# Practical Machine Learning with R and Python – Part 3

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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

- Practical Machine Learning with R and Python – Part 1
- Practical Machine Learning with R and Python – Part 2

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.

# 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 ML models have to be searched. This can be shown as follows

There are ways to choose single feature ML models among ‘n’ features, ways to choose 2 feature models among ‘n’ models and so on, or

= Total number of models in Best Fit. Since from Binomial theorem we have

When x=1 in the equation (1) above, this becomes

Hence there are models to search amongst in Best Fit. For 10 features this is or ~1000 models and for 40 features this becomes 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 ~ 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 which is of the order of . 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')

The table above shows the average score, 10 fold CV errors, the features included at every step, std. deviation and std. error

## 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,, 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 which is of the order of . 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 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 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')

The table above shows the average score, 10 fold CV errors, the features included at every step, std. deviation and std. error

## 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')

The table above shows the average score, 10 fold CV errors, the features included at every step, std. deviation and std. error

## 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

Ridge regularization =

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 to 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

## 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)

The plot below gives the training and test R squared error

## 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')

1. Natural language processing: What would Shakespeare say?

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