Chapter 09 (page 368): 5, 7, 8

We have seen that we can fit an SVM with a non-linear kernel in order to perform classification using a non-linear decision boundary.We will now see that we can also obtain a non-linear decision boundary by performing logistic regression using non-linear transformations of the features.

  1. Generate a data set with n = 500 and p = 2, such that the observations belong to two classes with a quadratic decision boundary between them. For instance, you can do this as follows:
library(tidyverse)
## Warning: package 'tidyverse' was built under R version 3.6.3
## -- Attaching packages --------------------------------------- tidyverse 1.3.0 --
## v ggplot2 3.3.2     v purrr   0.3.4
## v tibble  3.0.3     v dplyr   1.0.1
## v tidyr   1.1.1     v stringr 1.4.0
## v readr   1.3.1     v forcats 0.5.0
## Warning: package 'ggplot2' was built under R version 3.6.3
## Warning: package 'tibble' was built under R version 3.6.3
## Warning: package 'tidyr' was built under R version 3.6.3
## Warning: package 'readr' was built under R version 3.6.3
## Warning: package 'purrr' was built under R version 3.6.3
## Warning: package 'dplyr' was built under R version 3.6.3
## Warning: package 'stringr' was built under R version 3.6.3
## Warning: package 'forcats' was built under R version 3.6.3
## -- Conflicts ------------------------------------------ tidyverse_conflicts() --
## x dplyr::filter() masks stats::filter()
## x dplyr::lag()    masks stats::lag()
set.seed(421)
x1=runif(500) - 0.5 
x2=runif(500) - 0.5
y=as_factor(1*(x1*x1 - x2*x2 > 0))
df<-tibble(x1=x1,
           x2=x2,
           y=y)
df
## # A tibble: 500 x 3
##         x1      x2 y    
##      <dbl>   <dbl> <fct>
##  1  0.290  -0.384  0    
##  2 -0.355  -0.313  1    
##  3  0.215   0.0556 1    
##  4 -0.185  -0.459  0    
##  5  0.345  -0.143  1    
##  6  0.200   0.0506 1    
##  7  0.400   0.149  1    
##  8  0.183  -0.476  0    
##  9  0.0298 -0.388  0    
## 10 -0.291  -0.130  1    
## # ... with 490 more rows
  1. Plot the observations, colored according to their class labels. Your plot should display X1 on the x-axis, and X2 on the yaxis.
library(plotly)
## Warning: package 'plotly' was built under R version 3.6.3
## 
## Attaching package: 'plotly'
## The following object is masked from 'package:ggplot2':
## 
##     last_plot
## The following object is masked from 'package:stats':
## 
##     filter
## The following object is masked from 'package:graphics':
## 
##     layout
p<-ggplot(data=df,mapping=aes(x=x1,y=x2,color=y))+
  geom_point() +
  theme_light() +
  theme(legend.position = "none")
ggplotly(p)
  1. Fit a logistic regression model to the data, using X1 and X2 as predictors.

x1 and x2 are not significant

lm.fit = glm(y ~ x1 + x2, family = binomial)
summary(lm.fit)
## 
## Call:
## glm(formula = y ~ x1 + x2, family = binomial)
## 
## Deviance Residuals: 
##    Min      1Q  Median      3Q     Max  
## -1.278  -1.227   1.089   1.135   1.175  
## 
## Coefficients:
##             Estimate Std. Error z value Pr(>|z|)
## (Intercept)  0.11999    0.08971   1.338    0.181
## x1          -0.16881    0.30854  -0.547    0.584
## x2          -0.08198    0.31476  -0.260    0.795
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 691.35  on 499  degrees of freedom
## Residual deviance: 690.99  on 497  degrees of freedom
## AIC: 696.99
## 
## Number of Fisher Scoring iterations: 3
  1. Apply this model to the training data in order to obtain a predicted class label for each training observation. Plot the observations, colored according to the predicted class labels. The decision boundary should be linear.

