5. 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 obser- vations belong to two classes with a quadratic decision boundary between them.
x1 <- runif(500) - 0.5
x2 <- runif(500) - 0.5
y <- as.factor(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 y- axis.
plot(x1[y == 0], x2[y == 0], col = "mediumpurple2", xlab = "X1", ylab = "X2", pch = 20)
points(x1[y == 1], x2[y == 1], col = "lightpink", pch=20)

(c) Fit a logistic regression model to the data, using X1 and X2 as predictors.
logistic_model <- glm(y ~ x1 + x2, family = "binomial")
summary(logistic_model)
## 
## Call:
## glm(formula = y ~ x1 + x2, family = "binomial")
## 
## Coefficients:
##             Estimate Std. Error z value Pr(>|z|)  
## (Intercept) -0.01693    0.09000  -0.188   0.8508  
## x1          -0.19454    0.31412  -0.619   0.5357  
## x2          -0.72583    0.31841  -2.280   0.0226 *
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 693.08  on 499  degrees of freedom
## Residual deviance: 687.42  on 497  degrees of freedom
## AIC: 693.42
## 
## Number of Fisher Scoring iterations: 4
(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, x2, y)
data$pred <- ifelse(predict(logistic_model, type = "response") > 0.5, 1, 0)

x1_color <- "mediumpurple2"
x2_color <- "lightpink"
plot(data$x1, data$x2, col = ifelse(data$pred == 1, x1_color, x2_color), pch = 20, xlab = "X1", ylab = "X2")

(e) Now fit a logistic regression model to the data using non-linear functions of X1 and X2 as predictors (e.g. X12, X1 ×X2, log(X2), and so forth).
logistic_model_nonlinear <- glm(y ~ poly(x1, 2) + poly(x2, 2), data = data, 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.
data$pred_nonlinear <- ifelse(predict(logistic_model_nonlinear, type = "response") > 0.5, 1, 0)
plot(data$x1, data$x2, col = ifelse(data$pred_nonlinear == 1, x1_color, x2_color), pch = 20, xlab = "X1", ylab = "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.
svm_model <- svm(y ~ ., data = data, kernel = "linear")
svm_pred <- predict(svm_model, data)
custom_colors <- c(x1_color, x2_color)
plot(data$x1, data$x2, col = custom_colors[svm_pred], pch = 20, xlab = "X1", ylab = "X2")

(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_model_nonlinear <- svm(y ~ ., data = data, kernel = "radial", gamma= 1)
svm_pred_nonlinear <- predict(svm_model_nonlinear, data)
plot(data$x1, data$x2, col = custom_colors[svm_pred_nonlinear], pch = 20, xlab = "X1", ylab = "X2")

(i) Comment on your results.

Comparing all the plots, I can see that SVM with radial kernels are clearly showing the non-linear boundary.

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$mpg_binary <- ifelse(Auto$mpg > median(Auto$mpg), 1, 0)
Auto$mpg_binary <- as.factor(Auto$mpg_binary)
(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.
svm_auto <- tune(svm, mpg_binary ~ . - mpg, data = Auto, ranges = list(cost = c(0.1, 1, 10)))
summary(svm_auto)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##     1
## 
## - best performance: 0.08935897 
## 
## - Detailed performance results:
##   cost      error dispersion
## 1  0.1 0.16602564 0.05975575
## 2  1.0 0.08935897 0.04380206
## 3 10.0 0.09711538 0.05519876

Cost of 1 has the lowest error rate of 0.08423077.

