1. 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.

#(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. F

set.seed(1)
x1 = runif(500) - 0.5 
x2 = runif(500) - 0.5
y=1*(x1^2-x2^2>0)

#(b) Plot the observations, colored according to their class labels. Your plot should display X1 on the x-axis, and X2 on the yaxis.

plot(x1[y==0],x2[y==0],col="red",xlab="X1",ylab="X2")
points(x1[y==1],x2[y==1],col="blue")

#(c) Fit a logistic regression model to the data, using X1 and X2 as predictors.

df=data.frame(x1 = x1, x2 = x2, y = as.factor(y))
glm_fit=glm(y~.,data=df, family='binomial')

#(d) 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.

library(ggplot2)
glm_prob=predict(glm_fit,newdata=df,type='response')
glm_pred=ifelse(glm_prob>0.5,1,0)
ggplot(data = df, mapping = aes(x1, x2)) +
  geom_point(data = df, mapping = aes(colour = glm_pred))

#(e) 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).

glm_fit_2=glm(y~poly(x1,2)+poly(x2,2),data=df,family='binomial')
## Warning: glm.fit: algorithm did not converge
## Warning: glm.fit: fitted probabilities numerically 0 or 1 occurred

#(f) 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.

glm_prob_2=predict(glm_fit_2,newdata=df,type='response')
glm_pred_2=ifelse(glm_prob_2>0.5,1,0)
ggplot(data = df, mapping = aes(x1, x2)) +
  geom_point(data = df, mapping = aes(colour = glm_pred_2))

#(g) 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.

library("e1071")
svm_lin=svm(y~.,data=df,kernel='linear',cost=0.01)
plot(svm_lin,df)

#(h) 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.

svm_lin_2=svm(y~.,data=df,kernel='radial',gamma=1)
plot(svm_lin_2,data=df)

#(i) Comment on your results. Based upon the results we can gather compared to different models, the support vector machines with the radial kernels are vividly showing the non-linear boundry. The other methods did not find the linear boundy as profound as the svm.

  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.

#(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.

library(ISLR)
attach(Auto)
## The following object is masked from package:ggplot2:
## 
##     mpg
mpg_med = median(Auto$mpg)
bin.var = ifelse(Auto$mpg > mpg_med, 1, 0)
Auto$mpglevel = as.factor(bin.var)

#(b) 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.

library(e1071)
set.seed(1)
tune_out = tune(svm, 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
##   0.1
## 
## - best performance: 8.981009 
## 
## - Detailed performance results:
##    cost     error dispersion
## 1 1e-02 10.305990   5.295587
## 2 1e-01  8.981009   4.750742
## 3 1e+00  9.647184   4.313908
## 4 5e+00 10.149220   4.755080
## 5 1e+01 10.306219   4.953047
## 6 1e+02 10.684083   5.080506

#(c) 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.

set.seed(2)
tune_out = tune(svm, 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: 50.95606 
## 
## - Detailed performance results:
##    cost degree    error dispersion
## 1   0.1      2 61.59446   13.60292
## 2   1.0      2 60.15304   13.79293
## 3   5.0      2 55.06386   15.19391
## 4  10.0      2 50.95606   15.72388
## 5   0.1      3 61.71831   13.56940
## 6   1.0      3 61.39833   13.54758
## 7   5.0      3 59.99304   13.43208
## 8  10.0      3 58.28857   13.27760
## 9   0.1      4 61.75343   13.57197
## 10  1.0      4 61.74822   13.57317
## 11  5.0      4 61.72510   13.57851
## 12 10.0      4 61.69626   13.58520

#(d) Make some plots to back up your assertions in (b) and (c).

set.seed(33)
tune_out = tune(svm, mpg ~ ., data = Auto, kernel = "radial", 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: 7.294024 
## 
## - Detailed performance results:
##    cost degree     error dispersion
## 1   0.1      2 22.211472   8.440215
## 2   1.0      2 10.118616   4.286046
## 3   5.0      2  7.649506   2.993534
## 4  10.0      2  7.294024   2.818545
## 5   0.1      3 22.211472   8.440215
## 6   1.0      3 10.118616   4.286046
## 7   5.0      3  7.649506   2.993534
## 8  10.0      3  7.294024   2.818545
## 9   0.1      4 22.211472   8.440215
## 10  1.0      4 10.118616   4.286046
## 11  5.0      4  7.649506   2.993534
## 12 10.0      4  7.294024   2.818545
svm_lin = 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)
plotpairs = function(fit) {
    for (name in names(Auto)[!(names(Auto) %in% c("mpg", "mpglevel", "name"))]) {
        plot(fit, Auto, as.formula(paste("mpg~", name, sep = "")))
    }
}
plotpairs(svm_lin)

plotpairs(svm_poly)

plotpairs(svm_radial)

  1. 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.

attach(OJ)
set.seed(1)
data_Train = sample(nrow(OJ), 800)
oj_train = OJ[data_Train,]
oj_test = OJ[-data_Train,]

#(b) 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.

svc=svm(Purchase~.,data=oj_train,kernel='linear',cost=0.01)
summary(svc)
## 
## Call:
## svm(formula = Purchase ~ ., data = oj_train, kernel = "linear", cost = 0.01)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  linear 
##        cost:  0.01 
## 
## Number of Support Vectors:  435
## 
##  ( 219 216 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM

