.arange(1,13)
cvError=pd.DataFrame(cvError,index=maxFeatures)
oobMSE=pd.DataFrame(oobMSE,index=maxFeatures)
#Plot
fig8=df.plot()
fig8=plt.title('Random forest - CV Error and OOB Error vs max_features')
fig8.figure.savefig('fig8.png', bbox_inches='tight')

#Plot the OOB Error vs max_features
plt.plot(range(1,13),oobError)
fig2=plt.title("Random Forest - OOB Error vs max_features (variable no of features)")
fig2=plt.xlabel("max_features (variable no of features)")
fig2=plt.ylabel("OOB Error")
fig2.figure.savefig('fig7.png', bbox_inches='tight')


## 1.5a Boosting – R code

Here a Gradient Boosted ML Model is built with a n.trees=5000, with a learning rate of 0.01 and depth of 4. The feature importance plot also shows that rooms and status are the 2 most important features. The MSE vs the number of trees plateaus around 2000 trees

library(gbm)
# Perform gradient boosting on the Boston data set. The distribution is gaussian since we
# doing MSE. The interaction depth specifies the number of splits
boostBoston=gbm(medianValue~.,data=train,distribution="gaussian",n.trees=5000,
shrinkage=0.01,interaction.depth=4)
#The summary gives the variable importance. The 2 most significant variables are
# number of rooms and lower status
summary(boostBoston)

##                       var    rel.inf
## rooms               rooms 42.2267200
## status             status 27.3024671
## distances       distances  7.9447972
## crimeRate       crimeRate  5.0238827
## nox                   nox  4.0616548
## teacherRatio teacherRatio  3.1991999
## age                   age  2.7909772
## color               color  2.3436295
## tax                   tax  2.1386213
## charles           charles  1.3799109
## highways         highways  0.7644026
## indus               indus  0.7236082
## zone                 zone  0.1001287
# The plots below show how each variable relates to the median value of the home. As
# the number of roomd increase the median value increases and with increase in lower status
# the median value decreases
par(mfrow=c(1,2))
#Plot the relation between the top 2 features and the target
plot(boostBoston,i="rooms")
plot(boostBoston,i="status")

# Create a sequence of trees between 100-5000 incremented by 50
nTrees=seq(100,5000,by=50)
# Predict the values for the test data
pred <- predict(boostBoston,newdata=test,n.trees=nTrees)
# Compute the mean for each of the MSE for each of the number of trees
boostError <- apply((pred-test$medianValue)^2,2,mean) #Plot the MSE vs the number of trees plot(nTrees,boostError,pch=19,col="blue",ylab="Mean Squared Error", main="Boosting Test Error") ## 1.5b Cross Validation Boosting – R code Included below is a cross validation error vs the learning rate. The lowest error is when learning rate = 0.09 cvError <- NULL s <- c(.001,0.01,0.03,0.05,0.07,0.09,0.1) for(i in seq_along(s)){ cvBoost=gbm(medianValue~.,data=train,distribution="gaussian",n.trees=5000, shrinkage=s[i],interaction.depth=4,cv.folds=5) cvError[i] <- mean(cvBoost$cv.error)
}

# Create a data frame for plotting
a <- rbind(s,cvError)
b <- as.data.frame(t(a))
# It can be seen that a shrinkage parameter of 0,05 gives the lowes CV Error
ggplot(b,aes(s,cvError)) + geom_point() + geom_line(color="blue") +
xlab("Shrinkage") + ylab("Cross Validation Error") +
ggtitle("Gradient boosted trees - Cross Validation error vs Shrinkage")

## 1.5c Boosting – Python code

A gradient boost ML model in Python is created below. The Rsquared score is computed on the training and test data.

import numpy as np
import pandas as pd
import os
import matplotlib.pyplot as plt
from sklearn.model_selection import train_test_split
from sklearn.ensemble import GradientBoostingRegressor
df = pd.read_csv("Boston.csv",encoding = "ISO-8859-1")

X=df[['crimeRate','zone', 'indus','charles','nox','rooms', 'age','distances','highways','tax',
'teacherRatio','color','status']]
y=df['medianValue']

X_train, X_test, y_train, y_test = train_test_split(X, y, random_state = 0)

regr.fit(X_train, y_train)

print('R-squared score (training): {:.3f}'
.format(regr.score(X_train, y_train)))
print('R-squared score (test): {:.3f}'
.format(regr.score(X_test, y_test)))
## R-squared score (training): 0.983
## R-squared score (test): 0.821

## 1.5c Cross Validation Boosting – Python code

the cross validation error is computed as the learning rate is varied. The minimum CV eror occurs when lr = 0.04

import numpy as np
import pandas as pd
import os
import matplotlib.pyplot as plt
from sklearn.model_selection import train_test_split
from sklearn.ensemble import RandomForestRegressor
from sklearn.ensemble import GradientBoostingRegressor
from sklearn.model_selection import cross_val_score
df = pd.read_csv("Boston.csv",encoding = "ISO-8859-1")

X=df[['crimeRate','zone', 'indus','charles','nox','rooms', 'age','distances','highways','tax',
'teacherRatio','color','status']]
y=df['medianValue']

cvError=[]
learning_rate =[.001,0.01,0.03,0.05,0.07,0.09,0.1]
for lr in learning_rate:
regr = GradientBoostingRegressor(max_depth=4, n_estimators=400,learning_rate  =lr,random_state=0)
mse= np.mean(cross_val_score(regr, X, y, cv=10,scoring = 'neg_mean_squared_error'))
# Since this is neg_mean_squared_error I have inverted the sign to get MSE
cvError.append(-mse)
learning_rate =[.001,0.01,0.03,0.05,0.07,0.09,0.1]
plt.plot(learning_rate,cvError)
plt.title("Gradient Boosting - 5-fold CV- Mean Squared Error vs max_features (variable no of features)")
plt.xlabel("max_features (variable no of features)")
plt.ylabel("Mean Squared Error")
plt.savefig('fig6.png', bbox_inches='tight')

Conclusion This post covered Splines and Tree based ML models like Bagging, Random Forest and Boosting. Stay tuned for further updates.

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To see all posts see Index of posts

# Practical Machine Learning with R and Python – Part 4

This is the 4th installment of my ‘Practical Machine Learning with R and Python’ series. In this part I discuss classification with Support Vector Machines (SVMs), using both a Linear and a Radial basis kernel, and Decision Trees. Further, a closer look is taken at some of the metrics associated with binary classification, namely accuracy vs precision and recall. I also touch upon Validation curves, Precision-Recall, ROC curves and AUC with equivalent code in R and Python

This post is a continuation of my 3 earlier posts on Practical Machine Learning in R and Python
1. Practical Machine Learning with R and Python – Part 1
2. Practical Machine Learning with R and Python – Part 2
3. Practical Machine Learning with R and Python – Part 3

The RMarkdown file with the code and the associated data files can be downloaded from Github at MachineLearning-RandPython-Part4

The content of this post and much more is now available as a compact book  on Amazon in both formats – as Paperback ($9.99) and a Kindle version($6.99/Rs449/). see ‘Practical Machine Learning with R and Python – Machine Learning in stereo

Support Vector Machines (SVM) are another useful Machine Learning model that can be used for both regression and classification problems. SVMs used in classification, compute the hyperplane, that separates the 2 classes with the maximum margin. To do this the features may be transformed into a larger multi-dimensional feature space. SVMs can be used with different kernels namely linear, polynomial or radial basis to determine the best fitting model for a given classification problem.

In the 2nd part of this series Practical Machine Learning with R and Python – Part 2, I had mentioned the various metrics that are used in classification ML problems namely Accuracy, Precision, Recall and F1 score. Accuracy gives the fraction of data that were correctly classified as belonging to the +ve or -ve class. However ‘accuracy’ in itself is not a good enough measure because it does not take into account the fraction of the data that were incorrectly classified. This issue becomes even more critical in different domains. For e.g a surgeon who would like to detect cancer, would like to err on the side of caution, and classify even a possibly non-cancerous patient as possibly having cancer, rather than mis-classifying a malignancy as benign. Here we would like to increase recall or sensitivity which is  given by Recall= TP/(TP+FN) or we try reduce mis-classification by either increasing the (true positives) TP or reducing (false negatives) FN

On the other hand, search algorithms would like to increase precision which tries to reduce the number of irrelevant results in the search result. Precision= TP/(TP+FP). In other words we do not want ‘false positives’ or irrelevant results to come in the search results and there is a need to reduce the false positives.

When we try to increase ‘precision’, we do so at the cost of ‘recall’, and vice-versa. I found this diagram and explanation in Wikipedia very useful Source: Wikipedia

“Consider a brain surgeon tasked with removing a cancerous tumor from a patient’s brain. The surgeon needs to remove all of the tumor cells since any remaining cancer cells will regenerate the tumor. Conversely, the surgeon must not remove healthy brain cells since that would leave the patient with impaired brain function. The surgeon may be more liberal in the area of the brain she removes to ensure she has extracted all the cancer cells. This decision increases recall but reduces precision. On the other hand, the surgeon may be more conservative in the brain she removes to ensure she extracts only cancer cells. This decision increases precision but reduces recall. That is to say, greater recall increases the chances of removing healthy cells (negative outcome) and increases the chances of removing all cancer cells (positive outcome). Greater precision decreases the chances of removing healthy cells (positive outcome) but also decreases the chances of removing all cancer cells (negative outcome).”

## 1.1a. Linear SVM – R code

In R code below I use SVM with linear kernel

source('RFunctions-1.R')
library(dplyr)
library(e1071)
library(caret)
library(reshape2)
library(ggplot2)
# Read data. Data from SKLearn
cancer$target <- as.factor(cancer$target)

# Split into training and test sets
train_idx <- trainTestSplit(cancer,trainPercent=75,seed=5)
train <- cancer[train_idx, ]
test <- cancer[-train_idx, ]

# Fit a linear basis kernel. DO not scale the data
svmfit=svm(target~., data=train, kernel="linear",scale=FALSE)
ypred=predict(svmfit,test)
#Print a confusion matrix
confusionMatrix(ypred,test$target) ## Confusion Matrix and Statistics ## ## Reference ## Prediction 0 1 ## 0 54 3 ## 1 3 82 ## ## Accuracy : 0.9577 ## 95% CI : (0.9103, 0.9843) ## No Information Rate : 0.5986 ## P-Value [Acc > NIR] : <2e-16 ## ## Kappa : 0.9121 ## Mcnemar's Test P-Value : 1 ## ## Sensitivity : 0.9474 ## Specificity : 0.9647 ## Pos Pred Value : 0.9474 ## Neg Pred Value : 0.9647 ## Prevalence : 0.4014 ## Detection Rate : 0.3803 ## Detection Prevalence : 0.4014 ## Balanced Accuracy : 0.9560 ## ## 'Positive' Class : 0 ##  ## 1.1b Linear SVM – Python code The code below creates a SVM with linear basis in Python and also dumps the corresponding classification metrics import numpy as np import pandas as pd import os import matplotlib.pyplot as plt from sklearn.model_selection import train_test_split from sklearn.svm import LinearSVC from sklearn.datasets import make_classification, make_blobs from sklearn.metrics import confusion_matrix from matplotlib.colors import ListedColormap from sklearn.datasets import load_breast_cancer # Load the cancer data (X_cancer, y_cancer) = load_breast_cancer(return_X_y = True) X_train, X_test, y_train, y_test = train_test_split(X_cancer, y_cancer, random_state = 0) clf = LinearSVC().fit(X_train, y_train) print('Breast cancer dataset') print('Accuracy of Linear SVC classifier on training set: {:.2f}' .format(clf.score(X_train, y_train))) print('Accuracy of Linear SVC classifier on test set: {:.2f}' .format(clf.score(X_test, y_test))) ## Breast cancer dataset ## Accuracy of Linear SVC classifier on training set: 0.92 ## Accuracy of Linear SVC classifier on test set: 0.94 ## 1.2 Dummy classifier Often when we perform classification tasks using any ML model namely logistic regression, SVM, neural networks etc. it is very useful to determine how well the ML model performs agains at dummy classifier. A dummy classifier uses some simple computation like frequency of majority class, instead of fitting and ML model. It is essential that our ML model does much better that the dummy classifier. This problem is even more important in imbalanced classes where we have only about 10% of +ve samples. If any ML model we create has a accuracy of about 0.90 then it is evident that our classifier is not doing any better than a dummy classsfier which can just take a majority count of this imbalanced class and also come up with 0.90. We need to be able to do better than that. In the examples below (1.3a & 1.3b) it can be seen that SVMs with ‘radial basis’ kernel with unnormalized data, for both R and Python, do not perform any better than the dummy classifier. ## 1.2a Dummy classifier – R code R does not seem to have an explicit dummy classifier. I created a simple dummy classifier that predicts the majority class. SKlearn in Python also includes other strategies like uniform, stratified etc. but this should be possible to create in R also. # Create a simple dummy classifier that computes the ratio of the majority class to the totla DummyClassifierAccuracy <- function(train,test,type="majority"){ if(type=="majority"){ count <- sum(train$target==1)/dim(train)[1]
}
count
}

cancer$target <- as.factor(cancer$target)

# Create training and test sets
train_idx <- trainTestSplit(cancer,trainPercent=75,seed=5)
train <- cancer[train_idx, ]
test <- cancer[-train_idx, ]

#Dummy classifier majority class
acc=DummyClassifierAccuracy(train,test)
sprintf("Accuracy is %f",acc)
## [1] "Accuracy is 0.638498"

## 1.2b Dummy classifier – Python code

This dummy classifier uses the majority class.

import numpy as np
import pandas as pd
import os
import matplotlib.pyplot as plt
from sklearn.datasets import load_breast_cancer
from sklearn.model_selection import train_test_split
from sklearn.dummy import DummyClassifier
from sklearn.metrics import confusion_matrix
(X_cancer, y_cancer) = load_breast_cancer(return_X_y = True)
X_train, X_test, y_train, y_test = train_test_split(X_cancer, y_cancer,
random_state = 0)

# Negative class (0) is most frequent
dummy_majority = DummyClassifier(strategy = 'most_frequent').fit(X_train, y_train)
y_dummy_predictions = dummy_majority.predict(X_test)

print('Dummy classifier accuracy on test set: {:.2f}'
.format(dummy_majority.score(X_test, y_test)))

## Dummy classifier accuracy on test set: 0.63

## 1.3a – Radial SVM (un-normalized) – R code

SVMs perform better when the data is normalized or scaled. The 2 examples below show that SVM with radial basis kernel does not perform any better than the dummy classifier

library(dplyr)
library(e1071)
library(caret)
library(reshape2)
library(ggplot2)

# Radial SVM unnormalized
train_idx <- trainTestSplit(cancer,trainPercent=75,seed=5)
train <- cancer[train_idx, ]
test <- cancer[-train_idx, ]
# Unnormalized data
ypred=predict(svmfit,test)
confusionMatrix(ypred,test$target) ## Confusion Matrix and Statistics ## ## Reference ## Prediction 0 1 ## 0 0 0 ## 1 57 85 ## ## Accuracy : 0.5986 ## 95% CI : (0.5131, 0.6799) ## No Information Rate : 0.5986 ## P-Value [Acc > NIR] : 0.5363 ## ## Kappa : 0 ## Mcnemar's Test P-Value : 1.195e-13 ## ## Sensitivity : 0.0000 ## Specificity : 1.0000 ## Pos Pred Value : NaN ## Neg Pred Value : 0.5986 ## Prevalence : 0.4014 ## Detection Rate : 0.0000 ## Detection Prevalence : 0.0000 ## Balanced Accuracy : 0.5000 ## ## 'Positive' Class : 0 ##  ## 1.4b – Radial SVM (un-normalized) – Python code import numpy as np import pandas as pd import os import matplotlib.pyplot as plt from sklearn.datasets import load_breast_cancer from sklearn.model_selection import train_test_split from sklearn.svm import SVC # Load the cancer data (X_cancer, y_cancer) = load_breast_cancer(return_X_y = True) X_train, X_test, y_train, y_test = train_test_split(X_cancer, y_cancer, random_state = 0) clf = SVC(C=10).fit(X_train, y_train) print('Breast cancer dataset (unnormalized features)') print('Accuracy of RBF-kernel SVC on training set: {:.2f}' .format(clf.score(X_train, y_train))) print('Accuracy of RBF-kernel SVC on test set: {:.2f}' .format(clf.score(X_test, y_test))) ## Breast cancer dataset (unnormalized features) ## Accuracy of RBF-kernel SVC on training set: 1.00 ## Accuracy of RBF-kernel SVC on test set: 0.63 ## 1.5a – Radial SVM (Normalized) -R Code The data is scaled (normalized ) before using the SVM model. The SVM model has 2 paramaters a) C – Large C (less regularization), more regularization b) gamma – Small gamma has larger decision boundary with more misclassfication, and larger gamma has tighter decision boundary The R code below computes the accuracy as the regularization paramater is changed trainingAccuracy <- NULL testAccuracy <- NULL C1 <- c(.01,.1, 1, 10, 20) for(i in C1){ svmfit=svm(target~., data=train, kernel="radial",cost=i,scale=TRUE) ypredTrain <-predict(svmfit,train) ypredTest=predict(svmfit,test) a <-confusionMatrix(ypredTrain,train$target)
b <-confusionMatrix(ypredTest,test$target) trainingAccuracy <-c(trainingAccuracy,a$overall[1])
testAccuracy <-c(testAccuracy,b$overall[1]) } print(trainingAccuracy) ## Accuracy Accuracy Accuracy Accuracy Accuracy ## 0.6384977 0.9671362 0.9906103 0.9976526 1.0000000 print(testAccuracy) ## Accuracy Accuracy Accuracy Accuracy Accuracy ## 0.5985915 0.9507042 0.9647887 0.9507042 0.9507042 a <-rbind(C1,as.numeric(trainingAccuracy),as.numeric(testAccuracy)) b <- data.frame(t(a)) names(b) <- c("C1","trainingAccuracy","testAccuracy") df <- melt(b,id="C1") ggplot(df) + geom_line(aes(x=C1, y=value, colour=variable),size=2) + xlab("C (SVC regularization)value") + ylab("Accuracy") + ggtitle("Training and test accuracy vs C(regularization)") ## 1.5b – Radial SVM (normalized) – Python The Radial basis kernel is used on normalized data for a range of ‘C’ values and the result is plotted. import numpy as np import pandas as pd import os import matplotlib.pyplot as plt from sklearn.datasets import load_breast_cancer from sklearn.model_selection import train_test_split from sklearn.svm import SVC from sklearn.preprocessing import MinMaxScaler scaler = MinMaxScaler() # Load the cancer data (X_cancer, y_cancer) = load_breast_cancer(return_X_y = True) X_train, X_test, y_train, y_test = train_test_split(X_cancer, y_cancer, random_state = 0) X_train_scaled = scaler.fit_transform(X_train) X_test_scaled = scaler.transform(X_test) print('Breast cancer dataset (normalized with MinMax scaling)') trainingAccuracy=[] testAccuracy=[] for C1 in [.01,.1, 1, 10, 20]: clf = SVC(C=C1).fit(X_train_scaled, y_train) acctrain=clf.score(X_train_scaled, y_train) accTest=clf.score(X_test_scaled, y_test) trainingAccuracy.append(acctrain) testAccuracy.append(accTest) # Create a dataframe C1=[.01,.1, 1, 10, 20] trainingAccuracy=pd.DataFrame(trainingAccuracy,index=C1) testAccuracy=pd.DataFrame(testAccuracy,index=C1) # Plot training and test R squared as a function of alpha df=pd.concat([trainingAccuracy,testAccuracy],axis=1) df.columns=['trainingAccuracy','trainingAccuracy'] fig1=df.plot() fig1=plt.title('Training and test accuracy vs C (SVC)') fig1.figure.savefig('fig1.png', bbox_inches='tight') ## Breast cancer dataset (normalized with MinMax scaling) Output image: ## 1.6a Validation curve – R code Sklearn includes code creating validation curves by varying paramaters and computing and plotting accuracy as gamma or C or changd. I did not find this R but I think this is a useful function and so I have created the R equivalent of this. # The R equivalent of np.logspace seqLogSpace <- function(start,stop,len){ a=seq(log10(10^start),log10(10^stop),length=len) 10^a } # Read the data. This is taken the SKlearn cancer data cancer <- read.csv("cancer.csv") cancer$target <- as.factor(cancer$target) set.seed(6) # Create the range of C1 in log space param_range = seqLogSpace(-3,2,20) # Initialize the overall training and test accuracy to NULL overallTrainAccuracy <- NULL overallTestAccuracy <- NULL # Loop over the parameter range of Gamma for(i in param_range){ # Set no of folds noFolds=5 # Create the rows which fall into different folds from 1..noFolds folds = sample(1:noFolds, nrow(cancer), replace=TRUE) # Initialize the training and test accuracy of folds to 0 trainingAccuracy <- 0 testAccuracy <- 0 # Loop through the folds for(j in 1:noFolds){ # The training is all rows for which the row is != j (k-1 folds -> training) train <- cancer[folds!=j,] # The rows which have j as the index become the test set test <- cancer[folds==j,] # Create a SVM model for this svmfit=svm(target~., data=train, kernel="radial",gamma=i,scale=TRUE) # Add up all the fold accuracy for training and test separately ypredTrain <-predict(svmfit,train) ypredTest=predict(svmfit,test) # Create confusion matrix a <-confusionMatrix(ypredTrain,train$target)
b <-confusionMatrix(ypredTest,test$target) # Get the accuracy trainingAccuracy <-trainingAccuracy + a$overall[1]
testAccuracy <-testAccuracy+b$overall[1] } # Compute the average of accuracy for K folds for number of features 'i' overallTrainAccuracy=c(overallTrainAccuracy,trainingAccuracy/noFolds) overallTestAccuracy=c(overallTestAccuracy,testAccuracy/noFolds) } #Create a dataframe a <- rbind(param_range,as.numeric(overallTrainAccuracy), as.numeric(overallTestAccuracy)) b <- data.frame(t(a)) names(b) <- c("C1","trainingAccuracy","testAccuracy") df <- melt(b,id="C1") #Plot in log axis ggplot(df) + geom_line(aes(x=C1, y=value, colour=variable),size=2) + xlab("C (SVC regularization)value") + ylab("Accuracy") + ggtitle("Training and test accuracy vs C(regularization)") + scale_x_log10() ## 1.6b Validation curve – Python Compute and plot the validation curve as gamma is varied. import numpy as np import pandas as pd import os import matplotlib.pyplot as plt from sklearn.datasets import load_breast_cancer from sklearn.model_selection import train_test_split from sklearn.preprocessing import MinMaxScaler from sklearn.svm import SVC from sklearn.model_selection import validation_curve # Load the cancer data (X_cancer, y_cancer) = load_breast_cancer(return_X_y = True) scaler = MinMaxScaler() X_scaled = scaler.fit_transform(X_cancer) # Create a gamma values from 10^-3 to 10^2 with 20 equally spaced intervals param_range = np.logspace(-3, 2, 20) # Compute the validation curve train_scores, test_scores = validation_curve(SVC(), X_scaled, y_cancer, param_name='gamma', param_range=param_range, cv=10) #Plot the figure fig2=plt.figure() #Compute the mean train_scores_mean = np.mean(train_scores, axis=1) train_scores_std = np.std(train_scores, axis=1) test_scores_mean = np.mean(test_scores, axis=1) test_scores_std = np.std(test_scores, axis=1) fig2=plt.title('Validation Curve with SVM') fig2=plt.xlabel('$\gamma$(gamma)') fig2=plt.ylabel('Score') fig2=plt.ylim(0.0, 1.1) lw = 2 fig2=plt.semilogx(param_range, train_scores_mean, label='Training score', color='darkorange', lw=lw) fig2=plt.fill_between(param_range, train_scores_mean - train_scores_std, train_scores_mean + train_scores_std, alpha=0.2, color='darkorange', lw=lw) fig2=plt.semilogx(param_range, test_scores_mean, label='Cross-validation score', color='navy', lw=lw) fig2=plt.fill_between(param_range, test_scores_mean - test_scores_std, test_scores_mean + test_scores_std, alpha=0.2, color='navy', lw=lw) fig2.figure.savefig('fig2.png', bbox_inches='tight')  Output image: ## 1.7a Validation Curve (Preventing data leakage) – Python code In this course Applied Machine Learning in Python, the Professor states that when we apply the same data transformation to a entire dataset, it will cause a data leakage. “The proper way to do cross-validation when you need to scale the data is not to scale the entire dataset with a single transform, since this will indirectly leak information into the training data about the whole dataset, including the test data (see the lecture on data leakage later in the course). Instead, scaling/normalizing must be computed and applied for each cross-validation fold separately” So I apply separate scaling to the training and testing folds and plot. In the lecture the Prof states that this can be done using pipelines. import numpy as np import pandas as pd import os import matplotlib.pyplot as plt from sklearn.datasets import load_breast_cancer from sklearn.cross_validation import KFold from sklearn.preprocessing import MinMaxScaler from sklearn.svm import SVC # Read the data (X_cancer, y_cancer) = load_breast_cancer(return_X_y = True) # Set the parameter range param_range = np.logspace(-3, 2, 20) # Set number of folds folds=5 #Initialize overallTrainAccuracy=[] overallTestAccuracy=[] # Loop over the paramater range for c in param_range: trainingAccuracy=0 testAccuracy=0 kf = KFold(len(X_cancer),n_folds=folds) # Partition into training and test folds for train_index, test_index in kf: # Partition the data acccording the fold indices generated X_train, X_test = X_cancer[train_index], X_cancer[test_index] y_train, y_test = y_cancer[train_index], y_cancer[test_index] # Scale the X_train and X_test scaler = MinMaxScaler() X_train_scaled = scaler.fit_transform(X_train) X_test_scaled = scaler.transform(X_test) # Fit a SVC model for each C clf = SVC(C=c).fit(X_train_scaled, y_train) #Compute the training and test score acctrain=clf.score(X_train_scaled, y_train) accTest=clf.score(X_test_scaled, y_test) trainingAccuracy += np.sum(acctrain) testAccuracy += np.sum(accTest) # Compute the mean training and testing accuracy overallTrainAccuracy.append(trainingAccuracy/folds) overallTestAccuracy.append(testAccuracy/folds) overallTrainAccuracy=pd.DataFrame(overallTrainAccuracy,index=param_range) overallTestAccuracy=pd.DataFrame(overallTestAccuracy,index=param_range) # Plot training and test R squared as a function of alpha df=pd.concat([overallTrainAccuracy,overallTestAccuracy],axis=1) df.columns=['trainingAccuracy','testAccuracy'] fig3=plt.title('Validation Curve with SVM') fig3=plt.xlabel('$\gamma$(gamma)') fig3=plt.ylabel('Score') fig3=plt.ylim(0.5, 1.1) lw = 2 fig3=plt.semilogx(param_range, overallTrainAccuracy, label='Training score', color='darkorange', lw=lw) fig3=plt.semilogx(param_range, overallTestAccuracy, label='Cross-validation score', color='navy', lw=lw) fig3=plt.legend(loc='best') fig3.figure.savefig('fig3.png', bbox_inches='tight')  Output image: ## 1.8 a Decision trees – R code Decision trees in R can be plotted using RPart package library(rpart) library(rpart.plot) rpart = NULL # Create a decision tree m <-rpart(Species~.,data=iris) #Plot rpart.plot(m,extra=2,main="Decision Tree - IRIS") ## 1.8 b Decision trees – Python code from sklearn.datasets import load_iris from sklearn.tree import DecisionTreeClassifier from sklearn import tree from sklearn.model_selection import train_test_split import graphviz iris = load_iris() X_train, X_test, y_train, y_test = train_test_split(iris.data, iris.target, random_state = 3) clf = DecisionTreeClassifier().fit(X_train, y_train) print('Accuracy of Decision Tree classifier on training set: {:.2f}' .format(clf.score(X_train, y_train))) print('Accuracy of Decision Tree classifier on test set: {:.2f}' .format(clf.score(X_test, y_test))) dot_data = tree.export_graphviz(clf, out_file=None, feature_names=iris.feature_names, class_names=iris.target_names, filled=True, rounded=True, special_characters=True) graph = graphviz.Source(dot_data) graph ## Accuracy of Decision Tree classifier on training set: 1.00 ## Accuracy of Decision Tree classifier on test set: 0.97 ## 1.9a Feature importance – R code I found the following code which had a snippet for feature importance. Sklean has a nice method for this. For some reason the results in R and Python are different. Any thoughts? set.seed(3) # load the library library(mlbench) library(caret) # load the dataset cancer <- read.csv("cancer.csv") cancer$target <- as.factor(cancer$target) # Split as data data <- cancer[,1:31] target <- cancer[,32] # Train the model model <- train(data, target, method="rf", preProcess="scale", trControl=trainControl(method = "cv")) # Compute variable importance importance <- varImp(model) # summarize importance print(importance) # plot importance plot(importance) ## 1.9b Feature importance – Python code import numpy as np import pandas as pd import os import matplotlib.pyplot as plt from sklearn.tree import DecisionTreeClassifier from sklearn.model_selection import train_test_split from sklearn.datasets import load_breast_cancer import numpy as np # Read the data cancer= load_breast_cancer() (X_cancer, y_cancer) = load_breast_cancer(return_X_y = True) X_train, X_test, y_train, y_test = train_test_split(X_cancer, y_cancer, random_state = 0) # Use the DecisionTreClassifier clf = DecisionTreeClassifier(max_depth = 4, min_samples_leaf = 8, random_state = 0).fit(X_train, y_train) c_features=len(cancer.feature_names) print('Breast cancer dataset: decision tree') print('Accuracy of DT classifier on training set: {:.2f}' .format(clf.score(X_train, y_train))) print('Accuracy of DT classifier on test set: {:.2f}' .format(clf.score(X_test, y_test))) # Plot the feature importances fig4=plt.figure(figsize=(10,6),dpi=80) fig4=plt.barh(range(c_features), clf.feature_importances_) fig4=plt.xlabel("Feature importance") fig4=plt.ylabel("Feature name") fig4=plt.yticks(np.arange(c_features), cancer.feature_names) fig4=plt.tight_layout() plt.savefig('fig4.png', bbox_inches='tight')  ## Breast cancer dataset: decision tree ## Accuracy of DT classifier on training set: 0.96 ## Accuracy of DT classifier on test set: 0.94 Output image: ## 1.10a Precision-Recall, ROC curves & AUC- R code I tried several R packages for plotting the Precision and Recall and AUC curve. PRROC seems to work well. The Precision-Recall curves show the tradeoff between precision and recall. The higher the precision, the lower the recall and vice versa.AUC curves that hug the top left corner indicate a high sensitivity,specificity and an excellent accuracy. source("RFunctions-1.R") library(dplyr) library(caret) library(e1071) library(PRROC) # Read the data (this data is from sklearn!) d <- read.csv("digits.csv") digits <- d[2:66] digits$X64 <- as.factor(digits$X64) # Split as training and test sets train_idx <- trainTestSplit(digits,trainPercent=75,seed=5) train <- digits[train_idx, ] test <- digits[-train_idx, ] # Fit a SVM model with linear basis kernel with probabilities svmfit=svm(X64~., data=train, kernel="linear",scale=FALSE,probability=TRUE) ypred=predict(svmfit,test,probability=TRUE) head(attr(ypred,"probabilities")) ## 0 1 ## 6 7.395947e-01 2.604053e-01 ## 8 9.999998e-01 1.842555e-07 ## 12 1.655178e-05 9.999834e-01 ## 13 9.649997e-01 3.500032e-02 ## 15 9.994849e-01 5.150612e-04 ## 16 9.999987e-01 1.280700e-06 # Store the probability of 0s and 1s m0<-attr(ypred,"probabilities")[,1] m1<-attr(ypred,"probabilities")[,2] # Create a dataframe of scores scores <- data.frame(m1,test$X64)

