QUESTIONS

5.We have seen that we can fit an SVM with anon-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. (a) 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:

x1 <- runif (500) -0.5
x2 <- runif (500) - 0.5
y <- 1 * (x1^2 - x2^2 > 0)
y <- as.factor(y)
dataset <- data.frame(y, x1, x2)
  1. Plot the observations, colored according to their class labels. Your plot should display X1 on the x-axis, and X2 on the y-axis.
ggplot(dataset,aes(x1, x2, color=y))+geom_point()

  1. Fit a logistic regression model to the data, using X1 and X2 as predictors.
lm.model <- glm(y~., data = dataset, family="binomial")
summary(lm.model)
## 
## Call:
## glm(formula = y ~ ., family = "binomial", data = dataset)
## 
## Deviance Residuals: 
##    Min      1Q  Median      3Q     Max  
## -1.176  -1.142  -1.112   1.215   1.250  
## 
## Coefficients:
##             Estimate Std. Error z value Pr(>|z|)
## (Intercept) -0.08867    0.08956  -0.990    0.322
## x1          -0.08924    0.32582  -0.274    0.784
## x2          -0.09819    0.30629  -0.321    0.749
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 692.18  on 499  degrees of freedom
## Residual deviance: 692.00  on 497  degrees of freedom
## AIC: 698
## 
## 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.
lm.prob <-predict(lm.model, type="response") 
lm.pred <- rep("No", 500)
lm.pred[lm.prob>0.5]="Yes"

ggplot(dataset, aes(x1, x2, color=lm.pred)) + geom_point()

  1. Now fit a logistic regression model to the data using non-linear functions of X1 and X2 as predictors (e.g. X1^2 , X1×X2, log(X2), and so forth).
# USING Quadratic transformation
x1T <- x1^2
y1 <- 1 * (x1T - x2^2 > 0)
y1 <- as.factor(y1)
dataset2 <- data.frame(y1, x1T, x2)
new.model <- glm(y1~., data=dataset2, family="binomial")
summary(new.model)
## 
## Call:
## glm(formula = y1 ~ ., family = "binomial", data = dataset2)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -2.5467  -0.7052  -0.5374   0.6234   1.9369  
## 
## Coefficients:
##             Estimate Std. Error z value Pr(>|z|)    
## (Intercept)  -1.7132     0.1695 -10.109   <2e-16 ***
## x1T          23.6092     2.1565  10.948   <2e-16 ***
## x2           -0.5263     0.3817  -1.379    0.168    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 692.18  on 499  degrees of freedom
## Residual deviance: 481.58  on 497  degrees of freedom
## AIC: 487.58
## 
## Number of Fisher Scoring iterations: 5
  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 q 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.
lm.prob <-predict(new.model, type="response") 
lm.pred <- rep("No", 500)
lm.pred[lm.prob>0.5]="Yes"

ggplot(dataset2, aes(x1, x2, color=lm.pred)) + geom_point()

  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.
set.seed(1)

svmfit <- svm(y~x1+x2, data=dataset, kernel="linear", cost=5, scale=F)
plot(svmfit, dataset)

summary(svmfit)
## 
## Call:
## svm(formula = y ~ x1 + x2, data = dataset, kernel = "linear", cost = 5, 
##     scale = F)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  linear 
##        cost:  5 
## 
## Number of Support Vectors:  480
## 
##  ( 239 241 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  0 1
svm.predict <- predict(svmfit, dataset)

ggplot(dataset, aes(x1, x2, color=svm.predict)) + geom_point()

  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.
set.seed(1)
svmfit1 <- svm(y~., data=dataset, kernel="radial", cost=1000, gamma=0.5)
summary(svmfit1)
## 
## Call:
## svm(formula = y ~ ., data = dataset, kernel = "radial", cost = 1000, 
##     gamma = 0.5)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  radial 
##        cost:  1000 
## 
## Number of Support Vectors:  23
## 
##  ( 13 10 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  0 1
svm.predict <- predict(svmfit1, dataset)

ggplot(dataset, aes(x1, x2, color=svm.predict)) + geom_point()

7.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. (a) 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.