(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.
svm_radial <- tune(svm, mpg_binary ~ . - mpg, data = Auto, kernel = "radial", ranges = list(cost = c(0.1, 1, 10), gamma = c(0.1, 1, 10)))
summary(svm_radial)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost gamma
##     1     1
## 
## - best performance: 0.08429487 
## 
## - Detailed performance results:
##   cost gamma      error dispersion
## 1  0.1   0.1 0.08935897 0.05307427
## 2  1.0   0.1 0.08679487 0.05015153
## 3 10.0   0.1 0.08929487 0.03253208
## 4  0.1   1.0 0.52128205 0.14136381
## 5  1.0   1.0 0.08429487 0.04989877
## 6 10.0   1.0 0.08679487 0.05015153
## 7  0.1  10.0 0.55378205 0.04388459
## 8  1.0  10.0 0.51788462 0.03185187
## 9 10.0  10.0 0.51282051 0.03735835
svm_poly <- tune(svm, mpg_binary ~ . - mpg, data = Auto, kernel = "polynomial", degree = 2, ranges = list(cost = c(0.1, 1, 10)))
summary(svm_poly)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##    10
## 
## - best performance: 0.5077564 
## 
## - Detailed performance results:
##   cost     error dispersion
## 1  0.1 0.5462179 0.05788544
## 2  1.0 0.5462179 0.05788544
## 3 10.0 0.5077564 0.10993562

The best model for poly is cost of 10 with 2 degrees, and the best for radial is cost of 10 with 1 gamma.

(d) Make some plots to back up your assertions in (b) and (c).
best_linear = svm(mpg_binary ~ ., data = Auto, kernel = "linear", cost = 1)
best_poly = svm(mpg_binary ~ ., data = Auto, kernel = "polynomial", cost = 10, degree = 2)
best_radial = svm(mpg_binary ~ ., data = Auto, kernel = "radial", cost = 10, gamma = 1)
plotpairs = function(fit) {
    for (name in names(Auto)[!(names(Auto) %in% c("mpg", "mpg_binary", "name"))]) {
        plot(fit, Auto, as.formula(paste("mpg~", name, sep = "")))
    }
}
plotpairs(best_linear)

plotpairs(best_poly)

plotpairs(best_radial)

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(123)
index <- sample(1:nrow(OJ), 800)
train_data <- OJ[index, ]
test_data <- OJ[-index, ]
(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.
svm_OJ <- svm(Purchase ~ ., data = train_data, kernel = "linear", cost = 0.01)
summary(svm_OJ)
## 
## Call:
## svm(formula = Purchase ~ ., data = train_data, kernel = "linear", 
##     cost = 0.01)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  linear 
##        cost:  0.01 
## 
## Number of Support Vectors:  442
## 
##  ( 220 222 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM

The levels of the response variable “Purchase” are listed as “CH” and “MM”. The classifier is performing binary classification, as there are 2 classes. There are 442 support vectors in total. Among them, 220 belong to CH and 222 belong to MM.

(c) What are the training and test error rates?
train_pred <- predict(svm_OJ, train_data)
test_pred <- predict(svm_OJ, test_data)

train_error <- mean(train_pred != train_data$Purchase)
test_error <- mean(test_pred != test_data$Purchase)

train_error
## [1] 0.165
test_error
## [1] 0.1777778
(d) Use the tune() function to select an optimal cost. Consider values in the range 0.01 to 10.
tune_out <- tune(svm, Purchase ~ ., data = train_data, kernel = "linear", ranges = list(cost = c(0.01, 0.1, 1, 10)))
summary(tune_out)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##     1
## 
## - best performance: 0.16875 
## 
## - Detailed performance results:
##    cost   error dispersion
## 1  0.01 0.17625 0.03143004
## 2  0.10 0.17250 0.03425801
## 3  1.00 0.16875 0.03596391
## 4 10.00 0.17250 0.02751262
(e) Compute the training and test error rates using this new value for cost.
svm_tuned <- svm(Purchase ~ ., data = train_data, kernel = "linear", cost = 1)
train_tuned <- predict(svm_tuned, train_data)
test_tuned <- predict(svm_tuned, test_data)

train_error_tuned <- mean(train_tuned != train_data$Purchase)
test_error_tuned <- mean(test_tuned != test_data$Purchase)