# Class 0 is data points of +ve class (in this case, digit 1) and -ve class (digit 0)
#Compute Precision Recall
pr <- pr.curve(scores.class0=scores[scores$test.X64=="1",]$m1,
scores.class1=scores[scores$test.X64=="0",]$m1,
curve=T)

# Plot precision-recall curve
plot(pr)

#Plot the ROC curve
roc<-roc.curve(m0, m1,curve=TRUE)
plot(roc)

## 1.10b Precision-Recall, ROC curves & AUC- Python code

For Python Logistic Regression is used to plot Precision Recall, ROC curve and compute AUC

import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
from sklearn.model_selection import train_test_split
from sklearn.linear_model import LogisticRegression
from sklearn.datasets import load_digits
from sklearn.metrics import precision_recall_curve
from sklearn.metrics import roc_curve, auc
X, y = dataset.data, dataset.target
#Create 2 classes -i) Digit 1 (from digit 1) ii) Digit 0 (from all other digits)
# Make a copy of the target
z= y.copy()
# Replace all non 1's as 0
z[z != 1] = 0

X_train, X_test, y_train, y_test = train_test_split(X, z, random_state=0)
# Fit a LR model
lr = LogisticRegression().fit(X_train, y_train)

#Compute the decision scores
y_scores_lr = lr.fit(X_train, y_train).decision_function(X_test)
y_score_list = list(zip(y_test[0:20], y_scores_lr[0:20]))

#Show the decision_function scores for first 20 instances
y_score_list

precision, recall, thresholds = precision_recall_curve(y_test, y_scores_lr)
closest_zero = np.argmin(np.abs(thresholds))
closest_zero_p = precision[closest_zero]
closest_zero_r = recall[closest_zero]
#Plot
plt.figure()
plt.xlim([0.0, 1.01])
plt.ylim([0.0, 1.01])
plt.plot(precision, recall, label='Precision-Recall Curve')
plt.plot(closest_zero_p, closest_zero_r, 'o', markersize = 12, fillstyle = 'none', c='r', mew=3)
plt.xlabel('Precision', fontsize=16)
plt.ylabel('Recall', fontsize=16)
plt.axes().set_aspect('equal')
plt.savefig('fig5.png', bbox_inches='tight')

#Compute and plot the ROC
y_score_lr = lr.fit(X_train, y_train).decision_function(X_test)
fpr_lr, tpr_lr, _ = roc_curve(y_test, y_score_lr)
roc_auc_lr = auc(fpr_lr, tpr_lr)

plt.figure()
plt.xlim([-0.01, 1.00])
plt.ylim([-0.01, 1.01])
plt.plot(fpr_lr, tpr_lr, lw=3, label='LogRegr ROC curve (area = {:0.2f})'.format(roc_auc_lr))
plt.xlabel('False Positive Rate', fontsize=16)
plt.ylabel('True Positive Rate', fontsize=16)
plt.title('ROC curve (1-of-10 digits classifier)', fontsize=16)
plt.legend(loc='lower right', fontsize=13)
plt.plot([0, 1], [0, 1], color='navy', lw=3, linestyle='--')
plt.axes()
plt.savefig('fig6.png', bbox_inches='tight')


output

## 1.10c Precision-Recall, ROC curves & AUC- Python code

In the code below classification probabilities are used to compute and plot precision-recall, roc and AUC

import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
from sklearn.model_selection import train_test_split
from sklearn.datasets import load_digits
from sklearn.svm import LinearSVC
from sklearn.calibration import CalibratedClassifierCV

X, y = dataset.data, dataset.target
# Make a copy of the target
z= y.copy()
# Replace all non 1's as 0
z[z != 1] = 0

X_train, X_test, y_train, y_test = train_test_split(X, z, random_state=0)
svm = LinearSVC()
# Need to use CalibratedClassifierSVC to redict probabilities for lInearSVC
clf = CalibratedClassifierCV(svm)
clf.fit(X_train, y_train)
y_proba_lr = clf.predict_proba(X_test)
from sklearn.metrics import precision_recall_curve

precision, recall, thresholds = precision_recall_curve(y_test, y_proba_lr[:,1])
closest_zero = np.argmin(np.abs(thresholds))
closest_zero_p = precision[closest_zero]
closest_zero_r = recall[closest_zero]
#plt.figure(figsize=(15,15),dpi=80)
plt.figure()
plt.xlim([0.0, 1.01])
plt.ylim([0.0, 1.01])
plt.plot(precision, recall, label='Precision-Recall Curve')
plt.plot(closest_zero_p, closest_zero_r, 'o', markersize = 12, fillstyle = 'none', c='r', mew=3)
plt.xlabel('Precision', fontsize=16)
plt.ylabel('Recall', fontsize=16)
plt.axes().set_aspect('equal')
plt.savefig('fig7.png', bbox_inches='tight')

# Practical Machine Learning with R and Python – Part 3

In this post ‘Practical Machine Learning with R and Python – Part 3’,  I discuss ‘Feature Selection’ methods. This post is a continuation of my 2 earlier posts

While applying Machine Learning techniques, the data set will usually include a large number of predictors for a target variable. It is quite likely, that not all the predictors or feature variables will have an impact on the output. Hence it is becomes necessary to choose only those features which influence the output variable thus simplifying  to a reduced feature set on which to train the ML model on. The techniques that are used are the following

• Best fit
• Forward fit
• Backward fit
• Ridge Regression or L2 regularization
• Lasso or L1 regularization

This post includes the equivalent ML code in R and Python.

All these methods remove those features which do not sufficiently influence the output. As in my previous 2 posts on “Practical Machine Learning with R and Python’, this post is largely based on the topics in the following 2 MOOC courses
1. Statistical Learning, Prof Trevor Hastie & Prof Robert Tibesherani, Online Stanford
2. Applied Machine Learning in Python Prof Kevyn-Collin Thomson, University Of Michigan, Coursera

You can download this R Markdown file and the associated data from Github – Machine Learning-RandPython-Part3.

The content of this post and much more is now available as a compact book  on Amazon in both formats – as Paperback ($9.99) and a Kindle version($6.99/Rs449/). see ‘Practical Machine Learning with R and Python – Machine Learning in stereo

# 1.1 Best Fit

For a dataset with features f1,f2,f3…fn, the ‘Best fit’ approach, chooses all possible combinations of features and creates separate ML models for each of the different combinations. The best fit algotithm then uses some filtering criteria based on Adj Rsquared, Cp, BIC or AIC to pick out the best model among all models.

Since the Best Fit approach searches the entire solution space it is computationally infeasible. The number of models that have to be searched increase exponentially as the number of predictors increase. For ‘p’ predictors a total of $2^{p}$ ML models have to be searched. This can be shown as follows

There are $C_{1}$ ways to choose single feature ML models among ‘n’ features, $C_{2}$ ways to choose 2 feature models among ‘n’ models and so on, or
$1+C_{1} + C_{2} +... + C_{n}$
= Total number of models in Best Fit.  Since from Binomial theorem we have
$(1+x)^{n} = 1+C_{1}x + C_{2}x^{2} +... + C_{n}x^{n}$
When x=1 in the equation (1) above, this becomes
$2^{n} = 1+C_{1} + C_{2} +... + C_{n}$

Hence there are $2^{n}$ models to search amongst in Best Fit. For 10 features this is $2^{10}$ or ~1000 models and for 40 features this becomes $2^{40}$ which almost 1 trillion. Usually there are datasets with 1000 or maybe even 100000 features and Best fit becomes computationally infeasible.

Anyways I have included the Best Fit approach as I use the Boston crime datasets which is available both the MASS package in R and Sklearn in Python and it has 13 features. Even this small feature set takes a bit of time since the Best fit needs to search among ~$2^{13}= 8192$  models

Initially I perform a simple Linear Regression Fit to estimate the features that are statistically insignificant. By looking at the p-values of the features it can be seen that ‘indus’ and ‘age’ features have high p-values and are not significant

# 1.1a Linear Regression – R code

source('RFunctions-1.R')
#Read the Boston crime data
df=read.csv("Boston.csv",stringsAsFactors = FALSE) # Data from MASS - SL
# Rename the columns
names(df) <-c("no","crimeRate","zone","indus","charles","nox","rooms","age",
"distances","highways","tax","teacherRatio","color","status","cost")
# Select specific columns
df1 <- df %>% dplyr::select("crimeRate","zone","indus","charles","nox","rooms","age",
"distances","highways","tax","teacherRatio","color","status","cost")
dim(df1)
## [1] 506  14
# Linear Regression fit
fit <- lm(cost~. ,data=df1)
summary(fit)
##
## Call:
## lm(formula = cost ~ ., data = df1)
##
## Residuals:
##     Min      1Q  Median      3Q     Max
## -15.595  -2.730  -0.518   1.777  26.199
##
## Coefficients:
##                Estimate Std. Error t value Pr(>|t|)
## (Intercept)   3.646e+01  5.103e+00   7.144 3.28e-12 ***
## crimeRate    -1.080e-01  3.286e-02  -3.287 0.001087 **
## zone          4.642e-02  1.373e-02   3.382 0.000778 ***
## indus         2.056e-02  6.150e-02   0.334 0.738288
## charles       2.687e+00  8.616e-01   3.118 0.001925 **
## nox          -1.777e+01  3.820e+00  -4.651 4.25e-06 ***
## rooms         3.810e+00  4.179e-01   9.116  < 2e-16 ***
## age           6.922e-04  1.321e-02   0.052 0.958229
## distances    -1.476e+00  1.995e-01  -7.398 6.01e-13 ***
## highways      3.060e-01  6.635e-02   4.613 5.07e-06 ***
## tax          -1.233e-02  3.760e-03  -3.280 0.001112 **
## teacherRatio -9.527e-01  1.308e-01  -7.283 1.31e-12 ***
## color         9.312e-03  2.686e-03   3.467 0.000573 ***
## status       -5.248e-01  5.072e-02 -10.347  < 2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 4.745 on 492 degrees of freedom
## Multiple R-squared:  0.7406, Adjusted R-squared:  0.7338
## F-statistic: 108.1 on 13 and 492 DF,  p-value: < 2.2e-16

Next we apply the different feature selection models to automatically remove features that are not significant below

# 1.1a Best Fit – R code

The Best Fit requires the ‘leaps’ R package

library(leaps)
source('RFunctions-1.R')
#Read the Boston crime data
df=read.csv("Boston.csv",stringsAsFactors = FALSE) # Data from MASS - SL
# Rename the columns
names(df) <-c("no","crimeRate","zone","indus","charles","nox","rooms","age",
"distances","highways","tax","teacherRatio","color","status","cost")
# Select specific columns
df1 <- df %>% dplyr::select("crimeRate","zone","indus","charles","nox","rooms","age",
"distances","highways","tax","teacherRatio","color","status","cost")

# Perform a best fit
bestFit=regsubsets(cost~.,df1,nvmax=13)

# Generate a summary of the fit
bfSummary=summary(bestFit)

# Plot the Residual Sum of Squares vs number of variables
plot(bfSummary$rss,xlab="Number of Variables",ylab="RSS",type="l",main="Best fit RSS vs No of features") # Get the index of the minimum value a=which.min(bfSummary$rss)
# Mark this in red
points(a,bfSummary$rss[a],col="red",cex=2,pch=20) The plot below shows that the Best fit occurs with all 13 features included. Notice that there is no significant change in RSS from 11 features onward. # Plot the CP statistic vs Number of variables plot(bfSummary$cp,xlab="Number of Variables",ylab="Cp",type='l',main="Best fit Cp vs No of features")
# Find the lowest CP value
b=which.min(bfSummary$cp) # Mark this in red points(b,bfSummary$cp[b],col="red",cex=2,pch=20)

Based on Cp metric the best fit occurs at 11 features as seen below. The values of the coefficients are also included below

# Display the set of features which provide the best fit
coef(bestFit,b)
##   (Intercept)     crimeRate          zone       charles           nox
##  36.341145004  -0.108413345   0.045844929   2.718716303 -17.376023429
##         rooms     distances      highways           tax  teacherRatio
##   3.801578840  -1.492711460   0.299608454  -0.011777973  -0.946524570
##         color        status
##   0.009290845  -0.522553457
#  Plot the BIC value
plot(bfSummary$bic,xlab="Number of Variables",ylab="BIC",type='l',main="Best fit BIC vs No of Features") # Find and mark the min value c=which.min(bfSummary$bic)
points(c,bfSummary$bic[c],col="red",cex=2,pch=20) # R has some other good plots for best fit plot(bestFit,scale="r2",main="Rsquared vs No Features") R has the following set of really nice visualizations. The plot below shows the Rsquared for a set of predictor variables. It can be seen when Rsquared starts at 0.74- indus, charles and age have not been included. plot(bestFit,scale="Cp",main="Cp vs NoFeatures") The Cp plot below for value shows indus, charles and age as not included in the Best fit plot(bestFit,scale="bic",main="BIC vs Features") ## 1.1b Best fit (Exhaustive Search ) – Python code The Python package for performing a Best Fit is the Exhaustive Feature Selector EFS. import numpy as np import pandas as pd import os import matplotlib.pyplot as plt from sklearn.model_selection import train_test_split from sklearn.linear_model import LinearRegression from mlxtend.feature_selection import ExhaustiveFeatureSelector as EFS # Read the Boston crime data df = pd.read_csv("Boston.csv",encoding = "ISO-8859-1") #Rename the columns df.columns=["no","crimeRate","zone","indus","chasRiver","NO2","rooms","age", "distances","idxHighways","taxRate","teacherRatio","color","status","cost"] # Set X and y X=df[["crimeRate","zone","indus","chasRiver","NO2","rooms","age", "distances","idxHighways","taxRate","teacherRatio","color","status"]] y=df['cost'] # Perform an Exhaustive Search. The EFS and SFS packages use 'neg_mean_squared_error'. The 'mean_squared_error' seems to have been deprecated. I think this is just the MSE with the a negative sign. lr = LinearRegression() efs1 = EFS(lr, min_features=1, max_features=13, scoring='neg_mean_squared_error', print_progress=True, cv=5) # Create a efs fit efs1 = efs1.fit(X.as_matrix(), y.as_matrix()) print('Best negtive mean squared error: %.2f' % efs1.best_score_) ## Print the IDX of the best features print('Best subset:', efs1.best_idx_)  Features: 8191/8191Best negtive mean squared error: -28.92 ## ('Best subset:', (0, 1, 4, 6, 7, 8, 9, 10, 11, 12)) The indices for the best subset are shown above. # 1.2 Forward fit Forward fit is a greedy algorithm that tries to optimize the feature selected, by minimizing the selection criteria (adj Rsqaured, Cp, AIC or BIC) at every step. For a dataset with features f1,f2,f3…fn, the forward fit starts with the NULL set. It then pick the ML model with a single feature from n features which has the highest adj Rsquared, or minimum Cp, BIC or some such criteria. After picking the 1 feature from n which satisfies the criteria the most, the next feature from the remaining n-1 features is chosen. When the 2 feature model which satisfies the selection criteria the best is chosen, another feature from the remaining n-2 features are added and so on. The forward fit is a sub-optimal algorithm. There is no guarantee that the final list of features chosen will be the best among the lot. The computation required for this is of $n + n-1 + n -2 + .. 1 = n(n+1)/2$ which is of the order of $n^{2}$. Though forward fit is a sub optimal solution it is far more computationally efficient than best fit ## 1.2a Forward fit – R code Forward fit in R determines that 11 features are required for the best fit. The features are shown below library(leaps) # Read the data df=read.csv("Boston.csv",stringsAsFactors = FALSE) # Data from MASS - SL # Rename the columns names(df) <-c("no","crimeRate","zone","indus","charles","nox","rooms","age", "distances","highways","tax","teacherRatio","color","status","cost") # Select columns df1 <- df %>% dplyr::select("crimeRate","zone","indus","charles","nox","rooms","age", "distances","highways","tax","teacherRatio","color","status","cost") #Split as training and test train_idx <- trainTestSplit(df1,trainPercent=75,seed=5) train <- df1[train_idx, ] test <- df1[-train_idx, ] # Find the best forward fit fitFwd=regsubsets(cost~.,data=train,nvmax=13,method="forward") # Compute the MSE valErrors=rep(NA,13) test.mat=model.matrix(cost~.,data=test) for(i in 1:13){ coefi=coef(fitFwd,id=i) pred=test.mat[,names(coefi)]%*%coefi valErrors[i]=mean((test$cost-pred)^2)
}