Auto<-Auto|>mutate(mpg01=case_when(mpg>median(mpg)~1, mpg<median(mpg)~0))
Auto$mpg01 <- as.factor(Auto$mpg01)
Auto_new <- Auto|>select(!c(mpg,name))
Auto_new <- Auto_new|>relocate(mpg01, .before = 1)
  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. Note you will need to fit the classifier without the gas mileage variable to produce sensible results.
set.seed(2)
tuned <- tune(svm, mpg01~., data=Auto_new, kernel="linear", ranges = list (cost = c ( 0.1, 1, 5, 10, 100)))
summary(tuned)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##     5
## 
## - best performance: 0.08929487 
## 
## - Detailed performance results:
##    cost      error dispersion
## 1   0.1 0.09679487 0.06737312
## 2   1.0 0.09948718 0.05438007
## 3   5.0 0.08929487 0.03849957
## 4  10.0 0.08929487 0.03849957
## 5 100.0 0.09185897 0.04029356
bestmodL <- tuned$best.model
summary (bestmodL)
## 
## Call:
## best.tune(METHOD = svm, train.x = mpg01 ~ ., data = Auto_new, ranges = list(cost = c(0.1, 
##     1, 5, 10, 100)), kernel = "linear")
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  linear 
##        cost:  5 
## 
## Number of Support Vectors:  83
## 
##  ( 41 42 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  0 1
ypred <- predict(bestmodL, Auto_new)
table(predict=ypred, truth=Auto_new$mpg01)
##        truth
## predict   0   1
##       0 174  11
##       1  22 185

After performing cross validation, we see that the best choice of cost is 5 The test error rate is: 0.08

  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.

Ans:

#Polynomial

set.seed (1)
tune.out <- tune (svm , mpg01 ∼ ., data = Auto_new,
kernel = "polynomial",
ranges = list (
cost = c( 0.1, 1, 5, 10, 100),
degree=c(1,2,3,4)
))
bestmodP <- tune.out$best.model
summary(tune.out)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost degree
##   100      3
## 
## - best performance: 0.08423077 
## 
## - Detailed performance results:
##     cost degree      error dispersion
## 1    0.1      1 0.08673077 0.04551036
## 2    1.0      1 0.08929487 0.04229479
## 3    5.0      1 0.08429487 0.03229996
## 4   10.0      1 0.08435897 0.03662670
## 5  100.0      1 0.08948718 0.03898410
## 6    0.1      2 0.27846154 0.09486227
## 7    1.0      2 0.25307692 0.13751948
## 8    5.0      2 0.17621795 0.04945319
## 9   10.0      2 0.18647436 0.05598001
## 10 100.0      2 0.18128205 0.06251437
## 11   0.1      3 0.20192308 0.11347783
## 12   1.0      3 0.09448718 0.04180527
## 13   5.0      3 0.08429487 0.04016554
## 14  10.0      3 0.08435897 0.04544023
## 15 100.0      3 0.08423077 0.03636273
## 16   0.1      4 0.26564103 0.09977887
## 17   1.0      4 0.21205128 0.09560470
## 18   5.0      4 0.18371795 0.06175709
## 19  10.0      4 0.16589744 0.06962914
## 20 100.0      4 0.12756410 0.05208506
ypred <- predict(bestmodP, Auto_new)
table(predict=ypred, truth=Auto_new$mpg01)
##        truth
## predict   0   1
##       0 190   8
##       1   6 188

After performing cross validation, we see that the best choice of cost is 100, and degree=3 for the polynomial kernel. The test error rate is 0.04

#Radial

set.seed (1)
tune.out <- tune (svm , mpg01 ∼ ., data = Auto_new,
kernel = "radial",
ranges = list (
cost = c( 0.1, 1, 5, 10, 100),
gamma=c(0.5, 1, 2, 3, 4)
))
summary(tune.out)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost gamma
##     1     1
## 
## - best performance: 0.06634615 
## 
## - Detailed performance results:
##     cost gamma      error dispersion
## 1    0.1   0.5 0.08666667 0.04687413
## 2    1.0   0.5 0.06884615 0.02963114
## 3    5.0   0.5 0.07903846 0.03051601
## 4   10.0   0.5 0.08923077 0.02732003
## 5  100.0   0.5 0.10698718 0.03512661
## 6    0.1   1.0 0.08673077 0.04535158
## 7    1.0   1.0 0.06634615 0.03244101
## 8    5.0   1.0 0.08916667 0.02708952
## 9   10.0   1.0 0.08923077 0.02732003
## 10 100.0   1.0 0.10448718 0.04560852
## 11   0.1   2.0 0.14282051 0.07578262
## 12   1.0   2.0 0.08673077 0.04371113
## 13   5.0   2.0 0.09942308 0.04881948
## 14  10.0   2.0 0.09429487 0.05387705
## 15 100.0   2.0 0.09435897 0.05261602
## 16   0.1   3.0 0.31878205 0.13973969
## 17   1.0   3.0 0.08416667 0.04171436
## 18   5.0   3.0 0.09179487 0.05416218
## 19  10.0   3.0 0.09179487 0.05416218
## 20 100.0   3.0 0.09692308 0.05501861
## 21   0.1   4.0 0.51788462 0.06842176
## 22   1.0   4.0 0.08923077 0.03843042
## 23   5.0   4.0 0.09173077 0.05268076
## 24  10.0   4.0 0.09179487 0.05139393
## 25 100.0   4.0 0.09435897 0.05398656
bestmodelR <-tune.out$best.model