With the given model and a probability threshold of 0.5, all points are classified to single class and no decision boundary can be shown. So I shifted the probability threshold to 0.52 to show a meaningful decision boundary. This boundary is linear.

glm_prob = predict(lm.fit, df, type = "response")
df<-mutate(df, pred = as_factor(ifelse(glm_prob > 0.52, 1, 0)))
q<-ggplot(data=df,mapping=aes(x=x1,y=x2,color=pred))+
  geom_point() +
  theme_light() +
  theme(legend.position = "none")
ggplotly(q)
  1. Now fit a logistic regression model to the data using non-linear functions of X1 and X2 as predictors (e.g. X2 1 , X1×X2, log(X2),and so forth).

squares, product interaction terms are being used to fit the model.

glm_fit2 = glm(y ~ x1 + x2+ I(x1^2) + I(x2^2) + I(x1 * x2), data = df, family = binomial)
## Warning: glm.fit: algorithm did not converge
## Warning: glm.fit: fitted probabilities numerically 0 or 1 occurred
  1. Apply this model to the training data in order to obtain a predicted class label for each training observation. Plot the observations, colored according to the predicted class labels. The decision boundary should be obviously non-linear. If it is not, then repeat (a)-(e) until you come up with an example in which the predicted class labels are obviously non-linear.

This non-linear decision boundary closely resembles the true decision boundary.

  1. Fit a support vector classifier to the data with X1 and X2 as predictors. Obtain a class prediction for each training observation. Plot the observations, colored according to the predicted class labels.

A linear kernel, even with low cost fails to find non-linear decision boundary and classifies all points to a single class.

library(e1071)
## Warning: package 'e1071' was built under R version 3.6.3
svm_fit = svm(y ~ x1 + x2, df, kernel = "linear", cost = 0.1)
df3<-mutate(df, svm_pred = as_factor(predict(svm_fit, df)))
s<-ggplot(data=df3,mapping=aes(x=x1,y=x2,color=svm_pred))+
  geom_point() +
  theme_light() +
  theme(legend.position = "none")
ggplotly(s)

  1. Fit a SVM using a non-linear kernel to the data. Obtain a class prediction for each training observation. Plot the observations,colored according to the predicted class labels.

  2. Comment on your results.

Similar to the non-linear logistic regression model, the non-linear decision boundary on predicted labels closely resembles the true decision boundary.

  1. Comment on your results. SVMs with non-linear kernel are extremely powerful in finding non-linear boundary. Both, logistic regression with non-interactions and SVMs with linear kernels fail to find the decision boundary. Adding interaction terms to logistic regression seems to give them same power as radial-basis kernels. However, there is some manual efforts and tuning involved in picking right interaction terms. This effort can become prohibitive with large number of features. Radial basis kernels, on the other hand, only require tuning of one parameter, gamma, which can be easily done using cross-validation.
svm_fit2 = svm(y ~ x1 + x2, df, gamma = 1)
df4<-mutate(df, svm_pred = as_factor(predict(svm_fit2, df)))
t<-ggplot(data=df4,mapping=aes(x=x1,y=x2,color=svm_pred))+
  geom_point() +
  theme_light() +
  theme(legend.position = "none")
ggplotly(t)
  1. In this problem, you will use support vector approaches in order to predict whether a given car gets high or low gas mileage based on the Auto data set.
  1. Create a binary variable that takes on a 1 for cars with gas mileage above the median, and a 0 for cars with gas mileage below the median.
library(ISLR)
## Warning: package 'ISLR' was built under R version 3.6.3
data("Auto")
mpg.med = median(Auto$mpg)
bin.var = ifelse(Auto$mpg > mpg.med, 1, 0)
Auto$mpglevel = as.factor(bin.var)
  1. Fit a support vector classifier to the data with various values of cost, in order to predict whether a car gets high or low gas mileage. Report the cross-validation errors associated with different values of this parameter. Comment on your results.