# Plot the Residual Sum of Squares
plot(valErrors,xlab="Number of Variables",ylab="Validation Error",type="l",main="Forward fit RSS vs No of features")
# Gives the index of the minimum value
a<-which.min(valErrors)
print(a)
## [1] 11
# Highlight the smallest value
points(c,valErrors[a],col="blue",cex=2,pch=20)

Forward fit R selects 11 predictors as the best ML model to predict the ‘cost’ output variable. The values for these 11 predictors are included below

#Print the 11 ccoefficients
coefi=coef(fitFwd,id=i)
coefi
##   (Intercept)     crimeRate          zone         indus       charles
##  2.397179e+01 -1.026463e-01  3.118923e-02  1.154235e-04  3.512922e+00
##           nox         rooms           age     distances      highways
## -1.511123e+01  4.945078e+00 -1.513220e-02 -1.307017e+00  2.712534e-01
##           tax  teacherRatio         color        status
## -1.330709e-02 -8.182683e-01  1.143835e-02 -3.750928e-01

## 1.2b Forward fit with Cross Validation – R code

The Python package SFS includes N Fold Cross Validation errors for forward and backward fit so I decided to add this code to R. This is not available in the ‘leaps’ R package, however the implementation is quite simple. Another implementation is also available at Statistical Learning, Prof Trevor Hastie & Prof Robert Tibesherani, Online Stanford 2.

library(dplyr)
df=read.csv("Boston.csv",stringsAsFactors = FALSE) # Data from MASS - SL
names(df) <-c("no","crimeRate","zone","indus","charles","nox","rooms","age",
"distances","highways","tax","teacherRatio","color","status","cost")

# Select columns
df1 <- df %>% dplyr::select("crimeRate","zone","indus","charles","nox","rooms","age",
"distances","highways","tax","teacherRatio","color","status","cost")

set.seed(6)
# Set max number of features
nvmax<-13
cvError <- NULL
# Loop through each features
for(i in 1:nvmax){
# Set no of folds
noFolds=5
# Create the rows which fall into different folds from 1..noFolds
folds = sample(1:noFolds, nrow(df1), replace=TRUE)
cv<-0
# Loop through the folds
for(j in 1:noFolds){
# The training is all rows for which the row is != j (k-1 folds -> training)
train <- df1[folds!=j,]
# The rows which have j as the index become the test set
test <- df1[folds==j,]
# Create a forward fitting model for this
fitFwd=regsubsets(cost~.,data=train,nvmax=13,method="forward")
# Select the number of features and get the feature coefficients
coefi=coef(fitFwd,id=i)
#Get the value of the test data
test.mat=model.matrix(cost~.,data=test)
# Multiply the tes data with teh fitted coefficients to get the predicted value
# pred = b0 + b1x1+b2x2... b13x13
pred=test.mat[,names(coefi)]%*%coefi
# Compute mean squared error
rss=mean((test$cost - pred)^2) # Add all the Cross Validation errors cv=cv+rss } # Compute the average of MSE for K folds for number of features 'i' cvError[i]=cv/noFolds } a <- seq(1,13) d <- as.data.frame(t(rbind(a,cvError))) names(d) <- c("Features","CVError") #Plot the CV Error vs No of Features ggplot(d,aes(x=Features,y=CVError),color="blue") + geom_point() + geom_line(color="blue") + xlab("No of features") + ylab("Cross Validation Error") + ggtitle("Forward Selection - Cross Valdation Error vs No of Features") Forward fit with 5 fold cross validation indicates that all 13 features are required # This gives the index of the minimum value a=which.min(cvError) print(a) ## [1] 13 #Print the 13 coefficients of these features coefi=coef(fitFwd,id=a) coefi ## (Intercept) crimeRate zone indus charles ## 36.650645380 -0.107980979 0.056237669 0.027016678 4.270631466 ## nox rooms age distances highways ## -19.000715500 3.714720418 0.019952654 -1.472533973 0.326758004 ## tax teacherRatio color status ## -0.011380750 -0.972862622 0.009549938 -0.582159093 ## 1.2c Forward fit – Python code The Backward Fit in Python uses the Sequential feature selection (SFS) package (SFS)(https://rasbt.github.io/mlxtend/user_guide/feature_selection/SequentialFeatureSelector/) Note: The Cross validation error for SFS in Sklearn is negative, possibly because it computes the ‘neg_mean_squared_error’. The earlier ‘mean_squared_error’ in the package seems to have been deprecated. I have taken the -ve of this neg_mean_squared_error. I think this would give mean_squared_error. import numpy as np import pandas as pd import os import matplotlib.pyplot as plt from sklearn.model_selection import train_test_split from sklearn.linear_model import LinearRegression from sklearn.datasets import load_boston from mlxtend.plotting import plot_sequential_feature_selection as plot_sfs import matplotlib.pyplot as plt from mlxtend.feature_selection import SequentialFeatureSelector as SFS from sklearn.linear_model import LinearRegression df = pd.read_csv("Boston.csv",encoding = "ISO-8859-1") #Rename the columns df.columns=["no","crimeRate","zone","indus","chasRiver","NO2","rooms","age", "distances","idxHighways","taxRate","teacherRatio","color","status","cost"] X=df[["crimeRate","zone","indus","chasRiver","NO2","rooms","age", "distances","idxHighways","taxRate","teacherRatio","color","status"]] y=df['cost'] lr = LinearRegression() # Create a forward fit model sfs = SFS(lr, k_features=(1,13), forward=True, # Forward fit floating=False, scoring='neg_mean_squared_error', cv=5) # Fit this on the data sfs = sfs.fit(X.as_matrix(), y.as_matrix()) # Get all the details of the forward fits a=sfs.get_metric_dict() n=[] o=[] # Compute the mean cross validation scores for i in np.arange(1,13): n.append(-np.mean(a[i]['cv_scores'])) m=np.arange(1,13) # Get the index of the minimum CV score # Plot the CV scores vs the number of features fig1=plt.plot(m,n) fig1=plt.title('Mean CV Scores vs No of features') fig1.figure.savefig('fig1.png', bbox_inches='tight') print(pd.DataFrame.from_dict(sfs.get_metric_dict(confidence_interval=0.90)).T) idx = np.argmin(n) print "No of features=",idx #Get the features indices for the best forward fit and convert to list b=list(a[idx]['feature_idx']) print(b) # Index the column names. # Features from forward fit print("Features selected in forward fit") print(X.columns[b]) ## avg_score ci_bound cv_scores \ ## 1 -42.6185 19.0465 [-23.5582499971, -41.8215743748, -73.993608929... ## 2 -36.0651 16.3184 [-18.002498199, -40.1507894517, -56.5286659068... ## 3 -34.1001 20.87 [-9.43012884381, -25.9584955394, -36.184188174... ## 4 -33.7681 20.1638 [-8.86076528781, -28.650217633, -35.7246353855... ## 5 -33.6392 20.5271 [-8.90807628524, -28.0684679108, -35.827463022... ## 6 -33.6276 19.0859 [-9.549485942, -30.9724602876, -32.6689523347,... ## 7 -32.4082 19.1455 [-10.0177149635, -28.3780298492, -30.926917231... ## 8 -32.3697 18.533 [-11.1431684243, -27.5765510172, -31.168994094... ## 9 -32.4016 21.5561 [-10.8972555995, -25.739780653, -30.1837430353... ## 10 -32.8504 22.6508 [-12.3909282079, -22.1533250755, -33.385407342... ## 11 -34.1065 24.7019 [-12.6429253721, -22.1676650245, -33.956999528... ## 12 -35.5814 25.693 [-12.7303397453, -25.0145323483, -34.211898373... ## 13 -37.1318 23.2657 [-12.4603005692, -26.0486211062, -33.074137979... ## ## feature_idx std_dev std_err ## 1 (12,) 18.9042 9.45212 ## 2 (10, 12) 16.1965 8.09826 ## 3 (10, 12, 5) 20.7142 10.3571 ## 4 (10, 3, 12, 5) 20.0132 10.0066 ## 5 (0, 10, 3, 12, 5) 20.3738 10.1869 ## 6 (0, 3, 5, 7, 10, 12) 18.9433 9.47167 ## 7 (0, 2, 3, 5, 7, 10, 12) 19.0026 9.50128 ## 8 (0, 1, 2, 3, 5, 7, 10, 12) 18.3946 9.19731 ## 9 (0, 1, 2, 3, 5, 7, 10, 11, 12) 21.3952 10.6976 ## 10 (0, 1, 2, 3, 4, 5, 7, 10, 11, 12) 22.4816 11.2408 ## 11 (0, 1, 2, 3, 4, 5, 6, 7, 10, 11, 12) 24.5175 12.2587 ## 12 (0, 1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 12) 25.5012 12.7506 ## 13 (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12) 23.0919 11.546 ## No of features= 7 ## [0, 2, 3, 5, 7, 10, 12] ## ################################################################################# ## Features selected in forward fit ## Index([u'crimeRate', u'indus', u'chasRiver', u'rooms', u'distances', ## u'teacherRatio', u'status'], ## dtype='object') ## 1.3 Backward Fit Backward fit belongs to the class of greedy algorithms which tries to optimize the feature set, by dropping a feature at every stage which results in the worst performance for a given criteria of Adj RSquared, Cp, BIC or AIC. For a dataset with features f1,f2,f3…fn, the backward fit starts with the all the features f1,f2.. fn to begin with. It then pick the ML model with a n-1 features by dropping the feature,$f_{j}$, for e.g., the inclusion of which results in the worst performance in adj Rsquared, or minimum Cp, BIC or some such criteria. At every step 1 feature is dopped. There is no guarantee that the final list of features chosen will be the best among the lot. The computation required for this is of $n + n-1 + n -2 + .. 1 = n(n+1)/2$ which is of the order of $n^{2}$. Though backward fit is a sub optimal solution it is far more computationally efficient than best fit ## 1.3a Backward fit – R code library(dplyr) # Read the data df=read.csv("Boston.csv",stringsAsFactors = FALSE) # Data from MASS - SL # Rename the columns names(df) <-c("no","crimeRate","zone","indus","charles","nox","rooms","age", "distances","highways","tax","teacherRatio","color","status","cost") # Select columns df1 <- df %>% dplyr::select("crimeRate","zone","indus","charles","nox","rooms","age", "distances","highways","tax","teacherRatio","color","status","cost") set.seed(6) # Set max number of features nvmax<-13 cvError <- NULL # Loop through each features for(i in 1:nvmax){ # Set no of folds noFolds=5 # Create the rows which fall into different folds from 1..noFolds folds = sample(1:noFolds, nrow(df1), replace=TRUE) cv<-0 for(j in 1:noFolds){ # The training is all rows for which the row is != j train <- df1[folds!=j,] # The rows which have j as the index become the test set test <- df1[folds==j,] # Create a backward fitting model for this fitFwd=regsubsets(cost~.,data=train,nvmax=13,method="backward") # Select the number of features and get the feature coefficients coefi=coef(fitFwd,id=i) #Get the value of the test data test.mat=model.matrix(cost~.,data=test) # Multiply the tes data with teh fitted coefficients to get the predicted value # pred = b0 + b1x1+b2x2... b13x13 pred=test.mat[,names(coefi)]%*%coefi # Compute mean squared error rss=mean((test$cost - pred)^2)
# Add the Residual sum of square
}
# Compute the average of MSE for K folds for number of features 'i'
cvError[i]=cv/noFolds
}
a <- seq(1,13)
d <- as.data.frame(t(rbind(a,cvError)))
names(d) <- c("Features","CVError")
# Plot the Cross Validation Error vs Number of features
ggplot(d,aes(x=Features,y=CVError),color="blue") + geom_point() + geom_line(color="blue") +
xlab("No of features") + ylab("Cross Validation Error") +
ggtitle("Backward Selection - Cross Valdation Error vs No of Features")

# This gives the index of the minimum value
a=which.min(cvError)
print(a)
## [1] 13
#Print the 13 coefficients of these features
coefi=coef(fitFwd,id=a)
coefi
##   (Intercept)     crimeRate          zone         indus       charles
##  36.650645380  -0.107980979   0.056237669   0.027016678   4.270631466
##           nox         rooms           age     distances      highways
## -19.000715500   3.714720418   0.019952654  -1.472533973   0.326758004
##           tax  teacherRatio         color        status
##  -0.011380750  -0.972862622   0.009549938  -0.582159093

Backward selection in R also indicates the 13 features and the corresponding coefficients as providing the best fit

## 1.3b Backward fit – Python code

The Backward Fit in Python uses the Sequential feature selection (SFS) package (SFS)(https://rasbt.github.io/mlxtend/user_guide/feature_selection/SequentialFeatureSelector/)

import numpy as np
import pandas as pd
import os
import matplotlib.pyplot as plt
from sklearn.model_selection import train_test_split
from sklearn.linear_model import LinearRegression
from mlxtend.plotting import plot_sequential_feature_selection as plot_sfs
import matplotlib.pyplot as plt
from mlxtend.feature_selection import SequentialFeatureSelector as SFS
from sklearn.linear_model import LinearRegression

# Read the data
df = pd.read_csv("Boston.csv",encoding = "ISO-8859-1")
#Rename the columns
df.columns=["no","crimeRate","zone","indus","chasRiver","NO2","rooms","age",
"distances","idxHighways","taxRate","teacherRatio","color","status","cost"]
X=df[["crimeRate","zone","indus","chasRiver","NO2","rooms","age",
"distances","idxHighways","taxRate","teacherRatio","color","status"]]
y=df['cost']
lr = LinearRegression()

# Create the SFS model
sfs = SFS(lr,
k_features=(1,13),
forward=False, # Backward
floating=False,
scoring='neg_mean_squared_error',
cv=5)

# Fit the model
sfs = sfs.fit(X.as_matrix(), y.as_matrix())
a=sfs.get_metric_dict()
n=[]
o=[]

# Compute the mean of the validation scores
for i in np.arange(1,13):
n.append(-np.mean(a[i]['cv_scores']))
m=np.arange(1,13)

# Plot the Validation scores vs number of features
fig2=plt.plot(m,n)
fig2=plt.title('Mean CV Scores vs No of features')
fig2.figure.savefig('fig2.png', bbox_inches='tight')

print(pd.DataFrame.from_dict(sfs.get_metric_dict(confidence_interval=0.90)).T)

# Get the index of minimum cross validation error
idx = np.argmin(n)
print "No of features=",idx
#Get the features indices for the best forward fit and convert to list
b=list(a[idx]['feature_idx'])
# Index the column names.
# Features from backward fit
print("Features selected in bacward fit")
print(X.columns[b])

##    avg_score ci_bound                                          cv_scores  \
## 1   -42.6185  19.0465  [-23.5582499971, -41.8215743748, -73.993608929...
## 2   -36.0651  16.3184  [-18.002498199, -40.1507894517, -56.5286659068...
## 3   -35.4992  13.9619  [-17.2329292677, -44.4178648308, -51.633177846...
## 4    -33.463  12.4081  [-20.6415333292, -37.3247852146, -47.479302977...
## 5   -33.1038  10.6156  [-20.2872309863, -34.6367078466, -45.931870352...
## 6   -32.0638  10.0933  [-19.4463829372, -33.460638577, -42.726257249,...
## 7   -30.7133  9.23881  [-19.4425181917, -31.1742902259, -40.531266671...
## 8   -29.7432  9.84468  [-19.445277268, -30.0641187173, -40.2561247122...
## 9   -29.0878  9.45027  [-19.3545569877, -30.094768669, -39.7506036377...
## 10  -28.9225  9.39697  [-18.562171585, -29.968504938, -39.9586835965,...
## 11  -29.4301  10.8831  [-18.3346152225, -30.3312847532, -45.065432793...
## 12  -30.4589  11.1486  [-18.493389527, -35.0290639374, -45.1558231765...
## 13  -37.1318  23.2657  [-12.4603005692, -26.0486211062, -33.074137979...
##
##                                    feature_idx  std_dev  std_err
## 1                                        (12,)  18.9042  9.45212
## 2                                     (10, 12)  16.1965  8.09826
## 3                                  (10, 12, 7)  13.8576  6.92881
## 4                               (12, 10, 4, 7)  12.3154  6.15772
## 5                            (4, 7, 8, 10, 12)  10.5363  5.26816
## 6                         (4, 7, 8, 9, 10, 12)  10.0179  5.00896
## 7                      (1, 4, 7, 8, 9, 10, 12)  9.16981  4.58491
## 8                  (1, 4, 7, 8, 9, 10, 11, 12)  9.77116  4.88558
## 9               (0, 1, 4, 7, 8, 9, 10, 11, 12)  9.37969  4.68985
## 10           (0, 1, 4, 6, 7, 8, 9, 10, 11, 12)   9.3268   4.6634
## 11        (0, 1, 3, 4, 6, 7, 8, 9, 10, 11, 12)  10.8018  5.40092
## 12     (0, 1, 2, 3, 4, 6, 7, 8, 9, 10, 11, 12)  11.0653  5.53265
## 13  (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12)  23.0919   11.546
## No of features= 9
## Features selected in bacward fit
## Index([u'crimeRate', u'zone', u'NO2', u'distances', u'idxHighways', u'taxRate',
##        u'teacherRatio', u'color', u'status'],
##       dtype='object')

## 1.3c Sequential Floating Forward Selection (SFFS) – Python code

The Sequential Feature search also includes ‘floating’ variants which include or exclude features conditionally, once they were excluded or included. The SFFS can conditionally include features which were excluded from the previous step, if it results in a better fit. This option will tend to a better solution, than plain simple SFS. These variants are included below

import numpy as np
import pandas as pd
import os
import matplotlib.pyplot as plt
from sklearn.model_selection import train_test_split
from sklearn.linear_model import LinearRegression
from sklearn.datasets import load_boston
from mlxtend.plotting import plot_sequential_feature_selection as plot_sfs
import matplotlib.pyplot as plt
from mlxtend.feature_selection import SequentialFeatureSelector as SFS
from sklearn.linear_model import LinearRegression

df = pd.read_csv("Boston.csv",encoding = "ISO-8859-1")
#Rename the columns
df.columns=["no","crimeRate","zone","indus","chasRiver","NO2","rooms","age",
"distances","idxHighways","taxRate","teacherRatio","color","status","cost"]
X=df[["crimeRate","zone","indus","chasRiver","NO2","rooms","age",
"distances","idxHighways","taxRate","teacherRatio","color","status"]]
y=df['cost']
lr = LinearRegression()

# Create the floating forward search
sffs = SFS(lr,
k_features=(1,13),
forward=True,  # Forward
floating=True,  #Floating
scoring='neg_mean_squared_error',
cv=5)

# Fit a model
sffs = sffs.fit(X.as_matrix(), y.as_matrix())
a=sffs.get_metric_dict()
n=[]
o=[]
# Compute mean validation scores
for i in np.arange(1,13):
n.append(-np.mean(a[i]['cv_scores']))

m=np.arange(1,13)

# Plot the cross validation score vs number of features
fig3=plt.plot(m,n)
fig3=plt.title('SFFS:Mean CV Scores vs No of features')
fig3.figure.savefig('fig3.png', bbox_inches='tight')

print(pd.DataFrame.from_dict(sffs.get_metric_dict(confidence_interval=0.90)).T)
# Get the index of the minimum CV score
idx = np.argmin(n)
print "No of features=",idx
#Get the features indices for the best forward floating fit and convert to list
b=list(a[idx]['feature_idx'])
print(b)

print("#################################################################################")
# Index the column names.
# Features from forward fit
print("Features selected in forward fit")
print(X.columns[b])
##    avg_score ci_bound                                          cv_scores  \
## 1   -42.6185  19.0465  [-23.5582499971, -41.8215743748, -73.993608929...
## 2   -36.0651  16.3184  [-18.002498199, -40.1507894517, -56.5286659068...
## 3   -34.1001    20.87  [-9.43012884381, -25.9584955394, -36.184188174...
## 4   -33.7681  20.1638  [-8.86076528781, -28.650217633, -35.7246353855...
## 5   -33.6392  20.5271  [-8.90807628524, -28.0684679108, -35.827463022...
## 6   -33.6276  19.0859  [-9.549485942, -30.9724602876, -32.6689523347,...
## 7   -32.1834  12.1001  [-17.9491036167, -39.6479234651, -45.470227740...
## 8   -32.0908  11.8179  [-17.4389015788, -41.2453629843, -44.247557798...
## 9   -31.0671  10.1581  [-17.2689542913, -37.4379370429, -41.366372300...
## 10  -28.9225  9.39697  [-18.562171585, -29.968504938, -39.9586835965,...
## 11  -29.4301  10.8831  [-18.3346152225, -30.3312847532, -45.065432793...
## 12  -30.4589  11.1486  [-18.493389527, -35.0290639374, -45.1558231765...
## 13  -37.1318  23.2657  [-12.4603005692, -26.0486211062, -33.074137979...
##
##                                    feature_idx  std_dev  std_err
## 1                                        (12,)  18.9042  9.45212
## 2                                     (10, 12)  16.1965  8.09826
## 3                                  (10, 12, 5)  20.7142  10.3571
## 4                               (10, 3, 12, 5)  20.0132  10.0066
## 5                            (0, 10, 3, 12, 5)  20.3738  10.1869
## 6                         (0, 3, 5, 7, 10, 12)  18.9433  9.47167
## 7                      (0, 1, 2, 3, 7, 10, 12)  12.0097  6.00487
## 8                   (0, 1, 2, 3, 7, 8, 10, 12)  11.7297  5.86484
## 9                (0, 1, 2, 3, 7, 8, 9, 10, 12)  10.0822  5.04111
## 10           (0, 1, 4, 6, 7, 8, 9, 10, 11, 12)   9.3268   4.6634
## 11        (0, 1, 3, 4, 6, 7, 8, 9, 10, 11, 12)  10.8018  5.40092
## 12     (0, 1, 2, 3, 4, 6, 7, 8, 9, 10, 11, 12)  11.0653  5.53265
## 13  (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12)  23.0919   11.546
## No of features= 9
## [0, 1, 2, 3, 7, 8, 9, 10, 12]
## #################################################################################
## Features selected in forward fit
## Index([u'crimeRate', u'zone', u'indus', u'chasRiver', u'distances',
##        u'idxHighways', u'taxRate', u'teacherRatio', u'status'],
##       dtype='object')

## 1.3d Sequential Floating Backward Selection (SFBS) – Python code

The SFBS is an extension of the SBS. Here features that are excluded at any stage can be conditionally included if the resulting feature set gives a better fit.

