ISLR Assignment 8

Problem 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. For instance, you can do this as follows:

> x_1=runif (500) -0.5
> x_2=runif (500) -0.5
> y=1*(x1^2 - {x2}^2 > 0)

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 \(X_1\) on the x-axis, and \(X_2\) on the y-axis.

plot(x1, x2, xlab = "X1", ylab = "X2", col = ifelse(y, "darkmagenta", "darkcyan"))

(c) Fit a logistic regression model to the data, using \(X_1\) and \(X_2\) as predictors.

glm.x <- glm(y ~ ., data = data.frame(x1, x2, y), family = "binomial")
summary(glm.x)

## 
## Call:
## glm(formula = y ~ ., family = "binomial", data = data.frame(x1, 
##     x2, y))
## 
## 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.

pred.glm.x <- predict(glm.x, data.frame(x1, x2))

plot(x1, x2, xlab = "X1", ylab = "X2", col = ifelse(pred.glm.x > 0, "darkmagenta", "darkcyan"))

(e) Now fit a logistic regression model to the data using non-linear functions of \(X_1\) and \(X_2\) as predictors (e.g. \({X_1}^2\), \(X_1\) × \(X_2\), \(log(X_2)\), and so forth).

log.x <- glm(y ~ poly(x1, 2) + poly(x2, 2) + I(x1 * x2), data = data.frame(x1, x2, y), family = "binomial")
summary(log.x)

## 
## Call:
## glm(formula = y ~ poly(x1, 2) + poly(x2, 2) + I(x1 * x2), family = "binomial", 
##     data = data.frame(x1, x2, y))
## 
## Deviance Residuals: 
##        Min          1Q      Median          3Q         Max  
## -8.240e-04  -2.000e-08  -2.000e-08   2.000e-08   1.163e-03  
## 
## Coefficients:
##              Estimate Std. Error z value Pr(>|z|)
## (Intercept)    -102.2     4302.0  -0.024    0.981
## poly(x1, 2)1   2715.3   141109.5   0.019    0.985
## poly(x1, 2)2  27218.5   842987.2   0.032    0.974
## poly(x2, 2)1   -279.7    97160.4  -0.003    0.998
## poly(x2, 2)2 -28693.0   875451.3  -0.033    0.974
## I(x1 * x2)     -206.4    41802.8  -0.005    0.996
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 6.9218e+02  on 499  degrees of freedom
## Residual deviance: 3.5810e-06  on 494  degrees of freedom
## AIC: 12
## 
## Number of Fisher Scoring iterations: 25

(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.

train = data.frame(x1 = x1, x2 = x2, y = as.factor(y))

pred.log.x <- predict(log.x, data.frame(x1, x2))

plot(x1, x2, xlab = "X1", ylab = "X2", col = ifelse(pred.log.x > 0, "darkmagenta", "darkcyan"))

(g) Fit a support vector classifier to the data with \(X_1\) and \(X_2\) as predictors. Obtain a class prediction for each training observation. Plot the observations, colored according to the predicted class labels.

svm.x <- svm(y ~ ., data = train, kernel = "linear", cost = 0.01)
summary(svm.x)

## 
## Call:
## svm(formula = y ~ ., data = train, kernel = "linear", cost = 0.01)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  linear 
##        cost:  0.01 
## 
## Number of Support Vectors:  480
## 
##  ( 239 241 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  0 1

pred.svm.x <- predict(svm.x, data.frame(x1, x2), type = "response")

plot(x1, x2, xlab = "X1", ylab = "X2", col = ifelse(pred.svm.x == 0, "darkmagenta", "darkcyan"))

**(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.x <- svm(y ~ ., data = train, kernel = "radial", gamma = 1)
summary(svm.x)

## 
## Call:
## svm(formula = y ~ ., data = train, kernel = "radial", gamma = 1)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  radial 
##        cost:  1 
## 
## Number of Support Vectors:  147
## 
##  ( 73 74 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  0 1

pred.svm.x <- predict(svm.x, data.frame(x1, x2), type = "response")

plot(x1, x2, xlab = "X1", ylab = "X2", col = ifelse(pred.svm.x == 0, "darkmagenta", "darkcyan"))

(i) Comment on your results.

SVM models can help us find the non-linear models in the dataset. The radial model is a better fit for the data.

Problem 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.

mpg.binary <- ifelse(Auto$mpg > median(Auto$mpg), 1, 0)
Auto$mpg.class <- as.factor(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.

set.seed(123)
tune.mpg.lin <- tune(svm, mpg.class ~ ., data = Auto, kernel = "linear", ranges = list(cost = c(0.01, 0.1, 1, 5, 10)))

summary(tune.mpg.lin)

## 
## 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 0.07634615 0.03928191
## 2  0.10 0.04333333 0.03191738
## 3  1.00 0.01025641 0.01792836
## 4  5.00 0.01538462 0.01792836
## 5 10.00 0.01788462 0.01727588

plot(tune.mpg.lin)

Best performance is at cost = 1.