train_error_tuned
## [1] 0.16
test_error_tuned
## [1] 0.1555556
(f) Repeat parts (b) through (e) using a support vector machine with a radial kernel. Use the default value for gamma.
OJ_radial <- svm(Purchase ~ ., data = train_data, kernel = "radial")
summary(OJ_radial)
## 
## Call:
## svm(formula = Purchase ~ ., data = train_data, kernel = "radial")
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  radial 
##        cost:  1 
## 
## Number of Support Vectors:  367
## 
##  ( 181 186 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM
train_radial <- predict(OJ_radial, train_data)
test_radial <- predict(OJ_radial, test_data)

train_error_radial <- mean(train_radial != train_data$Purchase)
test_error_radial <- mean(test_radial != test_data$Purchase)

train_error_radial
## [1] 0.13875
test_error_radial
## [1] 0.1888889
tune_out1 <- tune(svm, Purchase ~ ., data = train_data, kernel = "radial", ranges = list(cost = c(0.01, 0.1, 1, 10)))
summary(tune_out1)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##     1
## 
## - best performance: 0.1625 
## 
## - Detailed performance results:
##    cost   error dispersion
## 1  0.01 0.39125 0.04411554
## 2  0.10 0.18125 0.03691676
## 3  1.00 0.16250 0.03486083
## 4 10.00 0.16500 0.03622844
svm_tuned1 <- svm(Purchase ~ ., data = train_data, kernel = "radial", cost = 1)
train_tuned1 <- predict(svm_tuned1, train_data)
test_tuned1 <- predict(svm_tuned1, test_data)

train_error_tuned1 <- mean(train_tuned1 != train_data$Purchase)
test_error_tuned1 <- mean(test_tuned1 != test_data$Purchase)

train_error_tuned1
## [1] 0.13875
test_error_tuned1
## [1] 0.1888889
(g) Repeat parts (b) through (e) using a support vector machine with a polynomial kernel. Set degree = 2.
OJ_poly <- svm(Purchase ~ ., data = train_data, kernel = "polynomial", degree = 2)
summary(OJ_poly)
## 
## Call:
## svm(formula = Purchase ~ ., data = train_data, kernel = "polynomial", 
##     degree = 2)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  polynomial 
##        cost:  1 
##      degree:  2 
##      coef.0:  0 
## 
## Number of Support Vectors:  445
## 
##  ( 219 226 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM
train_poly <- predict(OJ_poly, train_data)
test_poly <- predict(OJ_poly, test_data)

train_error_poly <- mean(train_poly != train_data$Purchase)
test_error_poly <- mean(test_poly != test_data$Purchase)

train_error_poly
## [1] 0.1725
test_error_poly
## [1] 0.2222222
poly_tune <- tune(svm, Purchase ~ ., data = train_data, kernel = "polynomial", degree = 2, ranges = list(cost = c(0.01, 0.1, 1, 10)))
summary(poly_tune)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##    10
## 
## - best performance: 0.17125 
## 
## - Detailed performance results:
##    cost   error dispersion
## 1  0.01 0.38750 0.04289846
## 2  0.10 0.30875 0.01772671
## 3  1.00 0.18375 0.03335936
## 4 10.00 0.17125 0.03729108
OJ_poly1 <- svm(Purchase ~ ., data = train_data, kernel = "polynomial", degree = 2, cost =10)
summary(OJ_poly1)
## 
## Call:
## svm(formula = Purchase ~ ., data = train_data, kernel = "polynomial", 
##     degree = 2, cost = 10)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  polynomial 
##        cost:  10 
##      degree:  2 
##      coef.0:  0 
## 
## Number of Support Vectors:  337
## 
##  ( 168 169 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM
train_poly1 <- predict(OJ_poly1, train_data)
test_poly1 <- predict(OJ_poly1, test_data)

train_error_poly1 <- mean(train_poly1 != train_data$Purchase)
test_error_poly1 <- mean(test_poly1 != test_data$Purchase)

train_error_poly1
## [1] 0.14375
test_error_poly1
## [1] 0.2037037
(h) Overall, which approach seems to give the best results on this data?

Based on the test error rates, the linear SVM has the lowest test error rate (17.8%), followed by the radial SVM (18.9%) and the polynomial SVM (20.4%). Therefore, the linear SVM appears to be the best model among the three for this dataset.