import numpy as np
import pandas as pd
import os
import matplotlib.pyplot as plt
from sklearn.model_selection import train_test_split
from sklearn.linear_model import LinearRegression
from sklearn.datasets import load_boston
from mlxtend.plotting import plot_sequential_feature_selection as plot_sfs
import matplotlib.pyplot as plt
from mlxtend.feature_selection import SequentialFeatureSelector as SFS
from sklearn.linear_model import LinearRegression

df = pd.read_csv("Boston.csv",encoding = "ISO-8859-1")
#Rename the columns
df.columns=["no","crimeRate","zone","indus","chasRiver","NO2","rooms","age",
"distances","idxHighways","taxRate","teacherRatio","color","status","cost"]
X=df[["crimeRate","zone","indus","chasRiver","NO2","rooms","age",
"distances","idxHighways","taxRate","teacherRatio","color","status"]]
y=df['cost']
lr = LinearRegression()

sffs = SFS(lr,
k_features=(1,13),
forward=False, # Backward
floating=True, # Floating
scoring='neg_mean_squared_error',
cv=5)

sffs = sffs.fit(X.as_matrix(), y.as_matrix())
a=sffs.get_metric_dict()
n=[]
o=[]
# Compute the mean cross validation score
for i in np.arange(1,13):
n.append(-np.mean(a[i]['cv_scores']))

m=np.arange(1,13)

fig4=plt.plot(m,n)
fig4=plt.title('SFBS: Mean CV Scores vs No of features')
fig4.figure.savefig('fig4.png', bbox_inches='tight')

print(pd.DataFrame.from_dict(sffs.get_metric_dict(confidence_interval=0.90)).T)

# Get the index of the minimum CV score
idx = np.argmin(n)
print "No of features=",idx
#Get the features indices for the best backward floating fit and convert to list
b=list(a[idx]['feature_idx'])
print(b)

print("#################################################################################")
# Index the column names.
# Features from forward fit
print("Features selected in backward floating fit")
print(X.columns[b])
##    avg_score ci_bound                                          cv_scores  \
## 1   -42.6185  19.0465  [-23.5582499971, -41.8215743748, -73.993608929...
## 2   -36.0651  16.3184  [-18.002498199, -40.1507894517, -56.5286659068...
## 3   -34.1001    20.87  [-9.43012884381, -25.9584955394, -36.184188174...
## 4    -33.463  12.4081  [-20.6415333292, -37.3247852146, -47.479302977...
## 5   -32.3699  11.2725  [-20.8771078371, -34.9825657934, -45.813447203...
## 6   -31.6742  11.2458  [-20.3082500364, -33.2288990522, -45.535507868...
## 7   -30.7133  9.23881  [-19.4425181917, -31.1742902259, -40.531266671...
## 8   -29.7432  9.84468  [-19.445277268, -30.0641187173, -40.2561247122...
## 9   -29.0878  9.45027  [-19.3545569877, -30.094768669, -39.7506036377...
## 10  -28.9225  9.39697  [-18.562171585, -29.968504938, -39.9586835965,...
## 11  -29.4301  10.8831  [-18.3346152225, -30.3312847532, -45.065432793...
## 12  -30.4589  11.1486  [-18.493389527, -35.0290639374, -45.1558231765...
## 13  -37.1318  23.2657  [-12.4603005692, -26.0486211062, -33.074137979...
##
##                                    feature_idx  std_dev  std_err
## 1                                        (12,)  18.9042  9.45212
## 2                                     (10, 12)  16.1965  8.09826
## 3                                  (10, 12, 5)  20.7142  10.3571
## 4                               (4, 10, 7, 12)  12.3154  6.15772
## 5                            (12, 10, 4, 1, 7)  11.1883  5.59417
## 6                        (4, 7, 8, 10, 11, 12)  11.1618  5.58088
## 7                      (1, 4, 7, 8, 9, 10, 12)  9.16981  4.58491
## 8                  (1, 4, 7, 8, 9, 10, 11, 12)  9.77116  4.88558
## 9               (0, 1, 4, 7, 8, 9, 10, 11, 12)  9.37969  4.68985
## 10           (0, 1, 4, 6, 7, 8, 9, 10, 11, 12)   9.3268   4.6634
## 11        (0, 1, 3, 4, 6, 7, 8, 9, 10, 11, 12)  10.8018  5.40092
## 12     (0, 1, 2, 3, 4, 6, 7, 8, 9, 10, 11, 12)  11.0653  5.53265
## 13  (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12)  23.0919   11.546
## No of features= 9
## [0, 1, 4, 7, 8, 9, 10, 11, 12]
## #################################################################################
## Features selected in backward floating fit
## Index([u'crimeRate', u'zone', u'NO2', u'distances', u'idxHighways', u'taxRate',
##        u'teacherRatio', u'color', u'status'],
##       dtype='object')

# 1.4 Ridge regression

In Linear Regression the Residual Sum of Squares (RSS) is given as

$RSS = \sum_{i=1}^{n} (y_{i} - \beta_{0} - \sum_{j=1}^{p}\beta_jx_{ij})^{2}$
Ridge regularization =$\sum_{i=1}^{n} (y_{i} - \beta_{0} - \sum_{j=1}^{p}\beta_jx_{ij})^{2} + \lambda \sum_{j=1}^{p}\beta^{2}$

where is the regularization or tuning parameter. Increasing increases the penalty on the coefficients thus shrinking them. However in Ridge Regression features that do not influence the target variable will shrink closer to zero but never become zero except for very large values of

Ridge regression in R requires the ‘glmnet’ package

## 1.4a Ridge Regression – R code

library(glmnet)
library(dplyr)
# Read the data
df=read.csv("Boston.csv",stringsAsFactors = FALSE) # Data from MASS - SL
#Rename the columns
names(df) <-c("no","crimeRate","zone","indus","charles","nox","rooms","age",
"distances","highways","tax","teacherRatio","color","status","cost")
# Select specific columns
df1 <- df %>% dplyr::select("crimeRate","zone","indus","charles","nox","rooms","age",
"distances","highways","tax","teacherRatio","color","status","cost")

# Set X and y as matrices
X=as.matrix(df1[,1:13])
y=df1$cost # Fit a Ridge model fitRidge <-glmnet(X,y,alpha=0) #Plot the model where the coefficient shrinkage is plotted vs log lambda plot(fitRidge,xvar="lambda",label=TRUE,main= "Ridge regression coefficient shrikage vs log lambda") The plot below shows how the 13 coefficients for the 13 predictors vary when lambda is increased. The x-axis includes log (lambda). We can see that increasing lambda from $10^{2}$ to $10^{6}$ significantly shrinks the coefficients. We can draw a vertical line from the x-axis and read the values of the 13 coefficients. Some of them will be close to zero # Compute the cross validation error cvRidge=cv.glmnet(X,y,alpha=0) #Plot the cross validation error plot(cvRidge, main="Ridge regression Cross Validation Error (10 fold)") This gives the 10 fold Cross Validation Error with respect to log (lambda) As lambda increase the MSE increases ## 1.4a Ridge Regression – Python code The coefficient shrinkage for Python can be plotted like R using Least Angle Regression model a.k.a. LARS package. This is included below import numpy as np import pandas as pd import os import matplotlib.pyplot as plt from sklearn.model_selection import train_test_split df = pd.read_csv("Boston.csv",encoding = "ISO-8859-1") #Rename the columns df.columns=["no","crimeRate","zone","indus","chasRiver","NO2","rooms","age", "distances","idxHighways","taxRate","teacherRatio","color","status","cost"] X=df[["crimeRate","zone","indus","chasRiver","NO2","rooms","age", "distances","idxHighways","taxRate","teacherRatio","color","status"]] y=df['cost'] from sklearn.preprocessing import MinMaxScaler scaler = MinMaxScaler() from sklearn.linear_model import Ridge X_train, X_test, y_train, y_test = train_test_split(X, y, random_state = 0) # Scale the X_train and X_test X_train_scaled = scaler.fit_transform(X_train) X_test_scaled = scaler.transform(X_test) # Fit a ridge regression with alpha=20 linridge = Ridge(alpha=20.0).fit(X_train_scaled, y_train) # Print the training R squared print('R-squared score (training): {:.3f}' .format(linridge.score(X_train_scaled, y_train))) # Print the test Rsquared print('R-squared score (test): {:.3f}' .format(linridge.score(X_test_scaled, y_test))) print('Number of non-zero features: {}' .format(np.sum(linridge.coef_ != 0))) trainingRsquared=[] testRsquared=[] # Plot the effect of alpha on the test Rsquared print('Ridge regression: effect of alpha regularization parameter\n') # Choose a list of alpha values for this_alpha in [0.001,.01,.1,0, 1, 10, 20, 50, 100, 1000]: linridge = Ridge(alpha = this_alpha).fit(X_train_scaled, y_train) # Compute training rsquared r2_train = linridge.score(X_train_scaled, y_train) # Compute test rsqaured r2_test = linridge.score(X_test_scaled, y_test) num_coeff_bigger = np.sum(abs(linridge.coef_) > 1.0) trainingRsquared.append(r2_train) testRsquared.append(r2_test) # Create a dataframe alpha=[0.001,.01,.1,0, 1, 10, 20, 50, 100, 1000] trainingRsquared=pd.DataFrame(trainingRsquared,index=alpha) testRsquared=pd.DataFrame(testRsquared,index=alpha) # Plot training and test R squared as a function of alpha df3=pd.concat([trainingRsquared,testRsquared],axis=1) df3.columns=['trainingRsquared','testRsquared'] fig5=df3.plot() fig5=plt.title('Ridge training and test squared error vs Alpha') fig5.figure.savefig('fig5.png', bbox_inches='tight') # Plot the coefficient shrinage using the LARS package from sklearn import linear_model # ############################################################################# # Compute paths n_alphas = 200 alphas = np.logspace(0, 8, n_alphas) coefs = [] for a in alphas: ridge = linear_model.Ridge(alpha=a, fit_intercept=False) ridge.fit(X_train_scaled, y_train) coefs.append(ridge.coef_) # ############################################################################# # Display results ax = plt.gca() fig6=ax.plot(alphas, coefs) fig6=ax.set_xscale('log') fig6=ax.set_xlim(ax.get_xlim()[::-1]) # reverse axis fig6=plt.xlabel('alpha') fig6=plt.ylabel('weights') fig6=plt.title('Ridge coefficients as a function of the regularization') fig6=plt.axis('tight') plt.savefig('fig6.png', bbox_inches='tight')  ## R-squared score (training): 0.620 ## R-squared score (test): 0.438 ## Number of non-zero features: 13 ## Ridge regression: effect of alpha regularization parameter The plot below shows the training and test error when increasing the tuning or regularization parameter ‘alpha’ For Python the coefficient shrinkage with LARS must be viewed from right to left, where you have increasing alpha. As alpha increases the coefficients shrink to 0. ## 1.5 Lasso regularization The Lasso is another form of regularization, also known as L1 regularization. Unlike the Ridge Regression where the coefficients of features which do not influence the target tend to zero, in the lasso regualrization the coefficients become 0. The general form of Lasso is as follows $\sum_{i=1}^{n} (y_{i} - \beta_{0} - \sum_{j=1}^{p}\beta_jx_{ij})^{2} + \lambda \sum_{j=1}^{p}|\beta|$ ## 1.5a Lasso regularization – R code library(glmnet) library(dplyr) df=read.csv("Boston.csv",stringsAsFactors = FALSE) # Data from MASS - SL names(df) <-c("no","crimeRate","zone","indus","charles","nox","rooms","age", "distances","highways","tax","teacherRatio","color","status","cost") df1 <- df %>% dplyr::select("crimeRate","zone","indus","charles","nox","rooms","age", "distances","highways","tax","teacherRatio","color","status","cost") # Set X and y as matrices X=as.matrix(df1[,1:13]) y=df1$cost

# Fit the lasso model
fitLasso <- glmnet(X,y)
# Plot the coefficient shrinkage as a function of log(lambda)
plot(fitLasso,xvar="lambda",label=TRUE,main="Lasso regularization - Coefficient shrinkage vs log lambda")

The plot below shows that in L1 regularization the coefficients actually become zero with increasing lambda

# Compute the cross validation error (10 fold)
cvLasso=cv.glmnet(X,y,alpha=0)
# Plot the cross validation error
plot(cvLasso)

This gives the MSE for the lasso model

## 1.5 b Lasso regularization – Python code

import numpy as np
import pandas as pd
import os
import matplotlib.pyplot as plt
from sklearn.model_selection import train_test_split
from sklearn.linear_model import Lasso
from sklearn.preprocessing import MinMaxScaler
from sklearn import linear_model

scaler = MinMaxScaler()
df = pd.read_csv("Boston.csv",encoding = "ISO-8859-1")
#Rename the columns
df.columns=["no","crimeRate","zone","indus","chasRiver","NO2","rooms","age",
"distances","idxHighways","taxRate","teacherRatio","color","status","cost"]
X=df[["crimeRate","zone","indus","chasRiver","NO2","rooms","age",
"distances","idxHighways","taxRate","teacherRatio","color","status"]]
y=df['cost']
X_train, X_test, y_train, y_test = train_test_split(X, y,
random_state = 0)

X_train_scaled = scaler.fit_transform(X_train)
X_test_scaled = scaler.transform(X_test)

linlasso = Lasso(alpha=0.1, max_iter = 10).fit(X_train_scaled, y_train)

print('Non-zero features: {}'
.format(np.sum(linlasso.coef_ != 0)))
print('R-squared score (training): {:.3f}'
.format(linlasso.score(X_train_scaled, y_train)))
print('R-squared score (test): {:.3f}\n'
.format(linlasso.score(X_test_scaled, y_test)))
print('Features with non-zero weight (sorted by absolute magnitude):')

for e in sorted (list(zip(list(X), linlasso.coef_)),
key = lambda e: -abs(e[1])):
if e[1] != 0:
print('\t{}, {:.3f}'.format(e[0], e[1]))

print('Lasso regression: effect of alpha regularization\n\
parameter on number of features kept in final model\n')

trainingRsquared=[]
testRsquared=[]
#for alpha in [0.01,0.05,0.1, 1, 2, 3, 5, 10, 20, 50]:
for alpha in [0.01,0.07,0.05, 0.1, 1,2, 3, 5, 10]:
linlasso = Lasso(alpha, max_iter = 10000).fit(X_train_scaled, y_train)
r2_train = linlasso.score(X_train_scaled, y_train)
r2_test = linlasso.score(X_test_scaled, y_test)
trainingRsquared.append(r2_train)
testRsquared.append(r2_test)

alpha=[0.01,0.07,0.05, 0.1, 1,2, 3, 5, 10]
#alpha=[0.01,0.05,0.1, 1, 2, 3, 5, 10, 20, 50]
trainingRsquared=pd.DataFrame(trainingRsquared,index=alpha)
testRsquared=pd.DataFrame(testRsquared,index=alpha)

df3=pd.concat([trainingRsquared,testRsquared],axis=1)
df3.columns=['trainingRsquared','testRsquared']

fig7=df3.plot()
fig7=plt.title('LASSO training and test squared error vs Alpha')
fig7.figure.savefig('fig7.png', bbox_inches='tight')


## Non-zero features: 7
## R-squared score (training): 0.726
## R-squared score (test): 0.561
##
## Features with non-zero weight (sorted by absolute magnitude):
##  status, -18.361
##  rooms, 18.232
##  teacherRatio, -8.628
##  taxRate, -2.045
##  color, 1.888
##  chasRiver, 1.670
##  distances, -0.529
## Lasso regression: effect of alpha regularization
## parameter on number of features kept in final model
##
## Computing regularization path using the LARS ...
## .C:\Users\Ganesh\ANACON~1\lib\site-packages\sklearn\linear_model\coordinate_descent.py:484: ConvergenceWarning: Objective did not converge. You might want to increase the number of iterations. Fitting data with very small alpha may cause precision problems.
##   ConvergenceWarning)

## 1.5c Lasso coefficient shrinkage – Python code

To plot the coefficient shrinkage for Lasso the Least Angle Regression model a.k.a. LARS package. This is shown below

import numpy as np
import pandas as pd
import os
import matplotlib.pyplot as plt
from sklearn.model_selection import train_test_split
from sklearn.linear_model import Lasso
from sklearn.preprocessing import MinMaxScaler
from sklearn import linear_model
scaler = MinMaxScaler()
df = pd.read_csv("Boston.csv",encoding = "ISO-8859-1")
#Rename the columns
df.columns=["no","crimeRate","zone","indus","chasRiver","NO2","rooms","age",
"distances","idxHighways","taxRate","teacherRatio","color","status","cost"]
X=df[["crimeRate","zone","indus","chasRiver","NO2","rooms","age",
"distances","idxHighways","taxRate","teacherRatio","color","status"]]
y=df['cost']
X_train, X_test, y_train, y_test = train_test_split(X, y,
random_state = 0)

X_train_scaled = scaler.fit_transform(X_train)
X_test_scaled = scaler.transform(X_test)

print("Computing regularization path using the LARS ...")
alphas, _, coefs = linear_model.lars_path(X_train_scaled, y_train, method='lasso', verbose=True)

xx = np.sum(np.abs(coefs.T), axis=1)
xx /= xx[-1]

fig8=plt.plot(xx, coefs.T)

ymin, ymax = plt.ylim()
fig8=plt.vlines(xx, ymin, ymax, linestyle='dashed')
fig8=plt.xlabel('|coef| / max|coef|')
fig8=plt.ylabel('Coefficients')
fig8=plt.title('LASSO Path - Coefficient Shrinkage vs L1')
fig8=plt.axis('tight')
plt.savefig('fig8.png', bbox_inches='tight')

This 3rd part of the series covers the main ‘feature selection’ methods. I hope these posts serve as a quick and useful reference to ML code both for R and Python!
Stay tuned for further updates to this series!
Watch this space!

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To see all posts see Index of posts

# Practical Machine Learning with R and Python – Part 2

In this 2nd part of the series “Practical Machine Learning with R and Python – Part 2”, I continue where I left off in my first post Practical Machine Learning with R and Python – Part 2. In this post I cover the some classification algorithmns and cross validation. Specifically I touch
-Logistic Regression
-K Nearest Neighbors (KNN) classification
-Leave out one Cross Validation (LOOCV)
-K Fold Cross Validation
in both R and Python.

As in my initial post the algorithms are based on the following courses.

You can download this R Markdown file along with the data from Github. I hope these posts can be used as a quick reference in R and Python and Machine Learning.I have tried to include the coolest part of either course in this post.

The content of this post and much more is now available as a compact book  on Amazon in both formats – as Paperback ($9.99) and a Kindle version($6.99/Rs449/). see ‘Practical Machine Learning with R and Python – Machine Learning in stereo

The following classification problem is based on Logistic Regression. The data is an included data set in Scikit-Learn, which I have saved as csv and use it also for R. The fit of a classification Machine Learning Model depends on how correctly classifies the data. There are several measures of testing a model’s classification performance. They are

Accuracy = TP + TN / (TP + TN + FP + FN) – Fraction of all classes correctly classified
Precision = TP / (TP + FP) – Fraction of correctly classified positives among those classified as positive
Recall = TP / (TP + FN) Also known as sensitivity, or True Positive Rate (True positive) – Fraction of correctly classified as positive among all positives in the data
F1 = 2 * Precision * Recall / (Precision + Recall)

## 1a. Logistic Regression – R code

The caret and e1071 package is required for using the confusionMatrix call

source("RFunctions.R")
library(dplyr)
library(caret)
library(e1071)
# Read the data (from sklearn)
# Rename the target variable
names(cancer) <- c(seq(1,30),"output")
# Split as training and test sets
train_idx <- trainTestSplit(cancer,trainPercent=75,seed=5)
train <- cancer[train_idx, ]
test <- cancer[-train_idx, ]

# Fit a generalized linear logistic model,
fit=glm(output~.,family=binomial,data=train,control = list(maxit = 50))
# Predict the output from the model
a=predict(fit,newdata=train,type="response")
# Set response >0.5 as 1 and <=0.5 as 0
b=ifelse(a>0.5,1,0)
# Compute the confusion matrix for training data
confusionMatrix(b,train$output) ## Confusion Matrix and Statistics ## ## Reference ## Prediction 0 1 ## 0 154 0 ## 1 0 272 ## ## Accuracy : 1 ## 95% CI : (0.9914, 1) ## No Information Rate : 0.6385 ## P-Value [Acc > NIR] : < 2.2e-16 ## ## Kappa : 1 ## Mcnemar's Test P-Value : NA ## ## Sensitivity : 1.0000 ## Specificity : 1.0000 ## Pos Pred Value : 1.0000 ## Neg Pred Value : 1.0000 ## Prevalence : 0.3615 ## Detection Rate : 0.3615 ## Detection Prevalence : 0.3615 ## Balanced Accuracy : 1.0000 ## ## 'Positive' Class : 0 ##  m=predict(fit,newdata=test,type="response") n=ifelse(m>0.5,1,0) # Compute the confusion matrix for test output confusionMatrix(n,test$output)
## Confusion Matrix and Statistics
##
##           Reference
## Prediction  0  1
##          0 52  4
##          1  5 81
##
##                Accuracy : 0.9366
##                  95% CI : (0.8831, 0.9706)
##     No Information Rate : 0.5986
##     P-Value [Acc > NIR] : <2e-16
##
##                   Kappa : 0.8677
##  Mcnemar's Test P-Value : 1
##
##             Sensitivity : 0.9123
##             Specificity : 0.9529
##          Pos Pred Value : 0.9286
##          Neg Pred Value : 0.9419
##              Prevalence : 0.4014
##          Detection Rate : 0.3662
##    Detection Prevalence : 0.3944
##       Balanced Accuracy : 0.9326
##
##        'Positive' Class : 0
## 

## 1b. Logistic Regression – Python code

import numpy as np
import pandas as pd
import os
import matplotlib.pyplot as plt
from sklearn.model_selection import train_test_split
from sklearn.linear_model import LogisticRegression
os.chdir("C:\\Users\\Ganesh\\RandPython")
from sklearn.datasets import make_classification, make_blobs

from sklearn.metrics import confusion_matrix
from matplotlib.colors import ListedColormap
from sklearn.datasets import load_breast_cancer
# Load the cancer data
(X_cancer, y_cancer) = load_breast_cancer(return_X_y = True)
X_train, X_test, y_train, y_test = train_test_split(X_cancer, y_cancer,
random_state = 0)
# Call the Logisitic Regression function
clf = LogisticRegression().fit(X_train, y_train)
fig, subaxes = plt.subplots(1, 1, figsize=(7, 5))
# Fit a model
clf = LogisticRegression().fit(X_train, y_train)

# Compute and print the Accuray scores
print('Accuracy of Logistic regression classifier on training set: {:.2f}'
.format(clf.score(X_train, y_train)))
print('Accuracy of Logistic regression classifier on test set: {:.2f}'
.format(clf.score(X_test, y_test)))
y_predicted=clf.predict(X_test)
# Compute and print confusion matrix
confusion = confusion_matrix(y_test, y_predicted)
from sklearn.metrics import accuracy_score, precision_score, recall_score, f1_score
print('Accuracy: {:.2f}'.format(accuracy_score(y_test, y_predicted)))
print('Precision: {:.2f}'.format(precision_score(y_test, y_predicted)))
print('Recall: {:.2f}'.format(recall_score(y_test, y_predicted)))
print('F1: {:.2f}'.format(f1_score(y_test, y_predicted)))
## Accuracy of Logistic regression classifier on training set: 0.96
## Accuracy of Logistic regression classifier on test set: 0.96
## Accuracy: 0.96
## Precision: 0.99
## Recall: 0.94
## F1: 0.97

## 2. Dummy variables

The following R and Python code show how dummy variables are handled in R and Python. Dummy variables are categorival variables which have to be converted into appropriate values before using them in Machine Learning Model For e.g. if we had currency as ‘dollar’, ‘rupee’ and ‘yen’ then the dummy variable will convert this as
dollar 0 0 0
rupee 0 0 1
yen 0 1 0

## 2a. Logistic Regression with dummy variables- R code

# Load the dummies library
library(dummies) 
df <- read.csv("adult1.csv",stringsAsFactors = FALSE,na.strings = c(""," "," ?"))