We see that cross-validation error is minimized for cost=1.

set.seed(12345)
tune_out = tune(svm, mpglevel ~ . -mpg, data = Auto, kernel = "linear", ranges = list(cost = c(0.01, 0.1, 1, 5, 10, 100)))
summary(tune_out)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##  0.01
## 
## - best performance: 0.08666667 
## 
## - Detailed performance results:
##    cost      error dispersion
## 1 1e-02 0.08666667 0.02443906
## 2 1e-01 0.09429487 0.03599454
## 3 1e+00 0.08903846 0.03765106
## 4 5e+00 0.10192308 0.04644181
## 5 1e+01 0.10705128 0.05214726
## 6 1e+02 0.12243590 0.04476075
  1. Now repeat (b), this time using SVMs with radial and polynomial basis kernels, with different values of gamma and degree and cost. Comment on your results.

Finally, for radial basis kernel, cost=10 and gamma=0.01. The lowest cross-validation error is obtained for cost=10 and degree=2.

set.seed(21)
tune.out = tune(svm, mpglevel ~ . -mpg, data = Auto, kernel = "polynomial", ranges = list(cost = c(0.1, 1, 5, 10), degree = c(2, 3, 4)))
summary(tune.out)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost degree
##    10      2
## 
## - best performance: 0.5435897 
## 
## - Detailed performance results:
##    cost degree     error dispersion
## 1   0.1      2 0.5587821 0.04538579
## 2   1.0      2 0.5587821 0.04538579
## 3   5.0      2 0.5587821 0.04538579
## 4  10.0      2 0.5435897 0.05611162
## 5   0.1      3 0.5587821 0.04538579
## 6   1.0      3 0.5587821 0.04538579
## 7   5.0      3 0.5587821 0.04538579
## 8  10.0      3 0.5587821 0.04538579
## 9   0.1      4 0.5587821 0.04538579
## 10  1.0      4 0.5587821 0.04538579
## 11  5.0      4 0.5587821 0.04538579
## 12 10.0      4 0.5587821 0.04538579
svm_poly<-tune_out$best.model
set.seed(463)
tune_out = tune(svm, mpglevel ~ .-mpg, data = Auto, kernel = "radial", ranges = list(cost = c(0.1, 1, 5, 10), gamma = c(0.01, 0.1, 1, 5, 10, 100)))
summary(tune_out)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost gamma
##    10   0.1
## 
## - best performance: 0.08160256 
## 
## - Detailed performance results:
##    cost gamma      error dispersion
## 1   0.1 1e-02 0.11467949 0.03621710
## 2   1.0 1e-02 0.09423077 0.04937793
## 3   5.0 1e-02 0.08653846 0.04145630
## 4  10.0 1e-02 0.08910256 0.03593430
## 5   0.1 1e-01 0.09679487 0.05051148
## 6   1.0 1e-01 0.08910256 0.04158825
## 7   5.0 1e-01 0.08410256 0.03390616
## 8  10.0 1e-01 0.08160256 0.04314552
## 9   0.1 1e+00 0.57660256 0.05479863
## 10  1.0 1e+00 0.09166667 0.04963524
## 11  5.0 1e+00 0.09435897 0.04681077
## 12 10.0 1e+00 0.09685897 0.04780259
## 13  0.1 5e+00 0.57660256 0.05479863
## 14  1.0 5e+00 0.50000000 0.07252377
## 15  5.0 5e+00 0.50012821 0.07730307
## 16 10.0 5e+00 0.50012821 0.07730307
## 17  0.1 1e+01 0.57660256 0.05479863
## 18  1.0 1e+01 0.52814103 0.05853644
## 19  5.0 1e+01 0.53064103 0.05772933
## 20 10.0 1e+01 0.53064103 0.05772933
## 21  0.1 1e+02 0.57660256 0.05479863
## 22  1.0 1e+02 0.57660256 0.05479863
## 23  5.0 1e+02 0.57660256 0.05479863
## 24 10.0 1e+02 0.57660256 0.05479863
  1. Make some plots to back up your assertions in (b) and (c).
svm_linear = svm(mpglevel ~ ., data = Auto, kernel = "linear", cost = 1)
svm_poly = svm(mpglevel ~ ., data = Auto, kernel = "polynomial", cost = 10, 
    degree = 2)
svm_radial = svm(mpglevel ~ ., data = Auto, kernel = "radial", cost = 10, gamma = 0.01)

