Chapter 9 (5,7,8)
Tu Thai
2023-04-27
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 observations belong to two classes with a quadratic decision boundary between them.
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 y-axis.
plot(x1,x2,xlab="x1", ylab="x2", col = (4-y), pch = (3-y))
(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))
logreg=glm(y~.,data=df, family='binomial')
summary(logreg)##
## Call:
## glm(formula = y ~ ., family = "binomial", data = df)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.179 -1.139 -1.112 1.206 1.257
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -0.087260 0.089579 -0.974 0.330
## x1 0.196199 0.316864 0.619 0.536
## x2 -0.002854 0.305712 -0.009 0.993
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 692.18 on 499 degrees of freedom
## Residual deviance: 691.79 on 497 degrees of freedom
## AIC: 697.79
##
## 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.
library(ggplot2)
probs <- predict(logreg, df, type = "response")
preds <- ifelse(probs > 0.5,1,0)
ggplot(data = df, mapping = aes(x1, x2))+
geom_point(data = df, mapping = aes(colour = preds))
(e) Now fit a logistic regression model to the data using non-linear functions of X1 and X2 as predictors (e.g. X21 , X1×X2, log(X2), and so forth).
logreg2=glm(y~poly(x1,2)+poly(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(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.
probs <- predict(logreg2, df, type = "response")
preds <- ifelse(probs > 0.5,1,0)
ggplot(data = df, mapping = aes(x1, x2))+
geom_point(data = df, mapping = aes(colour = preds))
(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=svm(y~., data=df, kernel ="linear", cost =0.1, scale=FALSE)
plot(svm, 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.
library(e1071)
svm=svm(y~., data=df, kernel ="radial", gamma=1)
plot(svm, df)
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.
library(ISLR)
Auto$mpglevel <- ifelse(Auto$mpg > median(Auto$mpg),1,0)
Auto$mpglevel <- as.factor(Auto$mpglevel)(b) Fit a support vector classifier to the data with various values of cost, in order to predict whether a car gets high or low gas mileage. Report the cross-validation errors associated with different values of this parameter. Comment on your results. Note you will need to fit the classifier without the gas mileage variable to produce sensible results.
library(e1071)
set.seed(1)
linear <- tune(svm, mpglevel ~ . -mpg, data = Auto, kernel = "linear", ranges = list(cost = c(0.001, 0.01, 0.1, 1, 5, 10, 100)))
summary(linear)We see that cost=0.1 results in the lowest cross-validation error rate.
(c) Now repeat (b), this time using SVMs with radial and polynomial basis kernels, with different values of gamma and degree and cost. Comment on your results.
# Polynomial
set.seed(2)
grid <- c(0.1, 1, 5, 10)
form <- mpg ~.
poly = tune.svm(form, data = Auto, kernel = "polynomial", cost = grid, gamma = grid)
best.poly <- poly$best.model
summary(best.poly)##
## Call:
## best.svm(x = form, data = Auto, gamma = grid, cost = grid, kernel = "polynomial")
##
##
## Parameters:
## SVM-Type: eps-regression
## SVM-Kernel: polynomial
## cost: 1
## degree: 3
## gamma: 0.1
## coef.0: 0
## epsilon: 0.1
##
##
## Number of Support Vectors: 264For polynomial, best cost is at 1, with degree of 3 and gamma = 0.1
#Radial
set.seed(2)
grid <- c(0.1, 1, 5, 10)
form <- mpg ~.
radial = tune.svm(form, data = Auto, kernel = "radial", cost = grid, gamma = grid)
best.radial <- radial$best.model
summary(best.radial)##
## Call:
## best.svm(x = form, data = Auto, gamma = grid, cost = grid, kernel = "radial")
##
##
## Parameters:
## SVM-Type: eps-regression
## SVM-Kernel: radial
## cost: 5
## gamma: 0.1
## epsilon: 0.1
##
##
## Number of Support Vectors: 275For radial, best cost is at 5, with gamma = 0.1 and epsilon = 0.1
(d) Make some plots to back up your assertions in (b) and (c).