# Remove rows which have NA
df1 <- df[complete.cases(df),]
dim(df1)
## [1] 30161    16
# Select specific columns
adult <- df1 %>% dplyr::select(age,occupation,education,educationNum,capitalGain,
capital.loss,hours.per.week,native.country,salary)
# Set the dummy data with appropriate values

#Split as training and test
train <- adult1[train_idx, ]
test <- adult1[-train_idx, ]

# Fit a binomial logistic regression
fit=glm(salary~.,family=binomial,data=train)
# Predict response
a=predict(fit,newdata=train,type="response")
# If response >0.5 then it is a 1 and 0 otherwise
b=ifelse(a>0.5,1,0)
confusionMatrix(b,train$salary) ## Confusion Matrix and Statistics ## ## Reference ## Prediction 0 1 ## 0 16065 3145 ## 1 968 2442 ## ## Accuracy : 0.8182 ## 95% CI : (0.8131, 0.8232) ## No Information Rate : 0.753 ## P-Value [Acc > NIR] : < 2.2e-16 ## ## Kappa : 0.4375 ## Mcnemar's Test P-Value : < 2.2e-16 ## ## Sensitivity : 0.9432 ## Specificity : 0.4371 ## Pos Pred Value : 0.8363 ## Neg Pred Value : 0.7161 ## Prevalence : 0.7530 ## Detection Rate : 0.7102 ## Detection Prevalence : 0.8492 ## Balanced Accuracy : 0.6901 ## ## 'Positive' Class : 0 ##  # Compute and display confusion matrix m=predict(fit,newdata=test,type="response") ## Warning in predict.lm(object, newdata, se.fit, scale = 1, type = ## ifelse(type == : prediction from a rank-deficient fit may be misleading n=ifelse(m>0.5,1,0) confusionMatrix(n,test$salary)
## Confusion Matrix and Statistics
##
##           Reference
## Prediction    0    1
##          0 5263 1099
##          1  357  822
##
##                Accuracy : 0.8069
##                  95% CI : (0.7978, 0.8158)
##     No Information Rate : 0.7453
##     P-Value [Acc > NIR] : < 2.2e-16
##
##                   Kappa : 0.4174
##  Mcnemar's Test P-Value : < 2.2e-16
##
##             Sensitivity : 0.9365
##             Specificity : 0.4279
##          Pos Pred Value : 0.8273
##          Neg Pred Value : 0.6972
##              Prevalence : 0.7453
##          Detection Rate : 0.6979
##    Detection Prevalence : 0.8437
##       Balanced Accuracy : 0.6822
##
##        'Positive' Class : 0
## 

## 2b. Logistic Regression with dummy variables- Python code

Pandas has a get_dummies function for handling dummies

import numpy as np
import pandas as pd
import os
import matplotlib.pyplot as plt
from sklearn.model_selection import train_test_split
from sklearn.linear_model import LogisticRegression
from sklearn.metrics import confusion_matrix
from sklearn.metrics import accuracy_score, precision_score, recall_score, f1_score
# Drop rows with NA
df1=df.dropna()
print(df1.shape)
# Select specific columns
'hours-per-week','native-country','salary']]

'hours-per-week','native-country']]
# Set approporiate values for dummy variables

random_state = 0)
clf = LogisticRegression().fit(X_adult_train, y_train)

# Compute and display Accuracy and Confusion matrix
print('Accuracy of Logistic regression classifier on training set: {:.2f}'
print('Accuracy of Logistic regression classifier on test set: {:.2f}'
confusion = confusion_matrix(y_test, y_predicted)
print('Accuracy: {:.2f}'.format(accuracy_score(y_test, y_predicted)))
print('Precision: {:.2f}'.format(precision_score(y_test, y_predicted)))
print('Recall: {:.2f}'.format(recall_score(y_test, y_predicted)))
print('F1: {:.2f}'.format(f1_score(y_test, y_predicted)))
## (30161, 16)
## Accuracy of Logistic regression classifier on training set: 0.82
## Accuracy of Logistic regression classifier on test set: 0.81
## Accuracy: 0.81
## Precision: 0.68
## Recall: 0.41
## F1: 0.51

## 3a – K Nearest Neighbors Classification – R code

The Adult data set is taken from UCI Machine Learning Repository

source("RFunctions.R")
df <- read.csv("adult1.csv",stringsAsFactors = FALSE,na.strings = c(""," "," ?"))
# Remove rows which have NA
df1 <- df[complete.cases(df),]
dim(df1)
## [1] 30161    16
# Select specific columns
adult <- df1 %>% dplyr::select(age,occupation,education,educationNum,capitalGain,
capital.loss,hours.per.week,native.country,salary)
# Set dummy variables

#Split train and test as required by KNN classsification model
train <- adult1[train_idx, ]
test <- adult1[-train_idx, ]
train.X <- train[,1:76]
train.y <- train[,77]
test.X <- test[,1:76]
test.y <- test[,77]

# Fit a model for 1,3,5,10 and 15 neighbors
cMat <- NULL
neighbors <-c(1,3,5,10,15)
for(i in seq_along(neighbors)){
fit =knn(train.X,test.X,train.y,k=i)
table(fit,test.y)
a<-confusionMatrix(fit,test.y)
cMat[i] <- a$overall[1] print(a$overall[1])
}
##  Accuracy
## 0.7835831
##  Accuracy
## 0.8162047
##  Accuracy
## 0.8089113
##  Accuracy
## 0.8209787
##  Accuracy
## 0.8184591
#Plot the Accuracy for each of the KNN models
df <- data.frame(neighbors,Accuracy=cMat)
ggplot(df,aes(x=neighbors,y=Accuracy)) + geom_point() +geom_line(color="blue") +
xlab("Number of neighbors") + ylab("Accuracy") +
ggtitle("KNN regression - Accuracy vs Number of Neighors (Unnormalized)")

## 3b – K Nearest Neighbors Classification – Python code

import numpy as np
import pandas as pd
import os
import matplotlib.pyplot as plt
from sklearn.model_selection import train_test_split
from sklearn.metrics import confusion_matrix
from sklearn.metrics import accuracy_score, precision_score, recall_score, f1_score
from sklearn.neighbors import KNeighborsClassifier
from sklearn.preprocessing import MinMaxScaler

df1=df.dropna()
print(df1.shape)
# Select specific columns
'hours-per-week','native-country','salary']]

'hours-per-week','native-country']]

#Set values for dummy variables

random_state = 0)

# KNN classification in Python requires the data to be scaled.
# Scale the data
scaler = MinMaxScaler()
# Apply scaling to test set also
# Compute the KNN model for 1,3,5,10 & 15 neighbors
accuracy=[]
neighbors=[1,3,5,10,15]
for i in neighbors:
knn = KNeighborsClassifier(n_neighbors = i)
knn.fit(X_train_scaled, y_train)
accuracy.append(knn.score(X_test_scaled, y_test))
print('Accuracy test score: {:.3f}'
.format(knn.score(X_test_scaled, y_test)))

# Plot the models with the Accuracy attained for each of these models
fig1=plt.plot(neighbors,accuracy)
fig1=plt.title("KNN regression - Accuracy vs Number of neighbors")
fig1=plt.xlabel("Neighbors")
fig1=plt.ylabel("Accuracy")
fig1.figure.savefig('foo1.png', bbox_inches='tight')
## (30161, 16)
## Accuracy test score: 0.749
## Accuracy test score: 0.779
## Accuracy test score: 0.793
## Accuracy test score: 0.804
## Accuracy test score: 0.803

Output image:

## 4 MPG vs Horsepower

The following scatter plot shows the non-linear relation between mpg and horsepower. This will be used as the data input for computing K Fold Cross Validation Error

## 4a MPG vs Horsepower scatter plot – R Code

df=read.csv("auto_mpg.csv",stringsAsFactors = FALSE) # Data from UCI
df1 <- as.data.frame(sapply(df,as.numeric))
df2 <- df1 %>% dplyr::select(cylinder,displacement, horsepower,weight, acceleration, year,mpg)
df3 <- df2[complete.cases(df2),]
ggplot(df3,aes(x=horsepower,y=mpg)) + geom_point() + xlab("Horsepower") +
ylab("Miles Per gallon") + ggtitle("Miles per Gallon vs Hosrsepower")

## 4b MPG vs Horsepower scatter plot – Python Code

import numpy as np
import pandas as pd
import os
import matplotlib.pyplot as plt
autoDF.shape
autoDF.columns
autoDF1=autoDF[['mpg','cylinder','displacement','horsepower','weight','acceleration','year']]
autoDF2 = autoDF1.apply(pd.to_numeric, errors='coerce')
autoDF3=autoDF2.dropna()
autoDF3.shape
#X=autoDF3[['cylinder','displacement','horsepower','weight']]
X=autoDF3[['horsepower']]
y=autoDF3['mpg']

fig11=plt.scatter(X,y)
fig11=plt.title("KNN regression - Accuracy vs Number of neighbors")
fig11=plt.xlabel("Neighbors")
fig11=plt.ylabel("Accuracy")
fig11.figure.savefig('foo11.png', bbox_inches='tight')


## 5 K Fold Cross Validation

K Fold Cross Validation is a technique in which the data set is divided into K Folds or K partitions. The Machine Learning model is trained on K-1 folds and tested on the Kth fold i.e.
we will have K-1 folds for training data and 1 for testing the ML model. Since we can partition this as $C_{1}^{K}$ or K choose 1, there will be K such partitions. The K Fold Cross
Validation estimates the average validation error that we can expect on a new unseen test data.

The formula for K Fold Cross validation is as follows

$MSE_{K} = \frac{\sum (y-yhat)^{2}}{n_{K}}$
and
$n_{K} = \frac{N}{K}$
and
$CV_{K} = \sum_{K=1}^{K} (\frac{n_{K}}{N}) MSE_{K}$

where $n_{K}$ is the number of elements in partition ‘K’ and N is the total number of elements
$CV_{K} =\sum_{K=1}^{K} MSE_{K}$

$CV_{K} =\frac{\sum_{K=1}^{K} MSE_{K}}{K}$
Leave Out one Cross Validation (LOOCV) is a special case of K Fold Cross Validation where N-1 data points are used to train the model and 1 data point is used to test the model. There are N such paritions of N-1 & 1 that are possible. The mean error is measured The Cross Valifation Error for LOOCV is

$CV_{N} = \frac{1}{n} *\frac{\sum_{1}^{n}(y-yhat)^{2}}{1-h_{i}}$
where $h_{i}$ is the diagonal hat matrix

see [Statistical Learning]

The above formula is also included in this blog post

It took me a day and a half to implement the K Fold Cross Validation formula. I think it is correct. In any case do let me know if you think it is off

## 5a. Leave out one cross validation (LOOCV) – R Code

R uses the package ‘boot’ for performing Cross Validation error computation

library(boot)
library(reshape2)
df=read.csv("auto_mpg.csv",stringsAsFactors = FALSE) # Data from UCI
df1 <- as.data.frame(sapply(df,as.numeric))
# Select complete cases
df2 <- df1 %>% dplyr::select(cylinder,displacement, horsepower,weight, acceleration, year,mpg)
df3 <- df2[complete.cases(df2),]
set.seed(17)
cv.error=rep(0,10)
# For polynomials 1,2,3... 10 fit a LOOCV model
for (i in 1:10){
glm.fit=glm(mpg~poly(horsepower,i),data=df3)
cv.error[i]=cv.glm(df3,glm.fit)$delta[1] } cv.error ## [1] 24.23151 19.24821 19.33498 19.42443 19.03321 18.97864 18.83305 ## [8] 18.96115 19.06863 19.49093 # Create and display a plot folds <- seq(1,10) df <- data.frame(folds,cvError=cv.error) ggplot(df,aes(x=folds,y=cvError)) + geom_point() +geom_line(color="blue") + xlab("Degree of Polynomial") + ylab("Cross Validation Error") + ggtitle("Leave one out Cross Validation - Cross Validation Error vs Degree of Polynomial") ## 5b. Leave out one cross validation (LOOCV) – Python Code In Python there is no available function to compute Cross Validation error and we have to compute the above formula. I have done this after several hours. I think it is now in reasonable shape. Do let me know if you think otherwise. For LOOCV I use the K Fold Cross Validation with K=N import numpy as np import pandas as pd import os import matplotlib.pyplot as plt from sklearn.linear_model import LinearRegression from sklearn.cross_validation import train_test_split, KFold from sklearn.preprocessing import PolynomialFeatures from sklearn.metrics import mean_squared_error # Read data autoDF =pd.read_csv("auto_mpg.csv",encoding="ISO-8859-1") autoDF.shape autoDF.columns autoDF1=autoDF[['mpg','cylinder','displacement','horsepower','weight','acceleration','year']] autoDF2 = autoDF1.apply(pd.to_numeric, errors='coerce') # Remove rows with NAs autoDF3=autoDF2.dropna() autoDF3.shape X=autoDF3[['horsepower']] y=autoDF3['mpg'] # For polynomial degree 1,2,3... 10 def computeCVError(X,y,folds): deg=[] mse=[] degree1=[1,2,3,4,5,6,7,8,9,10] nK=len(X)/float(folds) xval_err=0 # For degree 'j' for j in degree1: # Split as 'folds' kf = KFold(len(X),n_folds=folds) for train_index, test_index in kf: # Create the appropriate train and test partitions from the fold index X_train, X_test = X.iloc[train_index], X.iloc[test_index] y_train, y_test = y.iloc[train_index], y.iloc[test_index] # For the polynomial degree 'j' poly = PolynomialFeatures(degree=j) # Transform the X_train and X_test X_train_poly = poly.fit_transform(X_train) X_test_poly = poly.fit_transform(X_test) # Fit a model on the transformed data linreg = LinearRegression().fit(X_train_poly, y_train) # Compute yhat or ypred y_pred = linreg.predict(X_test_poly) # Compute MSE * n_K/N test_mse = mean_squared_error(y_test, y_pred)*float(len(X_train))/float(len(X)) # Add the test_mse for this partition of the data mse.append(test_mse) # Compute the mean of all folds for degree 'j' deg.append(np.mean(mse)) return(deg) df=pd.DataFrame() print(len(X)) # Call the function once. For LOOCV K=N. hence len(X) is passed as number of folds cvError=computeCVError(X,y,len(X)) # Create and plot LOOCV df=pd.DataFrame(cvError) fig3=df.plot() fig3=plt.title("Leave one out Cross Validation - Cross Validation Error vs Degree of Polynomial") fig3=plt.xlabel("Degree of Polynomial") fig3=plt.ylabel("Cross validation Error") fig3.figure.savefig('foo3.png', bbox_inches='tight') ## 6a K Fold Cross Validation – R code Here K Fold Cross Validation is done for 4, 5 and 10 folds using the R package boot and the glm package library(boot) library(reshape2) set.seed(17) #Read data df=read.csv("auto_mpg.csv",stringsAsFactors = FALSE) # Data from UCI df1 <- as.data.frame(sapply(df,as.numeric)) df2 <- df1 %>% dplyr::select(cylinder,displacement, horsepower,weight, acceleration, year,mpg) df3 <- df2[complete.cases(df2),] a=matrix(rep(0,30),nrow=3,ncol=10) set.seed(17) # Set the folds as 4,5 and 10 folds<-c(4,5,10) for(i in seq_along(folds)){ cv.error.10=rep(0,10) for (j in 1:10){ # Fit a generalized linear model glm.fit=glm(mpg~poly(horsepower,j),data=df3) # Compute K Fold Validation error a[i,j]=cv.glm(df3,glm.fit,K=folds[i])$delta[1]

}

}

# Create and display the K Fold Cross Validation Error
b <- t(a)
df <- data.frame(b)
df1 <- cbind(seq(1,10),df)
names(df1) <- c("PolynomialDegree","4-fold","5-fold","10-fold")

df2 <- melt(df1,id="PolynomialDegree")
ggplot(df2) + geom_line(aes(x=PolynomialDegree, y=value, colour=variable),size=2) +
xlab("Degree of Polynomial") + ylab("Cross Validation Error") +
ggtitle("K Fold Cross Validation - Cross Validation Error vs Degree of Polynomial")

## 6b. K Fold Cross Validation – Python code

The implementation of K-Fold Cross Validation Error has to be implemented and I have done this below. There is a small discrepancy in the shapes of the curves with the R plot above. Not sure why!

import numpy as np
import pandas as pd
import os
import matplotlib.pyplot as plt
from sklearn.linear_model import LinearRegression
from sklearn.cross_validation import train_test_split, KFold
from sklearn.preprocessing import PolynomialFeatures
from sklearn.metrics import mean_squared_error
autoDF.shape
autoDF.columns
autoDF1=autoDF[['mpg','cylinder','displacement','horsepower','weight','acceleration','year']]
autoDF2 = autoDF1.apply(pd.to_numeric, errors='coerce')
# Drop NA rows
autoDF3=autoDF2.dropna()
autoDF3.shape
#X=autoDF3[['cylinder','displacement','horsepower','weight']]
X=autoDF3[['horsepower']]
y=autoDF3['mpg']

# Create Cross Validation function
def computeCVError(X,y,folds):
deg=[]
mse=[]
# For degree 1,2,3,..10
degree1=[1,2,3,4,5,6,7,8,9,10]

nK=len(X)/float(folds)
xval_err=0
for j in degree1:
# Split the data into 'folds'
kf = KFold(len(X),n_folds=folds)
for train_index, test_index in kf:
# Partition the data acccording the fold indices generated
X_train, X_test = X.iloc[train_index], X.iloc[test_index]
y_train, y_test = y.iloc[train_index], y.iloc[test_index]

# Scale the X_train and X_test as per the polynomial degree 'j'
poly = PolynomialFeatures(degree=j)
X_train_poly = poly.fit_transform(X_train)
X_test_poly = poly.fit_transform(X_test)
# Fit a polynomial regression
linreg = LinearRegression().fit(X_train_poly, y_train)
# Compute yhat or ypred
y_pred = linreg.predict(X_test_poly)
# Compute MSE *(nK/N)
test_mse = mean_squared_error(y_test, y_pred)*float(len(X_train))/float(len(X))
# Append to list for different folds
mse.append(test_mse)
# Compute the mean for poylnomial 'j'
deg.append(np.mean(mse))

return(deg)

# Create and display a plot of K -Folds
df=pd.DataFrame()
for folds in [4,5,10]:
cvError=computeCVError(X,y,folds)
#print(cvError)
df1=pd.DataFrame(cvError)
df=pd.concat([df,df1],axis=1)
#print(cvError)

df.columns=['4-fold','5-fold','10-fold']
df=df.reindex([1,2,3,4,5,6,7,8,9,10])
df
fig2=df.plot()
fig2=plt.title("K Fold Cross Validation - Cross Validation Error vs Degree of Polynomial")
fig2=plt.xlabel("Degree of Polynomial")
fig2=plt.ylabel("Cross validation Error")
fig2.figure.savefig('foo2.png', bbox_inches='tight')


This concludes this 2nd part of this series. I will look into model tuning and model selection in R and Python in the coming parts. Comments, suggestions and corrections are welcome!
To be continued….
Watch this space!

Also see

To see all posts see Index of posts

# Introduction

This is the 1st part of a series of posts I intend to write on some common Machine Learning Algorithms in R and Python. In this first part I cover the following Machine Learning Algorithms

• Univariate Regression
• Multivariate Regression
• Polynomial Regression
• K Nearest Neighbors Regression

The code includes the implementation in both R and Python. This series of posts are based on the following 2 MOOC courses I did at Stanford Online and at Coursera

1. Statistical Learning, Prof Trevor Hastie & Prof Robert Tibesherani, Online Stanford
2. Applied Machine Learning in Python Prof Kevyn-Collin Thomson, University Of Michigan, Coursera

I have used the data sets from UCI Machine Learning repository(Communities and Crime and Auto MPG). I also use the Boston data set from MASS package

The content of this post and much more is now available as a compact book  on Amazon in both formats – as Paperback ($9.99) and a Kindle version($6.99/Rs449/). see ‘Practical Machine Learning with R and Python – Machine Learning in stereo

While coding in R and Python I found that there were some aspects that were more convenient in one language and some in the other. For example, plotting the fit in R is straightforward in R, while computing the R squared, splitting as Train & Test sets etc. are already available in Python. In any case, these minor inconveniences can be easily be implemented in either language.

R squared computation in R is computed as follows
$RSS=\sum (y-yhat)^{2}$
$TSS= \sum(y-mean(y))^{2}$
$Rsquared- 1-\frac{RSS}{TSS}$

Note: You can download this R Markdown file and the associated data sets from Github at MachineLearning-RandPython
Note 1: This post was created as an R Markdown file in RStudio which has a cool feature of including R and Python snippets. The plot of matplotlib needs a workaround but otherwise this is a real cool feature of RStudio!