library(RColorBrewer)
plotpairs<-function(fit){
  for(name in names(Auto)[!(names(Auto)) %in% c("mpg","name","mpglevel")]){
    plot(fit,Auto,as.formula(paste("mpg~",name,sep="")))
  }
}
plotpairs(svm_linear)

plotpairs(svm_poly)

  1. This problem involves the OJ data set which is part of the ISLR package.
  1. Create a training set containing a random sample of 800 observations, and a test set containing the remaining observations.
library(ISLR)
set.seed(9004)
inTrain = sample(nrow(OJ), 800)
train_oj = OJ[inTrain,]
test_oj = OJ[-inTrain,]
  1. Fit a support vector classifier to the training data using cost=0.01, with Purchase as the response and the other variables as predictors. Use the summary() function to produce summary statistics, and describe the results obtained.

Support vector classifier creates 442 support vectors out of 800 training points. Out of these, 222 belong to level CH and remaining 220 belong to level MM

library(e1071)
svm_linear = svm(Purchase ~ ., kernel = "linear", data = train_oj, cost = 0.01)
summary(svm_linear)
## 
## Call:
## svm(formula = Purchase ~ ., data = train_oj, kernel = "linear", cost = 0.01)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  linear 
##        cost:  0.01 
## 
## Number of Support Vectors:  442
## 
##  ( 222 220 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM
  1. What are the training and test error rates?

The training error rate is 16% and test error rate is about 17.37%.

train_pred = predict(svm_linear, train_oj)
(t<-table(train_oj$Purchase, train_pred))
##     train_pred
##       CH  MM
##   CH 432  51
##   MM  80 237
(train_error=(t[2]+t[3])/(t[1]+t[2]+t[3]+t[4]))
## [1] 0.16375
test_pred = predict(svm_linear, test_oj)
(t<-table(test_oj$Purchase, test_pred))
##     test_pred
##       CH  MM
##   CH 146  24
##   MM  22  78
(test_error=(t[2]+t[3])/(t[1]+t[2]+t[3]+t[4]))
## [1] 0.1703704
  1. Use the tune() function to select an optimal cost. Consider values in the range 0.01 to 10.

Tuning shows that optimal cost is 3.395258

set.seed(1554)
tune_out = tune(svm, Purchase ~ ., data = train_oj, kernel = "linear", ranges = list(cost = 10^seq(-2, 1, by = 0.25)))
summary(tune_out)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##      cost
##  3.162278
## 
## - best performance: 0.1625 
## 
## - Detailed performance results:
##           cost   error dispersion
## 1   0.01000000 0.16750 0.03395258
## 2   0.01778279 0.16875 0.02960973
## 3   0.03162278 0.16625 0.02638523
## 4   0.05623413 0.16875 0.03076005
## 5   0.10000000 0.16875 0.02901748
## 6   0.17782794 0.16750 0.02838231
## 7   0.31622777 0.17000 0.02898755
## 8   0.56234133 0.16875 0.02841288
## 9   1.00000000 0.16500 0.03106892
## 10  1.77827941 0.16500 0.03106892
## 11  3.16227766 0.16250 0.03118048
## 12  5.62341325 0.16375 0.02664713
## 13 10.00000000 0.16750 0.02581989
  1. Compute the training and test error rates using this new value for cost.

