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. 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)

(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 = "blue")
points(x1[y == 1], x2[y == 1], )

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

reg <- glm(y~x1 + x2, family = binomial)
summary(reg)
## 
## Call:
## glm(formula = y ~ x1 + x2, family = binomial)
## 
## Deviance Residuals: 
##    Min      1Q  Median      3Q     Max  
## -1.337  -1.174   1.017   1.144   1.343  
## 
## Coefficients:
##              Estimate Std. Error z value Pr(>|z|)  
## (Intercept)  0.007045   0.090154   0.078   0.9377  
## x1          -0.570872   0.308890  -1.848   0.0646 .
## x2          -0.250098   0.316155  -0.791   0.4289  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 693.14  on 499  degrees of freedom
## Residual deviance: 689.18  on 497  degrees of freedom
## AIC: 695.18
## 
## Number of Fisher Scoring iterations: 3

(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.

data = data.frame(x1 = x1, x2 = x2, y = y)
lm.prob = predict(reg, data, type = "response")
lm.pred = ifelse(lm.prob > 0.5, 1, 0)
data.pos = data[lm.pred == 1, ]
data.neg = data[lm.pred == 0, ]
plot(data.pos$x1, data.pos$x2, col = "red")
points(data.neg$x1, data.neg$x2)

(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).

glmfit = glm(y ~ I(x1^2) + I(x2^2) + I(x1 * x2), data = data, family = binomial)

(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.

glmprobs = predict(glmfit, data, type = "response")
glmpred = ifelse(glmprobs > 0.5, 1, 0)
positive = data[glmpred == 1, ]
negative = data[glmpred == 0, ]
plot(positive$x1, positive$x2, col = "red")
points(negative$x1, negative$x2,)

(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)
data$y <- as.factor(data$y)
svmfit <- svm(y ~ x1 + x2, data, kernel = "linear", cost = 0.01)
svmpred <- predict(svmfit, data)
data.pos <- data[svmpred == 1, ]
data.neg <-  data[svmpred == 0, ]
plot(data.pos$x1, data.pos$x2, col = "red")
points(data.neg$x1, data.neg$x2, col = "red")

(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.

svmfit = svm(y ~ x1 + x2, data, gamma = 1)
svmpred = predict(svmfit, data)
data.pos <- data[svmpred == 1, ]
data.neg <- data[svmpred == 0, ]
plot(data.pos$x1, data.pos$x2, col = "red")
points(data.neg$x1, data.neg$x2, col = "blue")

Number 7

(a) Create a binary variable that takes on a 1 for cars with gas mileage above the median, and a for cars with gas mileage below the median.

library(ISLR2)
auto <- Auto
median <- ifelse(auto$mpg > median(auto$mpg), 1, 0)
auto$mpg1 <- as.factor(median)

##(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.

options(scipen = 999)
set.seed(1)
tune <- tune(svm, mpg1 ~ ., data = auto, kernel = "linear", ranges = list(cost = c(0.01, 0.1, 1, 10, 100, 1000)))
summary(tune)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##     1
## 
## - best performance: 0.01025641 
## 
## - Detailed performance results:
##      cost      error dispersion
## 1    0.01 0.07653846 0.03617137
## 2    0.10 0.04596154 0.03378238
## 3    1.00 0.01025641 0.01792836
## 4   10.00 0.02051282 0.02648194
## 5  100.00 0.03076923 0.03151981
## 6 1000.00 0.03076923 0.03151981

(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(1)
tune <- tune(svm, mpg1 ~ ., data = auto, kernel = "polynomial", ranges = list(cost = c(0.01, 0.1, 10), degree = c(2, 3)))
summary(tune)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost degree
##    10      2
## 
## - best performance: 0.5130128 
## 
## - Detailed performance results:
##    cost degree     error dispersion
## 1  0.01      2 0.5511538 0.04366593
## 2  0.10      2 0.5511538 0.04366593
## 3 10.00      2 0.5130128 0.08963366
## 4  0.01      3 0.5511538 0.04366593
## 5  0.10      3 0.5511538 0.04366593
## 6 10.00      3 0.5511538 0.04366593
set.seed(1)
tune <- tune(svm, mpg1 ~ ., data = auto, kernel = "radial", gamma = c(10,.01), ranges = list(cost = c(0.01, 10)))
summary(tune)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##    10
## 
## - best performance: 0.5179487 
## 
## - Detailed performance results:
##    cost     error dispersion
## 1  0.01 0.5511538 0.04366593
## 2 10.00 0.5179487 0.04917316
# All the polynomial kernel svm with the different degrees and costs had the same error. The radial kernel had the best performance with gamma of .01 and cost = 10. However, the linear kernel outperformed both.

(d) Make some plots to back up your assertions in (b) and (c). Hint: In the lab, we used the plot() function for svm objects only in cases with p = 2. When p > 2, you can use the plot() function to create plots displaying pairs of variables at a time.

svmlinear <- svm(mpg1 ~., data = auto, kernel="linear", cost = 1)
svmpoly <- svm(mpg1 ~., data = auto, kernel="polynomial", cost = 10, degree = 2)
svmradial <- svm(mpg1 ~., data = auto, kernel="radial", gamma = 0.01, cost = 10)
plot(svmlinear , auto , mpg ~ cylinders)

plot(svmlinear , auto , mpg ~ displacement)

plot(svmlinear , auto , mpg ~ horsepower)

plot(svmlinear , auto , mpg ~ weight)

plot(svmlinear , auto , mpg ~ acceleration)

### Number 8 ## (a) Create a training set containing a random sample of 800 observations, and a test set containing the remaining observations.