## 1.1a Univariate Regression – R code

Here a simple linear regression line is fitted between a single input feature and the target variable

# Source in the R function library
source("RFunctions.R")
# Read the Boston data file
df=read.csv("Boston.csv",stringsAsFactors = FALSE) # Data from MASS - Statistical Learning

# Split the data into training and test sets (75:25)
train_idx <- trainTestSplit(df,trainPercent=75,seed=5)
train <- df[train_idx, ]
test <- df[-train_idx, ]

# Fit a linear regression line between 'Median value of owner occupied homes' vs 'lower status of
# population'
fit=lm(medv~lstat,data=df)
# Display details of fir
summary(fit)
##
## Call:
## lm(formula = medv ~ lstat, data = df)
##
## Residuals:
##     Min      1Q  Median      3Q     Max
## -15.168  -3.990  -1.318   2.034  24.500
##
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)
## (Intercept) 34.55384    0.56263   61.41   <2e-16 ***
## lstat       -0.95005    0.03873  -24.53   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 6.216 on 504 degrees of freedom
## Multiple R-squared:  0.5441, Adjusted R-squared:  0.5432
## F-statistic: 601.6 on 1 and 504 DF,  p-value: < 2.2e-16
# Display the confidence intervals
confint(fit)
##                 2.5 %     97.5 %
## (Intercept) 33.448457 35.6592247
## lstat       -1.026148 -0.8739505
plot(df$lstat,df$medv, xlab="Lower status (%)",ylab="Median value of owned homes ($1000)", main="Median value of homes ($1000) vs Lowe status (%)")
abline(fit)
abline(fit,lwd=3)
abline(fit,lwd=3,col="red")

rsquared=Rsquared(fit,test,test$medv) sprintf("R-squared for uni-variate regression (Boston.csv) is : %f", rsquared) ## [1] "R-squared for uni-variate regression (Boston.csv) is : 0.556964" ## 1.1b Univariate Regression – Python code import numpy as np import pandas as pd import os import matplotlib.pyplot as plt from sklearn.model_selection import train_test_split from sklearn.linear_model import LinearRegression #os.chdir("C:\\software\\machine-learning\\RandPython") # Read the CSV file df = pd.read_csv("Boston.csv",encoding = "ISO-8859-1") # Select the feature variable X=df['lstat'] # Select the target y=df['medv'] # Split into train and test sets (75:25) X_train, X_test, y_train, y_test = train_test_split(X, y,random_state = 0) X_train=X_train.values.reshape(-1,1) X_test=X_test.values.reshape(-1,1) # Fit a linear model linreg = LinearRegression().fit(X_train, y_train) # Print the training and test R squared score print('R-squared score (training): {:.3f}'.format(linreg.score(X_train, y_train))) print('R-squared score (test): {:.3f}'.format(linreg.score(X_test, y_test))) # Plot the linear regression line fig=plt.scatter(X_train,y_train) # Create a range of points. Compute yhat=coeff1*x + intercept and plot x=np.linspace(0,40,20) fig1=plt.plot(x, linreg.coef_ * x + linreg.intercept_, color='red') fig1=plt.title("Median value of homes ($1000) vs Lowe status (%)")
fig1=plt.xlabel("Lower status (%)")
fig1=plt.ylabel("Median value of owned homes ($1000)") fig.figure.savefig('foo.png', bbox_inches='tight') fig1.figure.savefig('foo1.png', bbox_inches='tight') print "Finished"  ## R-squared score (training): 0.571 ## R-squared score (test): 0.458 ## Finished ## 1.2a Multivariate Regression – R code # Read crimes data crimesDF <- read.csv("crimes.csv",stringsAsFactors = FALSE) # Remove the 1st 7 columns which do not impact output crimesDF1 <- crimesDF[,7:length(crimesDF)] # Convert all to numeric crimesDF2 <- sapply(crimesDF1,as.numeric) # Check for NAs a <- is.na(crimesDF2) # Set to 0 as an imputation crimesDF2[a] <-0 #Create as a dataframe crimesDF2 <- as.data.frame(crimesDF2) #Create a train/test split train_idx <- trainTestSplit(crimesDF2,trainPercent=75,seed=5) train <- crimesDF2[train_idx, ] test <- crimesDF2[-train_idx, ] # Fit a multivariate regression model between crimesPerPop and all other features fit <- lm(ViolentCrimesPerPop~.,data=train) # Compute and print R Squared rsquared=Rsquared(fit,test,test$ViolentCrimesPerPop)
sprintf("R-squared for multi-variate regression (crimes.csv)  is : %f", rsquared)
## [1] "R-squared for multi-variate regression (crimes.csv)  is : 0.653940"

## 1.2b Multivariate Regression – Python code

import numpy as np
import pandas as pd
import os
import matplotlib.pyplot as plt
from sklearn.model_selection import train_test_split
from sklearn.linear_model import LinearRegression
# Read the data
#Remove the 1st 7 columns
crimesDF1=crimesDF.iloc[:,7:crimesDF.shape[1]]
# Convert to numeric
crimesDF2 = crimesDF1.apply(pd.to_numeric, errors='coerce')
# Impute NA to 0s
crimesDF2.fillna(0, inplace=True)

# Select the X (feature vatiables - all)
X=crimesDF2.iloc[:,0:120]

# Set the target
y=crimesDF2.iloc[:,121]

X_train, X_test, y_train, y_test = train_test_split(X, y,random_state = 0)
# Fit a multivariate regression model
linreg = LinearRegression().fit(X_train, y_train)

# compute and print the R Square
print('R-squared score (training): {:.3f}'.format(linreg.score(X_train, y_train)))
print('R-squared score (test): {:.3f}'.format(linreg.score(X_test, y_test)))
## R-squared score (training): 0.699
## R-squared score (test): 0.677

## 1.3a Polynomial Regression – R

For Polynomial regression , polynomials of degree 1,2 & 3 are used and R squared is computed. It can be seen that the quadaratic model provides the best R squared score and hence the best fit

 # Polynomial degree 1
df=read.csv("auto_mpg.csv",stringsAsFactors = FALSE) # Data from UCI
df1 <- as.data.frame(sapply(df,as.numeric))

# Select key columns
df2 <- df1 %>% select(cylinder,displacement, horsepower,weight, acceleration, year,mpg)
df3 <- df2[complete.cases(df2),]

# Split as train and test sets
train_idx <- trainTestSplit(df3,trainPercent=75,seed=5)
train <- df3[train_idx, ]
test <- df3[-train_idx, ]

# Fit a model of degree 1
fit <- lm(mpg~. ,data=train)
rsquared1 <-Rsquared(fit,test,test$mpg) sprintf("R-squared for Polynomial regression of degree 1 (auto_mpg.csv) is : %f", rsquared1) ## [1] "R-squared for Polynomial regression of degree 1 (auto_mpg.csv) is : 0.763607" # Polynomial degree 2 - Quadratic x = as.matrix(df3[1:6]) # Make a polynomial of degree 2 for feature variables before split df4=as.data.frame(poly(x,2,raw=TRUE)) df5 <- cbind(df4,df3[7]) # Split into train and test set train_idx <- trainTestSplit(df5,trainPercent=75,seed=5) train <- df5[train_idx, ] test <- df5[-train_idx, ] # Fit the quadratic model fit <- lm(mpg~. ,data=train) # Compute R squared rsquared2=Rsquared(fit,test,test$mpg)
sprintf("R-squared for Polynomial regression of degree 2 (auto_mpg.csv)  is : %f", rsquared2)
## [1] "R-squared for Polynomial regression of degree 2 (auto_mpg.csv)  is : 0.831372"
#Polynomial degree 3
x = as.matrix(df3[1:6])
# Make polynomial of degree 4  of feature variables before split
df4=as.data.frame(poly(x,3,raw=TRUE))
df5 <- cbind(df4,df3[7])
train_idx <- trainTestSplit(df5,trainPercent=75,seed=5)

train <- df5[train_idx, ]
test <- df5[-train_idx, ]
# Fit a model of degree 3
fit <- lm(mpg~. ,data=train)
# Compute R squared
rsquared3=Rsquared(fit,test,test$mpg) sprintf("R-squared for Polynomial regression of degree 2 (auto_mpg.csv) is : %f", rsquared3) ## [1] "R-squared for Polynomial regression of degree 2 (auto_mpg.csv) is : 0.773225" df=data.frame(degree=c(1,2,3),Rsquared=c(rsquared1,rsquared2,rsquared3)) # Make a plot of Rsquared and degree ggplot(df,aes(x=degree,y=Rsquared)) +geom_point() + geom_line(color="blue") + ggtitle("Polynomial regression - R squared vs Degree of polynomial") + xlab("Degree") + ylab("R squared") ## 1.3a Polynomial Regression – Python For Polynomial regression , polynomials of degree 1,2 & 3 are used and R squared is computed. It can be seen that the quadaratic model provides the best R squared score and hence the best fit import numpy as np import pandas as pd import os import matplotlib.pyplot as plt from sklearn.model_selection import train_test_split from sklearn.linear_model import LinearRegression from sklearn.preprocessing import PolynomialFeatures autoDF =pd.read_csv("auto_mpg.csv",encoding="ISO-8859-1") autoDF.shape autoDF.columns # Select key columns autoDF1=autoDF[['mpg','cylinder','displacement','horsepower','weight','acceleration','year']] # Convert columns to numeric autoDF2 = autoDF1.apply(pd.to_numeric, errors='coerce') # Drop NAs autoDF3=autoDF2.dropna() autoDF3.shape X=autoDF3[['cylinder','displacement','horsepower','weight','acceleration','year']] y=autoDF3['mpg'] # Polynomial degree 1 X_train, X_test, y_train, y_test = train_test_split(X, y,random_state = 0) linreg = LinearRegression().fit(X_train, y_train) print('R-squared score - Polynomial degree 1 (training): {:.3f}'.format(linreg.score(X_train, y_train))) # Compute R squared rsquared1 =linreg.score(X_test, y_test) print('R-squared score - Polynomial degree 1 (test): {:.3f}'.format(linreg.score(X_test, y_test))) # Polynomial degree 2 poly = PolynomialFeatures(degree=2) X_poly = poly.fit_transform(X) X_train, X_test, y_train, y_test = train_test_split(X_poly, y,random_state = 0) linreg = LinearRegression().fit(X_train, y_train) # Compute R squared print('R-squared score - Polynomial degree 2 (training): {:.3f}'.format(linreg.score(X_train, y_train))) rsquared2 =linreg.score(X_test, y_test) print('R-squared score - Polynomial degree 2 (test): {:.3f}\n'.format(linreg.score(X_test, y_test))) #Polynomial degree 3 poly = PolynomialFeatures(degree=3) X_poly = poly.fit_transform(X) X_train, X_test, y_train, y_test = train_test_split(X_poly, y,random_state = 0) linreg = LinearRegression().fit(X_train, y_train) print('(R-squared score -Polynomial degree 3 (training): {:.3f}' .format(linreg.score(X_train, y_train))) # Compute R squared rsquared3 =linreg.score(X_test, y_test) print('R-squared score Polynomial degree 3 (test): {:.3f}\n'.format(linreg.score(X_test, y_test))) degree=[1,2,3] rsquared =[rsquared1,rsquared2,rsquared3] fig2=plt.plot(degree,rsquared) fig2=plt.title("Polynomial regression - R squared vs Degree of polynomial") fig2=plt.xlabel("Degree") fig2=plt.ylabel("R squared") fig2.figure.savefig('foo2.png', bbox_inches='tight') print "Finished plotting and saving"  ## R-squared score - Polynomial degree 1 (training): 0.811 ## R-squared score - Polynomial degree 1 (test): 0.799 ## R-squared score - Polynomial degree 2 (training): 0.861 ## R-squared score - Polynomial degree 2 (test): 0.847 ## ## (R-squared score -Polynomial degree 3 (training): 0.933 ## R-squared score Polynomial degree 3 (test): 0.710 ## ## Finished plotting and saving ## 1.4 K Nearest Neighbors The code below implements KNN Regression both for R and Python. This is done for different neighbors. The R squared is computed in each case. This is repeated after performing feature scaling. It can be seen the model fit is much better after feature scaling. Normalization refers to $X_{normalized} = \frac{X-min(X)}{max(X-min(X))}$ Another technique that is used is Standardization which is $X_{standardized} = \frac{X-mean(X)}{sd(X)}$ ## 1.4a K Nearest Neighbors Regression – R( Unnormalized) The R code below does not use feature scaling # KNN regression requires the FNN package df=read.csv("auto_mpg.csv",stringsAsFactors = FALSE) # Data from UCI df1 <- as.data.frame(sapply(df,as.numeric)) df2 <- df1 %>% select(cylinder,displacement, horsepower,weight, acceleration, year,mpg) df3 <- df2[complete.cases(df2),] # Split train and test train_idx <- trainTestSplit(df3,trainPercent=75,seed=5) train <- df3[train_idx, ] test <- df3[-train_idx, ] # Select the feature variables train.X=train[,1:6] # Set the target for training train.Y=train[,7] # Do the same for test set test.X=test[,1:6] test.Y=test[,7] rsquared <- NULL # Create a list of neighbors neighbors <-c(1,2,4,8,10,14) for(i in seq_along(neighbors)){ # Perform a KNN regression fit knn=knn.reg(train.X,test.X,train.Y,k=neighbors[i]) # Compute R sqaured rsquared[i]=knnRSquared(knn$pred,test.Y)
}

# Make a dataframe for plotting
df <- data.frame(neighbors,Rsquared=rsquared)
# Plot the number of neighors vs the R squared
ggplot(df,aes(x=neighbors,y=Rsquared)) + geom_point() +geom_line(color="blue") +
xlab("Number of neighbors") + ylab("R squared") +
ggtitle("KNN regression - R squared vs Number of Neighors (Unnormalized)")

## 1.4b K Nearest Neighbors Regression – Python( Unnormalized)

The Python code below does not use feature scaling

import numpy as np
import pandas as pd
import os
import matplotlib.pyplot as plt
from sklearn.model_selection import train_test_split
from sklearn.linear_model import LinearRegression
from sklearn.preprocessing import PolynomialFeatures
from sklearn.neighbors import KNeighborsRegressor
autoDF.shape
autoDF.columns
autoDF1=autoDF[['mpg','cylinder','displacement','horsepower','weight','acceleration','year']]
autoDF2 = autoDF1.apply(pd.to_numeric, errors='coerce')
autoDF3=autoDF2.dropna()
autoDF3.shape
X=autoDF3[['cylinder','displacement','horsepower','weight','acceleration','year']]
y=autoDF3['mpg']

# Perform a train/test split
X_train, X_test, y_train, y_test = train_test_split(X, y, random_state = 0)
# Create a list of neighbors
rsquared=[]
neighbors=[1,2,4,8,10,14]
for i in neighbors:
# Fit a KNN model
knnreg = KNeighborsRegressor(n_neighbors = i).fit(X_train, y_train)
# Compute R squared
rsquared.append(knnreg.score(X_test, y_test))
print('R-squared test score: {:.3f}'
.format(knnreg.score(X_test, y_test)))
# Plot the number of neighors vs the R squared
fig3=plt.plot(neighbors,rsquared)
fig3=plt.title("KNN regression - R squared vs Number of neighbors(Unnormalized)")
fig3=plt.xlabel("Neighbors")
fig3=plt.ylabel("R squared")
fig3.figure.savefig('foo3.png', bbox_inches='tight')
print "Finished plotting and saving"
## R-squared test score: 0.527
## R-squared test score: 0.678
## R-squared test score: 0.707
## R-squared test score: 0.684
## R-squared test score: 0.683
## R-squared test score: 0.670
## Finished plotting and saving

## 1.4c K Nearest Neighbors Regression – R( Normalized)

It can be seen that R squared improves when the features are normalized.

df=read.csv("auto_mpg.csv",stringsAsFactors = FALSE) # Data from UCI
df1 <- as.data.frame(sapply(df,as.numeric))
df2 <- df1 %>% select(cylinder,displacement, horsepower,weight, acceleration, year,mpg)
df3 <- df2[complete.cases(df2),]

# Perform MinMaxScaling of feature variables
train.X.scaled=MinMaxScaler(train.X)
test.X.scaled=MinMaxScaler(test.X)

# Create a list of neighbors
rsquared <- NULL
neighbors <-c(1,2,4,6,8,10,12,15,20,25,30)
for(i in seq_along(neighbors)){
# Fit a KNN model
knn=knn.reg(train.X.scaled,test.X.scaled,train.Y,k=i)
# Compute R ssquared
rsquared[i]=knnRSquared(knn$pred,test.Y) } df <- data.frame(neighbors,Rsquared=rsquared) # Plot the number of neighors vs the R squared ggplot(df,aes(x=neighbors,y=Rsquared)) + geom_point() +geom_line(color="blue") + xlab("Number of neighbors") + ylab("R squared") + ggtitle("KNN regression - R squared vs Number of Neighors(Normalized)") ## 1.4d K Nearest Neighbors Regression – Python( Normalized) R squared improves when the features are normalized with MinMaxScaling import numpy as np import pandas as pd import os import matplotlib.pyplot as plt from sklearn.model_selection import train_test_split from sklearn.linear_model import LinearRegression from sklearn.preprocessing import PolynomialFeatures from sklearn.neighbors import KNeighborsRegressor from sklearn.preprocessing import MinMaxScaler autoDF =pd.read_csv("auto_mpg.csv",encoding="ISO-8859-1") autoDF.shape autoDF.columns autoDF1=autoDF[['mpg','cylinder','displacement','horsepower','weight','acceleration','year']] autoDF2 = autoDF1.apply(pd.to_numeric, errors='coerce') autoDF3=autoDF2.dropna() autoDF3.shape X=autoDF3[['cylinder','displacement','horsepower','weight','acceleration','year']] y=autoDF3['mpg'] # Perform a train/ test split X_train, X_test, y_train, y_test = train_test_split(X, y, random_state = 0) # Use MinMaxScaling scaler = MinMaxScaler() X_train_scaled = scaler.fit_transform(X_train) # Apply scaling on test set X_test_scaled = scaler.transform(X_test) # Create a list of neighbors rsquared=[] neighbors=[1,2,4,6,8,10,12,15,20,25,30] for i in neighbors: # Fit a KNN model knnreg = KNeighborsRegressor(n_neighbors = i).fit(X_train_scaled, y_train) # Compute R squared rsquared.append(knnreg.score(X_test_scaled, y_test)) print('R-squared test score: {:.3f}' .format(knnreg.score(X_test_scaled, y_test))) # Plot the number of neighors vs the R squared fig4=plt.plot(neighbors,rsquared) fig4=plt.title("KNN regression - R squared vs Number of neighbors(Normalized)") fig4=plt.xlabel("Neighbors") fig4=plt.ylabel("R squared") fig4.figure.savefig('foo4.png', bbox_inches='tight') print "Finished plotting and saving" ## R-squared test score: 0.703 ## R-squared test score: 0.810 ## R-squared test score: 0.830 ## R-squared test score: 0.838 ## R-squared test score: 0.834 ## R-squared test score: 0.828 ## R-squared test score: 0.827 ## R-squared test score: 0.826 ## R-squared test score: 0.816 ## R-squared test score: 0.815 ## R-squared test score: 0.809 ## Finished plotting and saving # Conclusion In this initial post I cover the regression models when the output is continous. I intend to touch upon other Machine Learning algorithms. Comments, suggestions and corrections are welcome. Watch this this space! To be continued…. To see all posts see Index of posts # Using Linear Programming (LP) for optimizing bowling change or batting lineup in T20 cricket In my recent post, My travels through the realms of Data Science, Machine Learning, Deep Learning and (AI), I had recounted my journey in the domains of of Data Science, Machine Learning (ML), and more recently Deep Learning (DL) all of which are useful while analyzing data. Of late, I have come to the realization that there are many facets to data. And to glean insights from data, Data Science, ML and DL alone are not sufficient and one needs to also have a good handle on linear programming and optimization. My colleague at IBM Research also concurred with this view and told me he had arrived at this conclusion several years ago. While ML & DL are very useful and interesting to make inferences and predictions of outputs from input variables, optimization computes the choice of input which results in maximizing or minimizing the output. So I made a small course correction and started on a course from India’s own NPTEL Introduction to Linear Programming by Prof G. Srinivasan of IIT Madras (highly recommended!). The lectures are delivered with remarkable clarity by the Prof and I am just about halfway through the course (each lecture is of 50-55 min duration), when I decided that I needed to try to formulate and solve some real world Linear Programming problem. As usual, I turned towards cricket for some appropriate situations, and sure enough it was there in the open. For this LP formulation I take International T20 and IPL, though International ODI will also work equally well. You can download the associated code and data for this from Github at LP-cricket-analysis In T20 matches the captain has to make choice of how to rotate bowlers with the aim of restricting the batting side. Conversely, the batsmen need to take advantage of the bowling strength to maximize the runs scored. Note: a) A simple and obvious strategy would be – If the ith bowler’s economy rate is less than the economy rate of the jth bowler i.e. $er_{i}$ < $er_{j}$ then have bowler ‘i’ to bowl more overs as his/her economy rate is better b)A better strategy would be to consider the economy rate of each bowler against each batsman. How often have we witnessed bowlers with a great bowling average get thrashed time and again by the same batsman, or a bowler who is generally very poor being very effective against a particular batsman. i.e. $er_{ij}$ < $er_{ik}$ where the jth bowler is more effective than the kth bowler against the ith batsman. This now becomes a linear optimization problem as we can have several combinations of number of overs X economy rate for different bowlers and we will have to solve this algorithmically to determine the lowest score for bowling performance or highest score for batting order. This post uses the latter approach to optimize bowling change and batting lineup. Let is take a hypothetical situation Assume there are 3 bowlers – $bwlr_{1},bwlr_{2},bwlr_{3}$ and there are 4 batsmen – $bman_{1},bman_{2},bman_{3},bman_{4}$ Let the economy rate $er_{ij}$ be the Economy Rate of the jth bowler to the ith batsman. Also if remaining overs for the bowlers are $o_{1},o_{2},o_{3}$ and the total number of overs left to be bowled are $o_{1}+o_{2}+o_{3} = N$ then the question is a) Given the economy rate of each bowler per batsman, how many overs should each bowler bowl, so that the total runs scored by all the batsmen are minimum? b) Alternatively, if the know the individual strike rate of a batsman against the individual bowlers, how many overs should each batsman face with a bowler so that the total runs scored is maximized? ## 1. LP Formulation for bowling order Let the economy rate $er_{ij}$ be the Economy Rate of the jth bowler to the ith batsman. Objective function : Minimize – $er_{11}*o_{11} + er_{12}*o_{12} +..+er_{1n}*o_{1n}+ er_{21}*o_{21} + er_{22}*o_{22}+.. + er_{22}*o_{2n}+ er_{m1}*o_{m1}+..+ er_{mn}*o_{mn}$ i.e. $\sum_{i=1}^{i=m}\sum_{j=1}^{i=n}er_{ij}*o_{ij}$ Constraints Where $o_{j}$ is the number of overs remaining for the jth bowler against ‘k’ batsmen $o_{j1} + o_{j2} + .. o_{jk} < o_{j}$ and if the total number of overs remaining to be bowled is N then $o_{1} + o_{2} +...+ o_{k} = N$ or $\sum_{j=1}^{j=k} o_{j} =N$ The overs that any bowler can bowl is $o_{j} >=0$ ## 2. LP Formulation for batting lineup Let the strike rate $sr_{ij}$ be the Strike Rate of the ith batsman to the jth bowler Objective function : Maximize – $sr_{11}*o_{11} + sr_{12}*o_{12} +..+ sr_{1n}*o_{1n}+ sr_{21}*o_{21} + sr_{22}*o_{22}+.. sr_{2n}*o_{2n}+ sr_{m1}*o_{m1}+..+ sr_{mn}*o_{mn}$ i.e. $\sum_{i=1}^{i=4}\sum_{j=1}^{i=3}sr_{ij}*o_{ij}$ Constraints Where $o_{j}$ is the number of overs remaining for the jth bowler against ‘k’ batsmen $o_{j1} + o_{j2} + .. o_{jk} < o_{j}$ and the total number of overs remaining to be bowled is N then $o_{1} + o_{2} +...+ o_{k} = N$ or $\sum_{j=1}^{j=k} o_{j} =N$ The overs that any bowler can bowl is $o_{j} >=0$ lpSolveAPI– For this maximization and minimization problem I used lpSolveAPI. Below I take 2 simple examples (example1 & 2) to ensure that my LP formulation and solution is correct before applying it on real T20 cricket data (Intl. T20 and IPL) ## 3. LP formulation (Example 1) Initially I created a test example to ensure that I get the LP formulation and solution correct. Here the er1=4 and er2=3 and o1 & o2 are the overs bowled by bowlers 1 & 2. Also o1+o2=4 In this example as below o1 o2 Obj Fun(=4o1+3o2) 1 3 13 2 2 14 3 1 15 library(lpSolveAPI) library(dplyr) library(knitr) lprec <- make.lp(0, 2) a <-lp.control(lprec, sense="min") set.objfn(lprec, c(4, 3)) # Economy Rate of 4 and 3 for er1 and er2 add.constraint(lprec, c(1, 1), "=",4) # o1 + o2 =4 add.constraint(lprec, c(1, 0), ">",1) # o1 > 1 add.constraint(lprec, c(0, 1), ">",1) # o2 > 1 lprec ## Model name: ## C1 C2 ## Minimize 4 3 ## R1 1 1 = 4 ## R2 1 0 >= 1 ## R3 0 1 >= 1 ## Kind Std Std ## Type Real Real ## Upper Inf Inf ## Lower 0 0 b <-solve(lprec) get.objective(lprec) # 13 ## [1] 13 get.variables(lprec) # 1 3  ## [1] 1 3 Note 1: In the above example 13 runs is the minimum that can be scored and this requires LP solution: Minimum runs=13 • o1=1 • o2=3 Note 2:The numbers in the columns represent the number of overs that need to be bowled by a bowler to the corresponding batsman. ## 4. LP formulation (Example 2) In this formulation there are 2 bowlers and 2 batsmen o11,o12 are the oves bowled by bowler 1 to batsmen 1 & 2 and o21, o22 are the overs bowled by bowler 2 to batsmen 1 & 2 er11=4, er12=2,er21=2,er22=5 o11+o12+o21+o22=5 The solution for this manually computed is o11, o12, o21, o22 Runs where B11, B12 are the overs bowler 1 bowls to batsman 1 and B21 and B22 are overs bowler 2 bowls to batsman 2 o11 o12 o21 o22 Runs=(4*o11+2*o12+2*o21+5*o22) 1 1 1 2 18 1 2 1 1 15 2 1 1 1 17 1 1 2 1 15 lprec <- make.lp(0, 4) a <-lp.control(lprec, sense="min") set.objfn(lprec, c(4, 2,2,5)) add.constraint(lprec, c(1, 1,0,0), "<=",8) add.constraint(lprec, c(0, 0,1,1), "<=",7) add.constraint(lprec, c(1, 1,1,1), "=",5) add.constraint(lprec, c(1, 0,0,0), ">",1) add.constraint(lprec, c(0, 1,0,0), ">",1) add.constraint(lprec, c(0, 0,1,0), ">",1) add.constraint(lprec, c(0, 0,0,1), ">",1) lprec ## Model name: ## C1 C2 C3 C4 ## Minimize 4 2 2 5 ## R1 1 1 0 0 <= 8 ## R2 0 0 1 1 <= 7 ## R3 1 1 1 1 = 5 ## R4 1 0 0 0 >= 1 ## R5 0 1 0 0 >= 1 ## R6 0 0 1 0 >= 1 ## R7 0 0 0 1 >= 1 ## Kind Std Std Std Std ## Type Real Real Real Real ## Upper Inf Inf Inf Inf ## Lower 0 0 0 0 b<-solve(lprec) get.objective(lprec)  ## [1] 15 get.variables(lprec)  ## [1] 1 2 1 1 Note: In the above example 15 runs is the minimum that can be scored and this requires LP Solution: Minimum runs=15 • o11=1 • o12=2 • o21=1 • o22=1 It is possible to keep the minimum to other values and solves also. ## 5. LP formulation for International T20 India vs Australia (Batting lineup) To analyze batting and bowling lineups in the cricket world I needed to get the ball-by-ball details of runs scored by each batsman against each of the bowlers. Fortunately I had already created this with my R package yorkr. yorkr processes yaml data from Cricsheet. So I copied the data of all matches between Australia and India in International T20s. You can download my processed data for International T20 at Inswinger load("Australia-India-allMatches.RData") dim(matches) ## [1] 3541 25 The following functions compute the ‘Strike Rate’ of a batsman as SR=1/oversRunsScored Also the Economy Rate is computed as ER=1/oversRunsConceded Incidentally the SR=ER # Compute the Strike Rate of the batsman computeSR <- function(batsman1,bowler1){ a <- matches %>% filter(batsman==batsman1 & bowler==bowler1) a1 <- a %>% summarize(totalRuns=sum(runs),count=n()) %>% mutate(SR=(totalRuns/count)*6) a1 } # Compute the Economy Rate of the batsman computeER <- function(batsman1,bowler1){ a <- matches %>% filter(batsman==batsman1 & bowler==bowler1) a1 <- a %>% summarize(totalRuns=sum(runs),count=n()) %>% mutate(ER=(totalRuns/count)*6) a1 } Here I compute the Strike Rate of Virat Kohli, Yuvraj Singh and MS Dhoni against Shane Watson, Brett Lee and MA Starc  # Kohli kohliWatson<- computeSR("V Kohli","SR Watson") kohliWatson ## totalRuns count SR ## 1 45 37 7.297297 kohliLee <- computeSR("V Kohli","B Lee") kohliLee ## totalRuns count SR ## 1 10 7 8.571429 kohliStarc <- computeSR("V Kohli","MA Starc") kohliStarc ## totalRuns count SR ## 1 11 9 7.333333 # Yuvraj yuvrajWatson<- computeSR("Yuvraj Singh","SR Watson") yuvrajWatson ## totalRuns count SR ## 1 24 22 6.545455 yuvrajLee <- computeSR("Yuvraj Singh","B Lee") yuvrajLee ## totalRuns count SR ## 1 12 7 10.28571 yuvrajStarc <- computeSR("Yuvraj Singh","MA Starc") yuvrajStarc ## totalRuns count SR ## 1 12 8 9 # MS Dhoni dhoniWatson<- computeSR("MS Dhoni","SR Watson") dhoniWatson ## totalRuns count SR ## 1 33 28 7.071429 dhoniLee <- computeSR("MS Dhoni","B Lee") dhoniLee ## totalRuns count SR ## 1 26 20 7.8 dhoniStarc <- computeSR("MS Dhoni","MA Starc") dhoniStarc ## totalRuns count SR ## 1 11 8 8.25 When we consider the batting lineup, the problem is one of maximization. In the LP formulation below V Kohli has a SR of 7.29, 8.57, 7.33 against Watson, Lee & Starc Yuvraj has a SR of 6.5, 10.28, 9 against Watson, Lee & Starc and Dhoni has a SR of 7.07, 7.8, 8.25 against Watson, Lee and Starc The constraints are Watson, Lee and Starc have 3, 4 & 3 overs remaining respectively. The total number of overs remaining to be bowled is 9.The other constraints could be that a bowler bowls at least 1 over etc. Formulating and solving # 3 batsman x 3 bowlers lprec <- make.lp(0, 9) # Maximization a<-lp.control(lprec, sense="max") # Set the objective function set.objfn(lprec, c(kohliWatson$SR, kohliLee$SR,kohliStarc$SR,
yuvrajWatson$SR,yuvrajLee$SR,yuvrajStarc$SR, dhoniWatson$SR,dhoniLee$SR,dhoniStarc$SR))