ypred <- predict(bestmodelR, Auto_new)
table(predict=ypred, truth=Auto_new$mpg01)
##        truth
## predict   0   1
##       0 188   6
##       1   8 190

After performing cross validation, we see that the best choice of cost is 1, and gamma=1 for the radial kernel. The test error rate is 0.04

  1. Make some plots to back up your assertions in (b) and (c).
#LINEAR KERNEL
plot(bestmodL , Auto_new , acceleration~weight)

plot(bestmodL , Auto_new , acceleration~horsepower)

#POLYNOMIAL KERNEL
plot(bestmodP , Auto_new , acceleration~weight)

plot(bestmodP , Auto_new , acceleration~horsepower)

#RADIAL KERNEL
plot(bestmodelR , Auto_new , acceleration~weight)

plot(bestmodelR , Auto_new , acceleration~horsepower)

8.This problem involves the OJ data set which is part of the ISLR2 package. (a) Create a training set containing a random sample of 800 observations, and a test set containing the remaining observations.

 set.seed(2)
trainOJ <- sample(1:nrow(OJ), 800)
OJ.test <- OJ[-trainOJ, ]
OJtrain <- OJ[trainOJ,]
  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.
svmfit <- svm (Purchase ∼ ., data=OJ, subset=trainOJ, kernel = "linear",cost = 0.01, scale = FALSE)
summary(svmfit)
## 
## Call:
## svm(formula = Purchase ~ ., data = OJ, kernel = "linear", cost = 0.01, 
##     subset = trainOJ, scale = FALSE)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  linear 
##        cost:  0.01 
## 
## Number of Support Vectors:  614
## 
##  ( 306 308 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM
  1. What are the training and test error rates?
trainpred <- predict(svmfit, OJtrain)
testpred <- predict(svmfit, OJ.test)
table(predict=trainpred, truth=OJtrain$Purchase)
##        truth
## predict  CH  MM
##      CH 449 145
##      MM  41 165
table(predict=testpred, truth=OJ.test$Purchase)
##        truth
## predict  CH  MM
##      CH 146  59
##      MM  17  48
(146+48/270)
## [1] 146.1778

Train error rate: 0.234 Test error rate: 0.281

  1. Use the tune() function to select an optimal cost. Consider values in the range 0.01 to 10.
set.seed (1)
tune.out <-tune(svm , Purchase ∼ ., data =OJtrain, kernel = "linear",ranges = list (cost = c(0.01, 0.1,1, 5, 10)))

summary(tune.out)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##     5
## 
## - best performance: 0.1675 
## 
## - Detailed performance results:
##    cost   error dispersion
## 1  0.01 0.17500 0.04677072
## 2  0.10 0.17375 0.05050096
## 3  1.00 0.17000 0.04721405
## 4  5.00 0.16750 0.03961621
## 5 10.00 0.16750 0.04297932

optimal cost is 5.

  1. Compute the training and test error rates using this new value for cost.
best.mod <- tune.out$best.model
summary(best.mod)
## 
## Call:
## best.tune(METHOD = svm, train.x = Purchase ~ ., data = OJtrain, ranges = list(cost = c(0.01, 
##     0.1, 1, 5, 10)), kernel = "linear")
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  linear 
##        cost:  5 
## 
## Number of Support Vectors:  316
## 
##  ( 157 159 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM
trainpred <- predict(best.mod, OJtrain)
testpred <- predict(best.mod, OJ.test)
table(predict=trainpred, truth=OJtrain$Purchase)
##        truth
## predict  CH  MM
##      CH 433  68
##      MM  57 242
table(predict=testpred, truth=OJ.test$Purchase)
##        truth
## predict  CH  MM
##      CH 144  30
##      MM  19  77