The training error is the same to 16.32% but test error slightly increases to 17.4% by using best cost.

svm_linear = svm(Purchase ~ ., kernel = "linear", data = train_oj)
train_pred = predict(svm_linear, train_oj)
(t<-table(train_oj$Purchase, train_pred))
##     train_pred
##       CH  MM
##   CH 427  56
##   MM  75 242
(train_error_linear=(t[2]+t[3])/(t[1]+t[2]+t[3]+t[4]))
## [1] 0.16375
test_pred = predict(svm_linear, test_oj)
(t<-table(test_oj$Purchase, test_pred))
##     test_pred
##       CH  MM
##   CH 143  27
##   MM  20  80
(test_error_linear=(t[2]+t[3])/(t[1]+t[2]+t[3]+t[4]))
## [1] 0.1740741
  1. Repeat parts (b) through (e) using a support vector machine with a radial kernel. Use the default value for gamma.

The radial basis kernel with default gamma creates 371 support vectors, out of which, 188 belong to level CH and remaining 183 belong to level MM. The classifier has a training error of 14% and a test error of 18% which is an improvement over linear kernel. We now use cross validation to find optimal gamma.

set.seed(410)
svm_radial = svm(Purchase ~ ., kernel = "radial", data = train_oj)
summary(svm_radial)
## 
## Call:
## svm(formula = Purchase ~ ., data = train_oj, kernel = "radial")
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  radial 
##        cost:  1 
## 
## Number of Support Vectors:  371
## 
##  ( 188 183 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM
train_pred = predict(svm_radial, train_oj)
(t<-table(train_oj$Purchase, train_pred))
##     train_pred
##       CH  MM
##   CH 441  42
##   MM  74 243
(test_error=(t[2]+t[3])/(t[1]+t[2]+t[3]+t[4]))
## [1] 0.145
test_pred = predict(svm_radial, test_oj)
(t<-table(test_oj$Purchase, test_pred))
##     test_pred
##       CH  MM
##   CH 148  22
##   MM  27  73
(test_error=(t[2]+t[3])/(t[1]+t[2]+t[3]+t[4]))
## [1] 0.1814815
set.seed(755)
tune_out = tune(svm, Purchase ~ ., data = train_oj, kernel = "radial", ranges = list(cost = 10^seq(-2,1, by = 0.25)))
summary(tune_out)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##       cost
##  0.3162278
## 
## - best performance: 0.1675 
## 
## - Detailed performance results:
##           cost   error dispersion
## 1   0.01000000 0.39625 0.06615691
## 2   0.01778279 0.39625 0.06615691
## 3   0.03162278 0.35375 0.09754807
## 4   0.05623413 0.20000 0.04249183
## 5   0.10000000 0.17750 0.04073969
## 6   0.17782794 0.17125 0.03120831
## 7   0.31622777 0.16750 0.04216370
## 8   0.56234133 0.16750 0.03782269
## 9   1.00000000 0.17250 0.03670453
## 10  1.77827941 0.17750 0.03374743
## 11  3.16227766 0.18000 0.04005205
## 12  5.62341325 0.18000 0.03446012
## 13 10.00000000 0.18625 0.04427267
svm_radial = svm(Purchase ~ ., kernel = "radial", data = train_oj, cost = tune_out$best.parameters$cost)
train_pred = predict(svm_radial, train_oj)
(t<-table(train_oj$Purchase, train_pred))
##     train_pred
##       CH  MM
##   CH 440  43
##   MM  81 236
(train_error_radial=(t[2]+t[3])/(t[1]+t[2]+t[3]+t[4]))
## [1] 0.155
test_pred = predict(svm_radial, test_oj)
(t<-table(test_oj$Purchase, test_pred))
##     test_pred
##       CH  MM
##   CH 145  25
##   MM  28  72
(test_error_radial=(t[2]+t[3])/(t[1]+t[2]+t[3]+t[4]))
## [1] 0.1962963

Tuning slightly decreases training error to 15.5% and increases test error to 19.63% which is still not better than linear kernel.