#Assume the  bowlers have 3,4,3 overs left respectively
add.constraint(lprec, c(1, 1,1,0,0,0, 0,0,0), "<=",3)
#o11+o12+o13+o21+o22+o23+o31+o32+o33=8 (overs remaining)

add.constraint(lprec, c(1,0,0,0,0,0,0,0,0), ">=",1) #o11 >=1
add.constraint(lprec, c(0,1,0,0,0,0,0,0,0), ">=",0) #o12 >=0
add.constraint(lprec, c(0,0,1,0,0,0,0,0,0), ">=",0) #o13 >=0
add.constraint(lprec, c(0,0,0,1,0,0,0,0,0), ">=",1) #o21 >=1
add.constraint(lprec, c(0,0,0,0,1,0,0,0,0), ">=",1) #o22 >=1
add.constraint(lprec, c(0,0,0,0,0,1,0,0,0), ">=",0) #o23 >=0
add.constraint(lprec, c(0,0,0,0,0,0,1,0,0), ">=",1) #o31 >=1
add.constraint(lprec, c(0,0,0,0,0,0,0,1,0), ">=",0) #o32 >=0
add.constraint(lprec, c(0,0,0,0,0,0,0,0,1), ">=",0) #o33 >=0

lprec
## Model name:
##   a linear program with 9 decision variables and 13 constraints
b <-solve(lprec)
get.objective(lprec) #  
## [1] 77.16418
get.variables(lprec) # 
## [1] 1 2 0 1 3 0 1 0 1

This shows that the maximum runs that can be scored for the current strike rate is 77.16   runs in 9 overs The breakup is as follows

This is also shown below

get.variables(lprec) # 
## [1] 1 2 0 1 3 0 1 0 1

This is also shown below

e <- as.data.frame(rbind(c(1,2,0,3),c(1,3,0,4),c(1,0,1,2)))
names(e) <- c("S Watson","B Lee","MA Starc","Overs")
rownames(e) <- c("Kohli","Yuvraj","Dhoni")
e

LP Solution:
Maximum runs that can be scored by India against Australia is:77.164 if the 9 overs to be faced by the batsman are as below

##        S Watson B Lee MA Starc Overs
## Kohli         1     2        0     3
## Yuvraj        1     3        0     4
## Dhoni         1     0        1     2
#Total overs=9

Note: This assumes that the batsmen perform at their current Strike Rate. Howvever anything can happen in a real game, but nevertheless this is a fairly reasonable estimate of the performance

Note 2:The numbers in the columns represent the number of overs that need to be bowled by a bowler to the corresponding batsman.

Note 3:You could try other combinations of overs for the above SR. For the above constraints 77.16 is the highest score for the given number of overs

## 6. LP formulation for International T20 India vs Australia (Bowling lineup)

For this I compute how the bowling should be rotated between R Ashwin, RA Jadeja and JJ Bumrah when taking into account their performance against batsmen like Shane Watson, AJ Finch and David Warner. For the bowling performance I take the Economy rate of the bowlers. The data is the same as above

computeSR <- function(batsman1,bowler1){
a <- matches %>% filter(batsman==batsman1 & bowler==bowler1)
a1 <- a %>% summarize(totalRuns=sum(runs),count=n()) %>% mutate(SR=(totalRuns/count)*6)
a1
}
jadejaWatson
##   totalRuns count       ER
## 1        60    29 12.41379
jadejaFinch <- computeER("AJ Finch","RA Jadeja")
jadejaFinch
##   totalRuns count       ER
## 1        36    33 6.545455
jadejaWarner <- computeER("DA Warner","RA Jadeja")
jadejaWarner
##   totalRuns count       ER
## 1        23    11 12.54545
# Ashwin
ashwinWatson<- computeER("SR Watson","R Ashwin")
ashwinWatson
##   totalRuns count       ER
## 1        41    26 9.461538
ashwinFinch <- computeER("AJ Finch","R Ashwin")
ashwinFinch
##   totalRuns count   ER
## 1        63    36 10.5
ashwinWarner <- computeER("DA Warner","R Ashwin")
ashwinWarner
##   totalRuns count       ER
## 1        38    28 8.142857
# JJ Bunrah
bumrahWatson<- computeER("SR Watson","JJ Bumrah")
bumrahWatson
##   totalRuns count  ER
## 1        22    20 6.6
bumrahFinch <- computeER("AJ Finch","JJ Bumrah")
bumrahFinch
##   totalRuns count       ER
## 1        25    19 7.894737
bumrahWarner <- computeER("DA Warner","JJ Bumrah")
bumrahWarner
##   totalRuns count ER
## 1         2     4  3

As can be seen from above RA Jadeja has a ER of 12.4, 6.54, 12.54 against Watson, AJ Finch and Warner also Ashwin has a ER of 9.46, 10.5, 8.14 against Watson, Finch and Warner. Similarly Bumrah has an ER of 6.6,7.89, 3 against Watson, Finch and Warner
The constraints are Jadeja, Ashwin and Bumrah have 4, 3 & 4 overs remaining and the total overs remaining to be bowled is 10.

Formulating solving the bowling lineup is shown below

lprec <- make.lp(0, 9)
a <-lp.control(lprec, sense="min")

# Set the objective function
set.objfn(lprec, c(jadejaWatson$ER, jadejaFinch$ER,jadejaWarner$ER, ashwinWatson$ER,ashwinFinch$ER,ashwinWarner$ER,
bumrahWatson$ER,bumrahFinch$ER,bumrahWarner$ER)) add.constraint(lprec, c(1, 1,1,0,0,0, 0,0,0), "<=",4) # Jadeja has 4 overs add.constraint(lprec, c(0,0,0,1,1,1,0,0,0), "<=",3) # Ashwin has 3 overs left add.constraint(lprec, c(0,0,0,0,0,0,1,1,1), "<=",4) # Bumrah has 4 overs left add.constraint(lprec, c(1,1,1,1,1,1,1,1,1), "=",10) # Total overs = 10 add.constraint(lprec, c(1,0,0,0,0,0,0,0,0), ">=",1) add.constraint(lprec, c(0,1,0,0,0,0,0,0,0), ">=",0) add.constraint(lprec, c(0,0,1,0,0,0,0,0,0), ">=",1) add.constraint(lprec, c(0,0,0,1,0,0,0,0,0), ">=",0) add.constraint(lprec, c(0,0,0,0,1,0,0,0,0), ">=",1) add.constraint(lprec, c(0,0,0,0,0,1,0,0,0), ">=",0) add.constraint(lprec, c(0,0,0,0,0,0,1,0,0), ">=",0) add.constraint(lprec, c(0,0,0,0,0,0,0,1,0), ">=",1) add.constraint(lprec, c(0,0,0,0,0,0,0,0,1), ">=",0) lprec ## Model name: ## a linear program with 9 decision variables and 13 constraints b <-solve(lprec) get.objective(lprec) #  ## [1] 73.58775 get.variables(lprec) #  ## [1] 1 2 1 0 1 1 0 1 3 The minimum runs that will be conceded by these 3 bowlers in 10 overs is 73.58 assuming the bowling is rotated as follows e <- as.data.frame(rbind(c(1,0,0),c(2,1,1),c(1,1,3),c(4,2,4))) names(e) <- c("RA Jadeja","R Ashwin","JJ Bumrah") rownames(e) <- c("S Watson","AJ Finch","DA Warner","Overs") e  LP Solution: Minimum runs that will be conceded by India against Australia is 73.58 in 10 overs if the overs bowled are as follows ## RA Jadeja R Ashwin JJ Bumrah ## S Watson 1 0 0 ## AJ Finch 2 1 1 ## DA Warner 1 1 3 ## Overs 4 2 4 #Total overs=10  ## 7. LP formulation for IPL (Mumbai Indians – Kolkata Knight Riders – Bowling lineup) As in the case of International T20s I also have processed IPL data derived from my R package yorkr. yorkr. yorkr processes yaml data from Cricsheet. The processed data for all IPL matches can be downloaded from GooglyPlus load("Mumbai Indians-Kolkata Knight Riders-allMatches.RData") dim(matches) ## [1] 4237 25 # Compute the Economy Rate of the Mumbai Indian bowlers against Kolkata Knight Riders # Gambhir gambhirMalinga <- computeER("G Gambhir","SL Malinga") gambhirHarbhajan <- computeER("G Gambhir","Harbhajan Singh") gambhirPollard <- computeER("G Gambhir","KA Pollard") #Yusuf Pathan yusufMalinga <- computeER("YK Pathan","SL Malinga") yusufHarbhajan <- computeER("YK Pathan","Harbhajan Singh") yusufPollard <- computeER("YK Pathan","KA Pollard") #JH Kallis kallisMalinga <- computeER("JH Kallis","SL Malinga") kallisHarbhajan <- computeER("JH Kallis","Harbhajan Singh") kallisPollard <- computeER("JH Kallis","KA Pollard") #RV Uthappa uthappaMalinga <- computeER("RV Uthappa","SL Malinga") uthappaHarbhajan <- computeER("RV Uthappa","Harbhajan Singh") uthappaPollard <- computeER("RV Uthappa","KA Pollard") Here gambhirMalinga, yusufMalinga, kallisMalinga, uthappaMalinga is the ER of Malinga against Gambhir, Yusuf Pathan, Kallis and Uthappa gambhirHarbhajan, yusufHarbhajan, kallisHarbhajan, uthappaHarbhajan is the ER of Harbhajan against Gambhir, Yusuf Pathan, Kallis and Uthappa gambhirPollard, yusufPollard, kallisPollard, uthappaPollard is the ER of Kieron Pollard against Gambhir, Yusuf Pathan, Kallis and Uthappa The constraints are Malinga, Harbhajan and Pollard have 4 overs each and remaining overs to be bowled is 10. Formulating and solving this for the bowling lineup of Mumbai Indians against Kolkata Knight Riders  library("lpSolveAPI") lprec <- make.lp(0, 12) a=lp.control(lprec, sense="min") set.objfn(lprec, c(gambhirMalinga$ER, yusufMalinga$ER,kallisMalinga$ER,uthappaMalinga$ER, gambhirHarbhajan$ER,yusufHarbhajan$ER,kallisHarbhajan$ER,uthappaHarbhajan$ER, gambhirPollard$ER,yusufPollard$ER,kallisPollard$ER,uthappaPollard$ER)) add.constraint(lprec, c(1,1,1,1, 0,0,0,0, 0,0,0,0), "<=",4) add.constraint(lprec, c(0,0,0,0,1,1,1,1,0,0,0,0), "<=",4) add.constraint(lprec, c(0,0,0,0,0,0,0,0,1,1,1,1), "<=",4) add.constraint(lprec, c(1,1,1,1,1,1,1,1,1,1,1,1), "=",10) add.constraint(lprec, c(1,0,0,0,0,0,0,0,0,0,0,0), ">=",0) add.constraint(lprec, c(0,1,0,0,0,0,0,0,0,0,0,0), ">=",1) add.constraint(lprec, c(0,0,1,0,0,0,0,0,0,0,0,0), ">=",0) add.constraint(lprec, c(0,0,0,1,0,0,0,0,0,0,0,0), ">=",0) add.constraint(lprec, c(0,0,0,0,1,0,0,0,0,0,0,0), ">=",0) add.constraint(lprec, c(0,0,0,0,0,1,0,0,0,0,0,0), ">=",1) add.constraint(lprec, c(0,0,0,0,0,0,1,0,0,0,0,0), ">=",0) add.constraint(lprec, c(0,0,0,0,0,0,0,1,0,0,0,0), ">=",1) add.constraint(lprec, c(0,0,0,0,0,0,0,0,1,0,0,0), ">=",0) add.constraint(lprec, c(0,0,0,0,0,0,0,0,0,1,0,0), ">=",1) add.constraint(lprec, c(0,0,0,0,0,0,0,0,0,0,1,0), ">=",0) add.constraint(lprec, c(0,0,0,0,0,0,0,0,0,0,0,1), ">=",0) lprec ## Model name: ## a linear program with 12 decision variables and 16 constraints  b=solve(lprec) get.objective(lprec) #  ## [1] 55.57887  get.variables(lprec) #  ## [1] 3 1 0 0 0 1 0 1 3 1 0 0 e <- as.data.frame(rbind(c(3,1,0,0,4),c(0, 1, 0,1,2),c(3, 1, 0,0,4))) names(e) <- c("Gambhir","Yusuf","Kallis","Uthappa","Overs") rownames(e) <- c("Malinga","Harbhajan","Pollard") e LP Solution: Mumbai Indians can restrict Kolkata Knight Riders to 55.87 in 10 overs if the overs are bowled as below ## Gambhir Yusuf Kallis Uthappa Overs ## Malinga 3 1 0 0 4 ## Harbhajan 0 1 0 1 2 ## Pollard 3 1 0 0 4 #Total overs=10  ## 8. LP formulation for IPL (Mumbai Indians – Kolkata Knight Riders – Batting lineup) As I mentioned it is possible to perform a maximation with the same formulation since computeSR<==>computeER This just flips the problem around and computes the maximum runs that can be scored for the batsman’s Strike rate (this is same as the bowler’s Economy rate) i.e. gambhirMalinga, yusufMalinga, kallisMalinga, uthappaMalinga is the SR of Gambhir, Yusuf Pathan, Kallis and Uthappa against Malinga gambhirHarbhajan, yusufHarbhajan, kallisHarbhajan, uthappaHarbhajan is the SR of Gambhir, Yusuf Pathan, Kallis and Uthappa against Harbhajan gambhirPollard, yusufPollard, kallisPollard, uthappaPollard is the SR of Gambhir, Yusuf Pathan, Kallis and Uthappa against Kieron Pollard. The constraints are Malinga, Harbhajan and Pollard have 4 overs each and remaining overs to be bowled is 10.  library("lpSolveAPI") lprec <- make.lp(0, 12) a=lp.control(lprec, sense="max") a <-set.objfn(lprec, c(gambhirMalinga$ER, yusufMalinga$ER,kallisMalinga$ER,uthappaMalinga$ER, gambhirHarbhajan$ER,yusufHarbhajan$ER,kallisHarbhajan$ER,uthappaHarbhajan$ER, gambhirPollard$ER,yusufPollard$ER,kallisPollard$ER,uthappaPollard$ER)) add.constraint(lprec, c(1,1,1,1, 0,0,0,0, 0,0,0,0), "<=",4) add.constraint(lprec, c(0,0,0,0,1,1,1,1,0,0,0,0), "<=",4) add.constraint(lprec, c(0,0,0,0,0,0,0,0,1,1,1,1), "<=",4) add.constraint(lprec, c(1,1,1,1,1,1,1,1,1,1,1,1), "=",11) add.constraint(lprec, c(1,0,0,0,0,0,0,0,0,0,0,0), ">=",0) add.constraint(lprec, c(0,1,0,0,0,0,0,0,0,0,0,0), ">=",1) add.constraint(lprec, c(0,0,1,0,0,0,0,0,0,0,0,0), ">=",0) add.constraint(lprec, c(0,0,0,1,0,0,0,0,0,0,0,0), ">=",0) add.constraint(lprec, c(0,0,0,0,1,0,0,0,0,0,0,0), ">=",0) add.constraint(lprec, c(0,0,0,0,0,1,0,0,0,0,0,0), ">=",1) add.constraint(lprec, c(0,0,0,0,0,0,1,0,0,0,0,0), ">=",0) add.constraint(lprec, c(0,0,0,0,0,0,0,1,0,0,0,0), ">=",1) add.constraint(lprec, c(0,0,0,0,0,0,0,0,1,0,0,0), ">=",0) add.constraint(lprec, c(0,0,0,0,0,0,0,0,0,1,0,0), ">=",1) add.constraint(lprec, c(0,0,0,0,0,0,0,0,0,0,1,0), ">=",0) add.constraint(lprec, c(0,0,0,0,0,0,0,0,0,0,0,1), ">=",0) lprec ## Model name: ## a linear program with 12 decision variables and 16 constraints  b=solve(lprec) get.objective(lprec) #  ## [1] 94.22649  get.variables(lprec) #  ## [1] 0 3 0 0 0 1 0 3 0 1 3 0 e <- as.data.frame(rbind(c(0,3,0,0,3),c(0, 1, 0,3,4),c(0, 1, 3,0,4))) names(e) <- c("Gambhir","Yusuf","Kallis","Uthappa","Overs") rownames(e) <- c("Malinga","Harbhajan","Pollard") e LP Solution: Kolkata Knight Riders can score a maximum of 94.22 in 11 overs against Mumbai Indians if the the number of overs KKR face is as below ## Gambhir Yusuf Kallis Uthappa Overs ## Malinga 0 3 0 0 3 ## Harbhajan 0 1 0 3 4 ## Pollard 0 1 3 0 4 #Total overs=11  Conclusion: It is possible to thus determine the optimum no of overs to give to a specific bowler based on his/her Economy Rate with a particular batsman. Similarly one can determine the maximum runs that can be scored by a batsmen based on their strike rate with bowlers. Cricket like many other games is a game of strategy, skill, talent and some amount of luck. So while the LP formulation can provide some direction, one must be aware anything could happen in a game of cricket! Thoughts, comments, suggestions welcome! To see all posts see Index of Posts # R vs Python: Different similarities and similar differences A debate about which language is better suited for Datascience, R or Python, can set off diehard fans of these languages into a tizzy. This post tries to look at some of the different similarities and similar differences between these languages. To a large extent the ease or difficulty in learning R or Python is subjective. I have heard that R has a steeper learning curve than Python and also vice versa. This probably depends on the degree of familiarity with the languuge To a large extent both R an Python do the same thing in just slightly different ways and syntaxes. The ease or the difficulty in the R/Python construct’s largely is in the ‘eyes of the beholder’ nay, programmer’ we could say. I include my own experience with the languages below. ### 1. R data types R has the following data types 1. Character 2. Integer 3. Numeric 4. Logical 5. Complex 6. Raw Python has several data types 1. Int 2. float 3. Long 4. Complex and so on ### 2. R Vector vs Python List A common data type in R is the vector. Python has a similar data type, the list # R vectors a<-c(4,5,1,3,4,5) print(a[3]) ## [1] 1 print(a[3:4]) # R does not always need the explicit print.  ## [1] 1 3 #R type of variable print(class(a)) ## [1] "numeric" # Length of a print(length(a)) ## [1] 6 # Python lists a=[4,5,1,3,4,5] # print(a[2]) # Some python IDEs require the explicit print print(a[2:5]) print(type(a)) # Length of a print(len(a)) ## 1 ## [1, 3, 4] ## <class 'list'> ## 6 ### 2a. Other data types – Python Python also has certain other data types like the tuple, dictionary etc as shown below. R does not have as many of the data types, nevertheless we can do everything that Python does in R # Python tuple b = (4,5,7,8) print(b) #Python dictionary c={'name':'Ganesh','age':54,'Work':'Professional'} print(c) #Print type of variable c  ## (4, 5, 7, 8) ## {'name': 'Ganesh', 'age': 54, 'Work': 'Professional'} ### 2.Type of Variable To know the type of the variable in R we use ‘class’, In Python the corresponding command is ‘type’ #R - Type of variable a<-c(4,5,1,3,4,5) print(class(a)) ## [1] "numeric" #Python - Print type of tuple a a=[4,5,1,3,4,5] print(type(a)) b=(4,3,"the",2) print(type(b)) ## <class 'list'> ## <class 'tuple'> ### 3. Length To know length in R, use length() #R - Length of vector # Length of a a<-c(4,5,1,3,4,5) print(length(a)) ## [1] 6 To know the length of a list,tuple or dict we can use len() # Python - Length of list , tuple etc # Length of a a=[4,5,1,3,4,5] print(len(a)) # Length of b b = (4,5,7,8) print(len(b))  ## 6 ## 4 ### 4. Accessing help To access help in R we use the ‘?’ or the ‘help’ function #R - Help - To be done in R console or RStudio #?sapply #help(sapply) Help in python on any topic involves #Python help - This can be done on a (I)Python console #help(len) #?len ### 5. Subsetting The key difference between R and Python with regards to subsetting is that in R the index starts at 1. In Python it starts at 0, much like C,C++ or Java To subset a vector in R we use #R - Subset a<-c(4,5,1,3,4,8,12,18,1) print(a[3]) ## [1] 1 # To print a range or a slice. Print from the 3rd to the 5th element print(a[3:6]) ## [1] 1 3 4 8 Python also uses indices. The difference in Python is that the index starts from 0/ #Python - Subset a=[4,5,1,3,4,8,12,18,1] # Print the 4th element (starts from 0) print(a[3]) # Print a slice from 4 to 6th element print(a[3:6]) ## 3 ## [3, 4, 8] ### 6. Operations on vectors in R and operation on lists in Python In R we can do many operations on vectors for e.g. element by element addition, subtraction, exponentation,product etc. as show #R - Operations on vectors a<- c(5,2,3,1,7) b<- c(1,5,4,6,8) #Element wise Addition print(a+b) ## [1] 6 7 7 7 15 #Element wise subtraction print(a-b) ## [1] 4 -3 -1 -5 -1 #Element wise product print(a*b) ## [1] 5 10 12 6 56 # Exponentiating the elements of a vector print(a^2) ## [1] 25 4 9 1 49 In Python to do this on lists we need to use the ‘map’ and the ‘lambda’ function as follows # Python - Operations on list a =[5,2,3,1,7] b =[1,5,4,6,8] #Element wise addition with map & lambda print(list(map(lambda x,y: x+y,a,b))) #Element wise subtraction print(list(map(lambda x,y: x-y,a,b))) #Element wise product print(list(map(lambda x,y: x*y,a,b))) # Exponentiating the elements of a list print(list(map(lambda x: x**2,a)))  ## [6, 7, 7, 7, 15] ## [4, -3, -1, -5, -1] ## [5, 10, 12, 6, 56] ## [25, 4, 9, 1, 49] However if we create ndarrays from lists then we can do the element wise addition,subtraction,product, etc. like R. Numpy is really a powerful module with many, many functions for matrix manipulations import numpy as np a =[5,2,3,1,7] b =[1,5,4,6,8] a=np.array(a) b=np.array(b) #Element wise addition print(a+b) #Element wise subtraction print(a-b) #Element wise product print(a*b) # Exponentiating the elements of a list print(a**2)  ## [ 6 7 7 7 15] ## [ 4 -3 -1 -5 -1] ## [ 5 10 12 6 56] ## [25 4 9 1 49] ### 7. Getting the index of element To determine the index of an element which satisifies a specific logical condition in R use ‘which’. In the code below the index of element which is equal to 1 is 4 # R - Which a<- c(5,2,3,1,7) print(which(a == 1)) ## [1] 4 In Python array we can use np.where to get the same effect. The index will be 3 as the index starts from 0 # Python - np.where import numpy as np a =[5,2,3,1,7] a=np.array(a) print(np.where(a==1)) ## (array([3], dtype=int64),) ### 8. Data frames R, by default comes with a set of in-built datasets. There are some datasets which come with the SkiKit- Learn package # R # To check built datasets use #data() - In R console or in R Studio #iris - Don't print to console We can use the in-built data sets that come with Scikit package #Python import sklearn as sklearn import pandas as pd from sklearn import datasets # This creates a Sklearn bunch data = datasets.load_iris() # Convert to Pandas dataframe iris = pd.DataFrame(data.data, columns=data.feature_names) ### 9. Working with dataframes With R you can work with dataframes directly. For more complex dataframe operations in R there are convenient packages like dplyr, reshape2 etc. For Python we need to use the Pandas package. Pandas is quite comprehensive in the list of things we can do with data frames The most common operations on a dataframe are • Check the size of the dataframe • Take a look at the top 5 or bottom 5 rows of dataframe • Check the content of the dataframe #### a.Size In R use dim() #R - Size dim(iris) ## [1] 150 5 For Python use .shape #Python - size import sklearn as sklearn import pandas as pd from sklearn import datasets data = datasets.load_iris() # Convert to Pandas dataframe iris = pd.DataFrame(data.data, columns=data.feature_names) iris.shape #### b. Top & bottom 5 rows of dataframe To know the top and bottom rows of a data frame we use head() & tail as shown below for R and Python #R head(iris,5) ## Sepal.Length Sepal.Width Petal.Length Petal.Width Species ## 1 5.1 3.5 1.4 0.2 setosa ## 2 4.9 3.0 1.4 0.2 setosa ## 3 4.7 3.2 1.3 0.2 setosa ## 4 4.6 3.1 1.5 0.2 setosa ## 5 5.0 3.6 1.4 0.2 setosa tail(iris,5) ## Sepal.Length Sepal.Width Petal.Length Petal.Width Species ## 146 6.7 3.0 5.2 2.3 virginica ## 147 6.3 2.5 5.0 1.9 virginica ## 148 6.5 3.0 5.2 2.0 virginica ## 149 6.2 3.4 5.4 2.3 virginica ## 150 5.9 3.0 5.1 1.8 virginica #Python import sklearn as sklearn import pandas as pd from sklearn import datasets data = datasets.load_iris() # Convert to Pandas dataframe iris = pd.DataFrame(data.data, columns=data.feature_names) print(iris.head(5)) print(iris.tail(5)) ## sepal length (cm) sepal width (cm) petal length (cm) petal width (cm) ## 0 5.1 3.5 1.4 0.2 ## 1 4.9 3.0 1.4 0.2 ## 2 4.7 3.2 1.3 0.2 ## 3 4.6 3.1 1.5 0.2 ## 4 5.0 3.6 1.4 0.2 ## sepal length (cm) sepal width (cm) petal length (cm) petal width (cm) ## 145 6.7 3.0 5.2 2.3 ## 146 6.3 2.5 5.0 1.9 ## 147 6.5 3.0 5.2 2.0 ## 148 6.2 3.4 5.4 2.3 ## 149 5.9 3.0 5.1 1.8 #### c. Check the content of the dataframe #R summary(iris) ## Sepal.Length Sepal.Width Petal.Length Petal.Width ## Min. :4.300 Min. :2.000 Min. :1.000 Min. :0.100 ## 1st Qu.:5.100 1st Qu.:2.800 1st Qu.:1.600 1st Qu.:0.300 ## Median :5.800 Median :3.000 Median :4.350 Median :1.300 ## Mean :5.843 Mean :3.057 Mean :3.758 Mean :1.199 ## 3rd Qu.:6.400 3rd Qu.:3.300 3rd Qu.:5.100 3rd Qu.:1.800 ## Max. :7.900 Max. :4.400 Max. :6.900 Max. :2.500 ## Species ## setosa :50 ## versicolor:50 ## virginica :50 ## ## ##  str(iris) ## 'data.frame': 150 obs. of 5 variables: ##$ Sepal.Length: num  5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9 ...
##  $Sepal.Width : num 3.5 3 3.2 3.1 3.6 3.9 3.4 3.4 2.9 3.1 ... ##$ Petal.Length: num  1.4 1.4 1.3 1.5 1.4 1.7 1.4 1.5 1.4 1.5 ...
##  $Petal.Width : num 0.2 0.2 0.2 0.2 0.2 0.4 0.3 0.2 0.2 0.1 ... ##$ Species     : Factor w/ 3 levels "setosa","versicolor",..: 1 1 1 1 1 1 1 1 1 1 ...
#Python
import sklearn as sklearn
import pandas as pd
from sklearn import datasets
# Convert to Pandas dataframe
iris = pd.DataFrame(data.data, columns=data.feature_names)
print(iris.info())
## <class 'pandas.core.frame.DataFrame'>
## RangeIndex: 150 entries, 0 to 149
## Data columns (total 4 columns):
## sepal length (cm)    150 non-null float64
## sepal width (cm)     150 non-null float64
## petal length (cm)    150 non-null float64
## petal width (cm)     150 non-null float64
## dtypes: float64(4)
## memory usage: 4.8 KB
## None