#(c) What are the training and test error rates?

pred_train = predict(svc, oj_train)
(t<-table(oj_train$Purchase, pred_train))
##     pred_train
##       CH  MM
##   CH 420  65
##   MM  75 240
pred_test = predict(svc, oj_test)
table(oj_test$Purchase, pred_test)
##     pred_test
##       CH  MM
##   CH 153  15
##   MM  33  69

#(d) Use the tune() function to select an optimal cost. Consider values in the range 0.01 to 10.

set.seed(1)
tune_svc = tune(svm, Purchase ~ ., data = oj_train, kernel = "linear", ranges = list(cost = c(0.01,0.1,1,10)))
summary(tune_svc)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##   0.1
## 
## - best performance: 0.1725 
## 
## - Detailed performance results:
##    cost   error dispersion
## 1  0.01 0.17625 0.02853482
## 2  0.10 0.17250 0.03162278
## 3  1.00 0.17500 0.02946278
## 4 10.00 0.17375 0.03197764

the smallest error is at 0.10 cost, which intells the optimal cost is 0.0316227

#(e) Compute the training and test error rates using this new value for cost.

svm_lin_1 = svm(Purchase ~ ., kernel = "linear", data = oj_train, cost = tune_out$best.parameters$cost)
pred_train_1 = predict(svm_lin_1, oj_train)
table(oj_train$Purchase, pred_train_1)
##     pred_train_1
##       CH  MM
##   CH 423  62
##   MM  69 246
test_pred_1 = predict(svm_lin_1, oj_test)
(t<-table(oj_test$Purchase, test_pred_1))
##     test_pred_1
##       CH  MM
##   CH 156  12
##   MM  28  74

#(f) Repeat parts (b) through (e) using a support vector machine with a radial kernel. Use the default value for gamma.

set.seed(1)
svm_rad_1 = svm(Purchase ~ ., data = oj_train, kernel = "radial")
summary(svm_rad_1)
## 
## Call:
## svm(formula = Purchase ~ ., data = oj_train, kernel = "radial")
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  radial 
##        cost:  1 
## 
## Number of Support Vectors:  373
## 
##  ( 188 185 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM
pred_train_2 = predict(svm_rad_1, oj_train)
table(oj_train$Purchase, pred_train_2)
##     pred_train_2
##       CH  MM
##   CH 441  44
##   MM  77 238
test_pred_2 = predict(svm_rad_1, oj_test)
table(oj_test$Purchase, test_pred_2)
##     test_pred_2
##       CH  MM
##   CH 151  17
##   MM  33  69
svm_rad_1 = svm(Purchase ~ ., data = oj_train, kernel = "radial", cost = tune_svc$best.parameters$cost)
pred_train = predict(svm_rad_1, oj_train)
table(oj_train$Purchase, pred_train_1)
##     pred_train_1
##       CH  MM
##   CH 423  62
##   MM  69 246
test_pred_3 = predict(svm_rad_1, oj_test)
(t<-table(oj_test$Purchase, test_pred_3))
##     test_pred_3
##       CH  MM
##   CH 150  18
##   MM  37  65

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

svm_pol_1 = svm(Purchase ~ ., kernel = "poly", data = oj_train, degree=2)
summary(svm_pol_1)
## 
## Call:
## svm(formula = Purchase ~ ., data = oj_train, kernel = "poly", degree = 2)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  polynomial 
##        cost:  1 
##      degree:  2 
##      coef.0:  0 
## 
## Number of Support Vectors:  447
## 
##  ( 225 222 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM
pred_train_2 = predict(svm_pol_1, oj_train)
(t<-table(oj_train$Purchase, pred_train_2))
##     pred_train_2
##       CH  MM
##   CH 449  36
##   MM 110 205
test_pred_3 = predict(svm_pol_1, oj_test)
(t<-table(oj_test$Purchase, test_pred_3))
##     test_pred_3
##       CH  MM
##   CH 153  15
##   MM  45  57
set.seed(1)
tune_svc = tune(svm, Purchase ~ ., data = oj_train, kernel = "poly", degree = 2, ranges = list(cost = 10^seq(-2, 1, by = 0.25)))
summary(tune_svc)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##      cost
##  3.162278
## 
## - best performance: 0.1775 
## 
## - Detailed performance results:
##           cost   error dispersion
## 1   0.01000000 0.39125 0.04210189
## 2   0.01778279 0.37125 0.03537988
## 3   0.03162278 0.36500 0.03476109
## 4   0.05623413 0.33750 0.04714045
## 5   0.10000000 0.32125 0.05001736
## 6   0.17782794 0.24500 0.04758034
## 7   0.31622777 0.19875 0.03972562
## 8   0.56234133 0.20500 0.03961621
## 9   1.00000000 0.20250 0.04116363
## 10  1.77827941 0.18500 0.04199868
## 11  3.16227766 0.17750 0.03670453
## 12  5.62341325 0.18375 0.03064696
## 13 10.00000000 0.18125 0.02779513

#(h) Overall, which approach seems to give the best results on this data?

Based upon the results, the linear basis approach seems to provide the best data results.