  1. Repeat parts (b) through (e) using a support vector machine with a polynomial kernel. Set degree=2.

The polynomial kernel produces 456 support vectors, out of which, 232 belong to level CH and remaining 224 belong to level MM. This kernel produces a train error of 18% and a test error of 20.37% which are slightly higher than the errors produces by linear and radial kernel.

svm_poly = svm(Purchase ~ ., kernel = "poly", data = train_oj, degree=2)
summary(svm_poly)
## 
## Call:
## svm(formula = Purchase ~ ., data = train_oj, kernel = "poly", degree = 2)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  polynomial 
##        cost:  1 
##      degree:  2 
##      coef.0:  0 
## 
## Number of Support Vectors:  456
## 
##  ( 232 224 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM
train_pred = predict(svm_poly, train_oj)
(t<-table(train_oj$Purchase, train_pred))
##     train_pred
##       CH  MM
##   CH 450  33
##   MM 111 206
(train_error=(t[2]+t[3])/(t[1]+t[2]+t[3]+t[4]))
## [1] 0.18
test_pred = predict(svm_poly, test_oj)
(t<-table(test_oj$Purchase, test_pred))
##     test_pred
##       CH  MM
##   CH 149  21
##   MM  34  66
(test_error=(t[2]+t[3])/(t[1]+t[2]+t[3]+t[4]))
## [1] 0.2037037
set.seed(322)
tune_out = tune(svm, Purchase ~ ., data = train_oj, kernel = "poly", degree = 2, ranges = list(cost = 10^seq(-2, 1, by = 0.25)))
summary(tune_out)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##    10
## 
## - best performance: 0.18 
## 
## - Detailed performance results:
##           cost   error dispersion
## 1   0.01000000 0.39250 0.05749396
## 2   0.01778279 0.37500 0.05863020
## 3   0.03162278 0.36375 0.05756940
## 4   0.05623413 0.33875 0.06626179
## 5   0.10000000 0.30375 0.05172376
## 6   0.17782794 0.24000 0.04440971
## 7   0.31622777 0.21000 0.04362084
## 8   0.56234133 0.20250 0.03987829
## 9   1.00000000 0.20375 0.03634805
## 10  1.77827941 0.19500 0.04866267
## 11  3.16227766 0.18750 0.04409586
## 12  5.62341325 0.18875 0.04185375
## 13 10.00000000 0.18000 0.03593976
svm_poly = svm(Purchase ~ ., kernel = "poly", data = train_oj, degree=2, cost = tune_out$best.parameters$cost)
train_pred = predict(svm_poly, train_oj)
(t<-table(train_oj$Purchase, train_pred))
##     train_pred
##       CH  MM
##   CH 447  36
##   MM  85 232
(train_error_poly=(t[2]+t[3])/(t[1]+t[2]+t[3]+t[4]))
## [1] 0.15125
test_pred = predict(svm_poly, test_oj)
(t<-table(test_oj$Purchase, test_pred))
##     test_pred
##       CH  MM
##   CH 148  22
##   MM  28  72
(test_error_poly=(t[2]+t[3])/(t[1]+t[2]+t[3]+t[4]))
## [1] 0.1851852

Tuning reduces the training error to 15.12% and test error to 18.52% which is about the same as the radial kernel but worse than linear kernel.

  1. Overall, which approach seems to give the best results on thisdata?
df<-tibble(`SVM kernel`=c("Linear","Radial","Polynomial"),
           `Training Error`=c(train_error_linear,train_error_radial,train_error_poly),
           `Test Error`=c(test_error_linear,test_error_radial,test_error_poly))
df
## # A tibble: 3 x 3
##   `SVM kernel` `Training Error` `Test Error`
##   <chr>                   <dbl>        <dbl>
## 1 Linear                  0.164        0.174
## 2 Radial                  0.155        0.196
## 3 Polynomial              0.151        0.185

Overall, the linear basis kernel seems to be producing minimum misclassification error on test data.