#### d. Check column names

#R
names(iris)
## [1] "Sepal.Length" "Sepal.Width"  "Petal.Length" "Petal.Width"
## [5] "Species"
colnames(iris)
## [1] "Sepal.Length" "Sepal.Width"  "Petal.Length" "Petal.Width"
## [5] "Species"
#Python
import sklearn as sklearn
import pandas as pd
from sklearn import datasets
# Convert to Pandas dataframe
iris = pd.DataFrame(data.data, columns=data.feature_names)
#Get column names
print(iris.columns)
## Index(['sepal length (cm)', 'sepal width (cm)', 'petal length (cm)',
##        'petal width (cm)'],
##       dtype='object')

#### e. Rename columns

In R we can assign a vector to column names

#R
colnames(iris) <- c("lengthOfSepal","widthOfSepal","lengthOfPetal","widthOfPetal","Species")
colnames(iris)
## [1] "lengthOfSepal" "widthOfSepal"  "lengthOfPetal" "widthOfPetal"
## [5] "Species"

In Python we can assign a list to s.columns

#Python
import sklearn as sklearn
import pandas as pd
from sklearn import datasets
# Convert to Pandas dataframe
iris = pd.DataFrame(data.data, columns=data.feature_names)
iris.columns = ["lengthOfSepal","widthOfSepal","lengthOfPetal","widthOfPetal"]
print(iris.columns)
## Index(['lengthOfSepal', 'widthOfSepal', 'lengthOfPetal', 'widthOfPetal'], dtype='object')

### f.Details of dataframe

#Python
import sklearn as sklearn
import pandas as pd
from sklearn import datasets
# Convert to Pandas dataframe
iris = pd.DataFrame(data.data, columns=data.feature_names)
print(iris.info())
## <class 'pandas.core.frame.DataFrame'>
## RangeIndex: 150 entries, 0 to 149
## Data columns (total 4 columns):
## sepal length (cm)    150 non-null float64
## sepal width (cm)     150 non-null float64
## petal length (cm)    150 non-null float64
## petal width (cm)     150 non-null float64
## dtypes: float64(4)
## memory usage: 4.8 KB
## None

#### g. Subsetting dataframes

# R
#To subset a dataframe 'df' in R we use df[row,column] or df[row vector,column vector]
#df[row,column]
iris[3,4]
## [1] 0.2
#df[row vector, column vector]
iris[2:5,1:3]
##   lengthOfSepal widthOfSepal lengthOfPetal
## 2           4.9          3.0           1.4
## 3           4.7          3.2           1.3
## 4           4.6          3.1           1.5
## 5           5.0          3.6           1.4
#If we omit the row vector, then it implies all rows or if we omit the column vector
# then implies all columns for that row
iris[2:5,]
##   lengthOfSepal widthOfSepal lengthOfPetal widthOfPetal Species
## 2           4.9          3.0           1.4          0.2  setosa
## 3           4.7          3.2           1.3          0.2  setosa
## 4           4.6          3.1           1.5          0.2  setosa
## 5           5.0          3.6           1.4          0.2  setosa
# In R we can all specific columns by column names
iris$Sepal.Length[2:5] ## NULL #Python # To select an entire row we use .iloc. The index can be used with the ':'. If # .iloc[start row: end row]. If start row is omitted then it implies the beginning of # data frame, if end row is omitted then it implies all rows till end #Python import sklearn as sklearn import pandas as pd from sklearn import datasets data = datasets.load_iris() # Convert to Pandas dataframe iris = pd.DataFrame(data.data, columns=data.feature_names) print(iris.iloc[3]) print(iris[:5]) # In python we can select columns by column name as follows print(iris['sepal length (cm)'][2:6]) #If you want to select more than 2 columns then you must use the double '[[]]' since the # index is a list itself print(iris[['sepal length (cm)','sepal width (cm)']][4:7]) ## sepal length (cm) 4.6 ## sepal width (cm) 3.1 ## petal length (cm) 1.5 ## petal width (cm) 0.2 ## Name: 3, dtype: float64 ## sepal length (cm) sepal width (cm) petal length (cm) petal width (cm) ## 0 5.1 3.5 1.4 0.2 ## 1 4.9 3.0 1.4 0.2 ## 2 4.7 3.2 1.3 0.2 ## 3 4.6 3.1 1.5 0.2 ## 4 5.0 3.6 1.4 0.2 ## 2 4.7 ## 3 4.6 ## 4 5.0 ## 5 5.4 ## Name: sepal length (cm), dtype: float64 ## sepal length (cm) sepal width (cm) ## 4 5.0 3.6 ## 5 5.4 3.9 ## 6 4.6 3.4 #### h. Computing Mean, Standard deviation #R #Mean mean(iris$lengthOfSepal)
## [1] 5.843333
#Standard deviation
sd(iris$widthOfSepal) ## [1] 0.4358663 #Python #Mean import sklearn as sklearn import pandas as pd from sklearn import datasets data = datasets.load_iris() # Convert to Pandas dataframe iris = pd.DataFrame(data.data, columns=data.feature_names) # Convert to Pandas dataframe print(iris['sepal length (cm)'].mean()) #Standard deviation print(iris['sepal width (cm)'].std()) ## 5.843333333333335 ## 0.4335943113621737 #### i. Boxplot Boxplot can be produced in R using baseplot #R boxplot(iris$lengthOfSepal)

Matplotlib is a popular package in Python for plots

#Python
import sklearn as sklearn
import pandas as pd
import matplotlib.pyplot as plt
from sklearn import datasets
# Convert to Pandas dataframe
iris = pd.DataFrame(data.data, columns=data.feature_names)
img=plt.boxplot(iris['sepal length (cm)'])
plt.show(img)

#### j.Scatter plot

#R
plot(iris$widthOfSepal,iris$lengthOfSepal)

#Python
import matplotlib.pyplot as plt
import sklearn as sklearn
import pandas as pd
from sklearn import datasets
# Convert to Pandas dataframe
iris = pd.DataFrame(data.data, columns=data.feature_names)
img=plt.scatter(iris['sepal width (cm)'],iris['sepal length (cm)'])
#plt.show(img)

#### k. Read from csv file

#R
tendulkar= read.csv("tendulkar.csv",stringsAsFactors = FALSE,na.strings=c(NA,"-"))
#Dimensions of dataframe
dim(tendulkar)
## [1] 347  13
names(tendulkar)
##  [1] "X"          "Runs"       "Mins"       "BF"         "X4s"
##  [6] "X6s"        "SR"         "Pos"        "Dismissal"  "Inns"
## [11] "Opposition" "Ground"     "Start.Date"

Use pandas.read_csv() for Python

#Python
import pandas as pd
print(tendulkar.shape)
print(tendulkar.columns)
## (347, 13)
## Index(['Unnamed: 0', 'Runs', 'Mins', 'BF', '4s', '6s', 'SR', 'Pos',
##        'Dismissal', 'Inns', 'Opposition', 'Ground', 'Start Date'],
##       dtype='object')

#### l. Clean the dataframe in R and Python.

The following steps are done for R and Python
1.Remove rows with ‘DNB’
2.Remove rows with ‘TDNB’
3.Remove rows with absent
4.Remove the “*” indicating not out
5.Remove incomplete rows with NA for R or NaN in Python
6.Do a scatter plot

#R
# Remove rows with 'DNB'
a <- tendulkar$Runs != "DNB" tendulkar <- tendulkar[a,] dim(tendulkar) ## [1] 330 13 # Remove rows with 'TDNB' b <- tendulkar$Runs != "TDNB"
tendulkar <- tendulkar[b,]

# Remove rows with absent
c <- tendulkar$Runs != "absent" tendulkar <- tendulkar[c,] dim(tendulkar) ## [1] 329 13 # Remove the "* indicating not out tendulkar$Runs <- as.numeric(gsub("\\*","",tendulkar$Runs)) dim(tendulkar) ## [1] 329 13 # Select only complete rows - complete.cases() c <- complete.cases(tendulkar) #Subset the rows which are complete tendulkar <- tendulkar[c,] dim(tendulkar) ## [1] 327 13 # Do some base plotting - Scatter plot plot(tendulkar$BF,tendulkar$Runs) #Python import pandas as pd import matplotlib.pyplot as plt #Read csv tendulkar= pd.read_csv("tendulkar.csv",na_values=["-"]) print(tendulkar.shape) # Remove rows with 'DNB' a=tendulkar.Runs !="DNB" tendulkar=tendulkar[a] print(tendulkar.shape) # Remove rows with 'TDNB' b=tendulkar.Runs !="TDNB" tendulkar=tendulkar[b] print(tendulkar.shape) # Remove rows with absent c= tendulkar.Runs != "absent" tendulkar=tendulkar[c] print(tendulkar.shape) # Remove the "* indicating not out tendulkar.Runs= tendulkar.Runs.str.replace(r"[*]","") #Select only complete rows - dropna() tendulkar=tendulkar.dropna() print(tendulkar.shape) tendulkar.Runs = tendulkar.Runs.astype(int) tendulkar.BF = tendulkar.BF.astype(int) #Scatter plot plt.scatter(tendulkar.BF,tendulkar.Runs) ## (347, 13) ## (330, 13) ## (329, 13) ## (329, 13) ## (327, 13) #### m.Chaining operations on dataframes To chain a set of operations we need to use an R package like dplyr. Pandas does this The following operations are done on tendulkar data frame by dplyr for R and Pandas for Python below 1. Group by ground 2. Compute average runs in each ground 3. Arrange in descending order #R library(dplyr) tendulkar1 <- tendulkar %>% group_by(Ground) %>% summarise(meanRuns= mean(Runs)) %>% arrange(desc(meanRuns)) head(tendulkar1,10) ## # A tibble: 10 × 2 ## Ground meanRuns ## ## 1 Multan 194.00000 ## 2 Leeds 193.00000 ## 3 Colombo (RPS) 143.00000 ## 4 Lucknow 142.00000 ## 5 Dhaka 132.75000 ## 6 Manchester 93.50000 ## 7 Sydney 87.22222 ## 8 Bloemfontein 85.00000 ## 9 Georgetown 81.00000 ## 10 Colombo (SSC) 77.55556 #Python import pandas as pd #Read csv tendulkar= pd.read_csv("tendulkar.csv",na_values=["-"]) print(tendulkar.shape) # Remove rows with 'DNB' a=tendulkar.Runs !="DNB" tendulkar=tendulkar[a] # Remove rows with 'TDNB' b=tendulkar.Runs !="TDNB" tendulkar=tendulkar[b] # Remove rows with absent c= tendulkar.Runs != "absent" tendulkar=tendulkar[c] # Remove the "* indicating not out tendulkar.Runs= tendulkar.Runs.str.replace(r"[*]","") #Select only complete rows - dropna() tendulkar=tendulkar.dropna() tendulkar.Runs = tendulkar.Runs.astype(int) tendulkar.BF = tendulkar.BF.astype(int) tendulkar1= tendulkar.groupby('Ground').mean()['Runs'].sort_values(ascending=False) print(tendulkar1.head(10)) ## (347, 13) ## Ground ## Multan 194.000000 ## Leeds 193.000000 ## Colombo (RPS) 143.000000 ## Lucknow 142.000000 ## Dhaka 132.750000 ## Manchester 93.500000 ## Sydney 87.222222 ## Bloemfontein 85.000000 ## Georgetown 81.000000 ## Colombo (SSC) 77.555556 ## Name: Runs, dtype: float64 ### 9. Functions product <- function(a,b){ c<- a*b c } product(5,7) ## [1] 35 def product(a,b): c = a*b return c print(product(5,7))  ## 35  ## Conclusion Personally, I took to R, much like a ‘duck takes to water’. I found the R syntax very simple and mostly intuitive. R packages like dplyr, ggplot2, reshape2, make the language quite irrestible. R is weakly typed and has only numeric and character types as opposed to the full fledged data types in Python. Python, has too many bells and whistles, which can be a little bewildering to the novice. It is possible that they may be useful as one becomes more experienced with the language. Also I found that installing Python packages sometimes gives errors with Python versions 2.7 or 3.6. This will leave you scrambling to google to find how to fix these problems. These can be quite frustrating. R on the other hand makes installing R packages a breeze. Anyway, this is my current opinion, and like all opinions, may change in the course of time. Let’s see! I may write a follow up post with more advanced features of R and Python. So do keep checking! Long live R! Viva la Python! Note: This post was created using RStudio’s RMarkdown which allows you to embed R and Python code snippets. It works perfectly, except that matplotlib’s pyplot does not display. # More book, more cricket! 2nd edition of my books now on Amazon a) Cricket analytics with cricketr b) Beaten by sheer pace – Cricket analytics with yorkr is now available on Amazon, both as Paperback and Kindle versions. The Kindle versions are just$4.99 for both books. Pick up your copies today!!!

A) Cricket analytics with cricketr: Second Edition

Click hereCricket analytics with cricketr: Second Edition

B) Beaten by sheer pace: Cricket analytics with yorkr(2nd edition)

Pick up your copies today!!!

# My 3 video presentations on “Essential R”

In this post I include my  3 video presentations on the topic “Essential R”. In these 3 presentations I cover the entire landscape of R. I cover the following

• R Language – The essentials
• Key R Packages (dplyr, lubridate, ggplot2, etc.)
• How to create R Markdown and share reports
• A look at Shiny apps
• How to create a simple R package

You can download the relevant slide deck and practice code at Essential R

Essential R – Part 1
This video cover basic R data types – character, numeric, vectors, matrices, lists and data frames. It also touches on how to subset these data types

Essential R – Part 2
This video continues on how to subset dataframes (the most important data type) and some important packages. It also presents one of the most important job of a Data Scientist – that of cleaning and shaping the data. This is done with an example unclean data frame. It also  touches on some  key operations of dplyr like select, filter, arrange, summarise and mutate. Other packages like lubridate, quantmod are also included. This presentation also shows how to use base plot and ggplot2

Essential R – Part 3
This final session covers R Markdown , and  touches on some of the key markdown elements. There is a brief overview of a simple Shiny app. Finally this presentation also shows the key steps to create an R package

These 3 R sessions cover most of the basic R topics that we tend to use in a our day-to-day R way of life. With this you should be able to hit the ground running!

Hope you enjoy these video presentation and also hope you have an even greater time with R!

Check out my 2 books on cricket, a) Cricket analytics with cricketr b) Beaten by sheer pace – Cricket analytics with yorkr, now available in both paperback & kindle versions on Amazon!!! Pick up your copies today!

To see all my posts click – Index of posts