Train error rate: 0.156 Test error rate: 0.181

  1. Repeat parts (b) through (e) using a support vector machine with a radial kernel. Use the default value for gamma.
set.seed(2)
svmfitR <- svm (Purchase ∼ ., data=OJtrain, kernel = "radial",cost = 0.01, gamma=1)
tune.out <-tune(svm , Purchase ∼ ., data =OJtrain, kernel = "radial",ranges = list (cost = c(0.01, 0.1,1, 5, 10)), gamma=1)

summary(tune.out)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##     1
## 
## - best performance: 0.20125 
## 
## - Detailed performance results:
##    cost   error dispersion
## 1  0.01 0.38750 0.08799463
## 2  0.10 0.33500 0.07835106
## 3  1.00 0.20125 0.04767147
## 4  5.00 0.21000 0.04816061
## 5 10.00 0.21500 0.04706674
bestmod <- tune.out$best.model
best.mod <- tune.out$best.model
summary(best.mod)
## 
## Call:
## best.tune(METHOD = svm, train.x = Purchase ~ ., data = OJtrain, ranges = list(cost = c(0.01, 
##     0.1, 1, 5, 10)), kernel = "radial", gamma = 1)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  radial 
##        cost:  1 
## 
## Number of Support Vectors:  468
## 
##  ( 215 253 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM
trainpred <- predict(best.mod, OJtrain)
testpred <- predict(best.mod, OJ.test)
table(predict=trainpred, truth=OJtrain$Purchase)
##        truth
## predict  CH  MM
##      CH 456  52
##      MM  34 258
table(predict=testpred, truth=OJ.test$Purchase)
##        truth
## predict  CH  MM
##      CH 139  33
##      MM  24  74

A. For the tuned model: Train error rate: 0.1075 Test error rate: 0.21

  1. Repeat parts (b) through (e) using a support vector machine with a polynomial kernel. Set degree = 2.
set.seed(1)
svmfitP <- svm (Purchase ∼ ., data=OJtrain, kernel = "polynomial",cost = 0.01, degree=2)
summary(svmfit)
## 
## Call:
## svm(formula = Purchase ~ ., data = OJ, kernel = "linear", cost = 0.01, 
##     subset = trainOJ, scale = FALSE)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  linear 
##        cost:  0.01 
## 
## Number of Support Vectors:  614
## 
##  ( 306 308 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM
trainpred <- predict(svmfitP, OJtrain)
testpred <- predict(svmfitP, OJ.test)
table(predict=trainpred, truth=OJtrain$Purchase)
##        truth
## predict  CH  MM
##      CH 489 292
##      MM   1  18
table(predict=testpred, truth=OJ.test$Purchase)
##        truth
## predict  CH  MM
##      CH 162 100
##      MM   1   7
tune.out <-tune(svm , Purchase ∼ ., data =OJtrain, kernel = "polynomial",ranges = list (cost = c(0.01, 0.1,1, 5, 10)), degree=2)

summary(tune.out)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##    10
## 
## - best performance: 0.17 
## 
## - Detailed performance results:
##    cost   error dispersion
## 1  0.01 0.38250 0.06157651
## 2  0.10 0.31000 0.03944053
## 3  1.00 0.19375 0.02144923
## 4  5.00 0.18000 0.03016160
## 5 10.00 0.17000 0.02958040
bestmod <- tune.out$best.model
best.mod <- tune.out$best.model
summary(best.mod)
## 
## Call:
## best.tune(METHOD = svm, train.x = Purchase ~ ., data = OJtrain, ranges = list(cost = c(0.01, 
##     0.1, 1, 5, 10)), kernel = "polynomial", degree = 2)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  polynomial 
##        cost:  10 
##      degree:  2 
##      coef.0:  0 
## 
## Number of Support Vectors:  329
## 
##  ( 163 166 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM
trainpred <- predict(best.mod, OJtrain)
testpred <- predict(best.mod, OJ.test)
table(predict=trainpred, truth=OJtrain$Purchase)
##        truth
## predict  CH  MM
##      CH 452  66
##      MM  38 244
table(predict=testpred, truth=OJ.test$Purchase)
##        truth
## predict  CH  MM
##      CH 141  32
##      MM  22  75

Training error rate: 0.13 Test error rate: 0.2

  1. Overall, which approach seems to give the best results on this data?

Ans: The linear kernel approach seems to give the best results as it had the lowest test error rate.