library(ISLR2)
oj <- OJ
set.seed(1)
sample <- sample(dim(oj)[1], 800)
train <- oj[sample, ]
test <- oj[-sample, ]

(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.

library(e1071)
svmlinear = svm(Purchase ~ ., kernel = "linear", data =train, cost = 0.01)
summary(svmlinear)
## 
## Call:
## svm(formula = Purchase ~ ., data = 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?

svmpred = predict(svmlinear, train)
table(train$Purchase, svmpred)
##     svmpred
##       CH  MM
##   CH 420  65
##   MM  75 240
(65+75)/800
## [1] 0.175
# the train data error rate is 0.175.

svmpred = predict(svmlinear, test)
table(test$Purchase, svmpred)
##     svmpred
##       CH  MM
##   CH 153  15
##   MM  33  69
(69+153)/270
## [1] 0.8222222
#the test error rate was 0.822.

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

options(scipen = 999)
set.seed(1)
tune <- tune(svm, Purchase ~ ., data = train, kernel = "linear", ranges = list(cost = c(0.01, 0.1, 1, 5, 10)))
summary(tune)
## 
## 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  5.00 0.17250 0.03162278
## 5 10.00 0.17375 0.03197764
# optimal cost is 5.

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

svmlinear = svm(Purchase ~ ., kernel = "linear", data =train, cost = 5)

svmpred = predict(svmlinear, train)
table(train$Purchase, svmpred)
##     svmpred
##       CH  MM
##   CH 423  62
##   MM  71 244
(62+71)/800
## [1] 0.16625
# the train data error rate is 0.166.

svmpred = predict(svmlinear, test)
table(test$Purchase, svmpred)
##     svmpred
##       CH  MM
##   CH 155  13
##   MM  29  73
(13+29)/270
## [1] 0.1555556
#the test error rate was 0.156.

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

svmradial <- svm(Purchase ~., data = train, kernel="radial", cost=0.01)

svmpred = predict(svmradial, train)
table(train$Purchase, svmpred)
##     svmpred
##       CH  MM
##   CH 485   0
##   MM 315   0
(315+0)/800
## [1] 0.39375
# the train data error rate is 0.39.

svmpred = predict(svmradial, test)
table(test$Purchase, svmpred)
##     svmpred
##       CH  MM
##   CH 168   0
##   MM 102   0
(102+0)/270
## [1] 0.3777778
# the test error is 0.37.

set.seed(1)
tune <- tune(svm, Purchase ~ ., data = train, kernel = "radial", ranges = list(cost = c(0.01, 0.1, 1, 5, 10)))
summary(tune)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##     1
## 
## - best performance: 0.17125 
## 
## - Detailed performance results:
##    cost   error dispersion
## 1  0.01 0.39375 0.04007372
## 2  0.10 0.18625 0.02853482
## 3  1.00 0.17125 0.02128673
## 4  5.00 0.18000 0.02220485
## 5 10.00 0.18625 0.02853482
# the best cost was 1. 

svmradial <- svm(Purchase ~., data = train, kernel="radial", cost = 1)

svmpred = predict(svmradial, train)
table(train$Purchase, svmpred)
##     svmpred
##       CH  MM
##   CH 441  44
##   MM  77 238
(44+77)/800
## [1] 0.15125
#the train error rate with cost = 1 was .15.

svmpred = predict(svmradial, test)
table(test$Purchase, svmpred)
##     svmpred
##       CH  MM
##   CH 151  17
##   MM  33  69
(17+33)/270
## [1] 0.1851852
# the test error rate with cost = 1 is .185

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

svmpoly <- svm(Purchase ~., data = train, kernel="polynomial", cost=0.01, degree = 2)

svmpred = predict(svmpoly, train)
table(train$Purchase, svmpred)
##     svmpred
##       CH  MM
##   CH 484   1
##   MM 297  18
(297+1)/800
## [1] 0.3725
# the train data error rate is 0.39.

svmpred = predict(svmpoly, test)
table(test$Purchase, svmpred)
##     svmpred
##       CH  MM
##   CH 167   1
##   MM  98   4
(98+1)/270
## [1] 0.3666667
# the test error is 0.37.

set.seed(1)
tune <- tune(svm, Purchase ~ ., data = train, kernel = "polynomial", ranges = list(cost = c(0.01, 0.1, 1, 5, 10, degree = 2)))
summary(tune)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##     1
## 
## - best performance: 0.185 
## 
## - Detailed performance results:
##    cost   error dispersion
## 1  0.01 0.37125 0.03537988
## 2  0.10 0.28750 0.05068969
## 3  1.00 0.18500 0.02415229
## 4  5.00 0.18875 0.02913689
## 5 10.00 0.19500 0.03184162
## 6  2.00 0.19000 0.02188988
# the best cost was 1. 

svmpoly <- svm(Purchase ~., data = train, kernel="polynomial", cost = 1)

svmpred = predict(svmpoly, train)
table(train$Purchase, svmpred)
##     svmpred
##       CH  MM
##   CH 453  32
##   MM  91 224
(91+32)/800
## [1] 0.15375
#the train error rate with cost = 1 was .15.

svmpred = predict(svmpoly, test)
table(test$Purchase, svmpred)
##     svmpred
##       CH  MM
##   CH 155  13
##   MM  47  55
(47+13)/270
## [1] 0.2222222
#the test error rate was 0.22.

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

# Overall, the radial kernel with a cost of 1 seemed to give the best results on the training set. The test error was 0.15. However, the results for all three tests were relatively the same.