8. This problem involves the OJ data set which is part of the ISLR2 package.
(a) Create a training set containing a random sample of 800 observations, and a test set containing the remaining observations.
library(ISLR)
set.seed(1)
tr.idx <- sample(1:nrow(OJ), size = 800, replace=FALSE)
train <- OJ[tr.idx,]
test <- OJ[-tr.idx,](b) Fit a support vector classifier to the training data using cost = 0.01, with Purchase as the response and the other variables as predictors. Use the summary() function to produce summary statistics, and describe the results obtained.
library(e1071)
OJ.svm=svm(Purchase~., data=train, kernel='linear', cost=0.01)
summary(OJ.svm)##
## Call:
## svm(formula = Purchase ~ ., data = train, kernel = "linear", cost = 0.01)
##
##
## Parameters:
## SVM-Type: C-classification
## SVM-Kernel: linear
## cost: 0.01
##
## Number of Support Vectors: 435
##
## ( 219 216 )
##
##
## Number of Classes: 2
##
## Levels:
## CH MMThere are 435 support vectors created, 219 belongs to CH and 216 for MM.
(c) What are the training and test error rates?
# Training error rate:
pred_train = predict(OJ.svm, train)
table(train$Purchase, pred_train)## pred_train
## CH MM
## CH 420 65
## MM 75 240(72+65)/800## [1] 0.17125# Testing error rate:
pred_test = predict(OJ.svm, test)
table(test$Purchase, pred_test)## pred_test
## CH MM
## CH 153 15
## MM 33 69(33+15)/nrow(test)## [1] 0.1777778training error ratge is 17.12% and testing error rate is at 17.78%
(d) Use the tune() function to select an optimal cost. Consider values in the range 0.01 to 10.
set.seed(1)
cost.grid <- c(0.01,0.1,1,10)
form <- Purchase ~ .
OJ.tune <- tune.svm(form, data = train, kernel = "linear", cost = cost.grid)
summary(OJ.tune)##
## Parameter tuning of 'svm':
##
## - sampling method: 10-fold cross validation
##
## - best parameters:
## cost
## 0.1
##
## - best performance: 0.1725
##
## - Detailed performance results:
## cost error dispersion
## 1 0.01 0.17625 0.02853482
## 2 0.10 0.17250 0.03162278
## 3 1.00 0.17500 0.02946278
## 4 10.00 0.17375 0.03197764cost of 0.1 has lowest error rate, meaning optimal cost to sell the product is 0.03
(e) Compute the training and test error rates using this new value for cost.
svm2=svm(Purchase~., data=train, kernel='linear', cost=0.1)
# Training error rate:
pred_train = predict(svm2, train)
table(train$Purchase, pred_train)## pred_train
## CH MM
## CH 422 63
## MM 69 246(63+69)/nrow(train)## [1] 0.165# Testing error rate:
pred_test = predict(svm2, test)
table(test$Purchase, pred_test)## pred_test
## CH MM
## CH 155 13
## MM 31 71(31+13)/nrow(test)## [1] 0.162963Both the train and test error rate decrease. New train error rate is 16.5% while new test error rate is 16.29%
(f) Repeat parts (b) through (e) using a support vector machine with a radial kernel. Use the default value for gamma.
set.seed(1)
svm_radial = svm(Purchase ~ ., data = train, kernel = "radial", cost=0.1)
# Training error rate:
pred_train_radial = predict(svm_radial, train)
table(train$Purchase, pred_train_radial)## pred_train_radial
## CH MM
## CH 433 52
## MM 87 228(87+52)/nrow(train)## [1] 0.17375# Testing error rate:
pred_test_radial = predict(svm_radial, test)
table(test$Purchase, pred_test_radial)## pred_test_radial
## CH MM
## CH 150 18
## MM 37 65(37+18)/nrow(test)## [1] 0.2037037For radial kernel, train error rate is 17.38 while test error rate is 0.20
(g) Repeat parts (b) through (e) using a support vector machine with a polynomial kernel. Set degree = 2.
set.seed(1)
svm_poly = svm(Purchase ~ ., data = train, kernel = "polynomial", cost=0.1, degree=2)
# Training error rate:
pred_train_poly = predict(svm_poly, train)
table(train$Purchase, pred_train_poly)## pred_train_poly
## CH MM
## CH 465 20
## MM 225 90(225+20)/nrow(train)## [1] 0.30625# Testing error rate:
pred_test_poly = predict(svm_poly, test)
table(test$Purchase, pred_test_poly)## pred_test_poly
## CH MM
## CH 161 7
## MM 73 29(73+7)/nrow(test)## [1] 0.2962963For polynomial kernel, train error rate is 30.625 while test error rate is 0.296
(h) Overall, which approach seems to give the best results on this data?
Overall, linear approach gives the best results