**(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(123)
tune.mpg.rad <- tune(svm, mpg.class ~ ., data = Auto, kernel = "radial", ranges = list(cost = c(0.01, 0.1, 1, 5, 10, 20, 30, 40, 50)))

summary(tune.mpg.rad)

## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##    40
## 
## - best performance: 0.01788462 
## 
## - Detailed performance results:
##    cost      error dispersion
## 1  0.01 0.58173077 0.04740051
## 2  0.10 0.10692308 0.05900981
## 3  1.00 0.07891026 0.03828837
## 4  5.00 0.06608974 0.04785032
## 5 10.00 0.05602564 0.03551922
## 6 20.00 0.03820513 0.02463148
## 7 30.00 0.02288462 0.01427008
## 8 40.00 0.01788462 0.01234314
## 9 50.00 0.01788462 0.01234314

The best performance with a radial kernel has an error of 0.01788, which happens around cost = 40.

set.seed(123)
tune.mpg.poly <- tune(svm, mpg.class ~ ., data = Auto, kernel = "linear", ranges = list(cost = c(0.01, 0.1, 1, 5, 10)))

tune.mpg.poly

## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##     1
## 
## - best performance: 0.01025641

The best performance has an error of 0.01026, which is at cost = 1.

(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. Essentially, instead of typing

> plot(svmfit , dat)

where `svmfit` contains your fitted model and `dat` is a data frame containing your data, you can type

> plot(svmfit , dat, x1∼x4)

in order to plot just the first and fourth variables. However, you must replace `x1` and `x4` with the correct variable names. To find out more, type `?plot.svm`.

plot(tune.mpg.rad)

Confirmed that best cost is around cost = 40.

plot(tune.mpg.poly)

Confirmed that best performance is when cost = 1.

detach(Auto)

Problem 8

This problem involves the `OJ` data set which is part of the `ISLR` package.

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

set.seed(21)
set.training.oj <- sample(nrow(OJ), 800)
training.oj <- OJ[set.training.oj,]
test.oj <- OJ[-set.training.oj, ]

(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.lin <- svm(Purchase ~ ., data = training.oj, cost = 0.01, kernel = "linear")

summary(svm.oj.lin)

## 
## Call:
## svm(formula = Purchase ~ ., data = training.oj, cost = 0.01, kernel = "linear")
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  linear 
##        cost:  0.01 
## 
## Number of Support Vectors:  444
## 
##  ( 223 221 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM

There are 444 support vectors out of 800 training points.
Of the 2 levels, 223 of those support vectors are for CH, and 221 are for MM.

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

Training Error

train.pred.oj.lin <- predict(svm.oj.lin, training.oj)
mean(train.pred.oj.lin != training.oj$Purchase)

## [1] 0.1675

Test Error

test.pred.oj.lin <- predict(svm.oj.lin, test.oj)
mean(test.pred.oj.lin != test.oj$Purchase)

## [1] 0.1703704

Training Error Rate: 0.1675
Test Error Rate: 0.1703704

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

set.seed(123)
tune.oj.lin <- tune(svm, Purchase ~ ., data = OJ, kernel = "linear", ranges = list(cost = c(0.01, 0.1, 1, 5, 10)))

tune.oj.lin

## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##     5
## 
## - best performance: 0.164486

The optimal cost is 5.

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

Training Error

train.cost.oj <- svm(Purchase ~ ., data = training.oj,  kernel = "linear", cost = 5)
pred.train.cost <- predict(train.cost.oj, training.oj)
mean(pred.train.cost != training.oj$Purchase)

## [1] 0.165

Test Error

test.cost.oj <- svm(Purchase ~ ., data = test.oj,  kernel = "linear", cost = 5)
pred.test.cost <- predict(test.cost.oj, test.oj)
mean(pred.test.cost != test.oj$Purchase)

## [1] 0.1296296

Training Error: 0.165
Test Error: 0.1296

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

Radial Kernel

Summary Radial

svm.oj.rad <- svm(Purchase ~ ., data = training.oj, cost = 0.01, kernel = "radial")

summary(svm.oj.rad)

## 
## Call:
## svm(formula = Purchase ~ ., data = training.oj, cost = 0.01, kernel = "radial")
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  radial 
##        cost:  0.01 
## 
## Number of Support Vectors:  627
## 
##  ( 316 311 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM

Radial Training Error

train.pred.oj.rad <- predict(svm.oj.rad, training.oj)
mean(train.pred.oj.rad != training.oj$Purchase)

## [1] 0.38875

Radial Test Error

test.pred.oj.rad <- predict(svm.oj.rad, test.oj)
mean(test.pred.oj.rad != test.oj$Purchase)

## [1] 0.3925926

Tune Radial

set.seed(123)
tune.oj.rad<- tune(svm, Purchase ~ ., data = OJ, kernel = "radial", ranges = list(cost = c(0.01, 0.1, 1, 5, 10)))

tune.oj.rad

## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##     1
## 
## - best performance: 0.1691589

New Radial Training Error

train.cost.oj.rad <- svm(Purchase ~ ., data = training.oj,  kernel = "radial", cost = tune.oj.rad$best.parameter$cost)
pred.train.cost.rad <- predict(train.cost.oj.rad, training.oj)
mean(pred.train.cost.rad != training.oj$Purchase)

## [1] 0.15125

New Radial Test Error

test.cost.oj.rad <- svm(Purchase ~ ., data = test.oj,  kernel = "radial", cost = tune.oj.rad$best.parameter$cost)
pred.test.cost.rad <- predict(test.cost.oj.rad, test.oj)
mean(pred.test.cost.rad != test.oj$Purchase)

## [1] 0.1296296

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

Polynomial Kernel

Summary Polynomial

svm.oj.poly <- svm(Purchase ~ ., data = training.oj, cost = 0.01, kernel = "polynomial")

summary(svm.oj.poly)

## 
## Call:
## svm(formula = Purchase ~ ., data = training.oj, cost = 0.01, kernel = "polynomial")
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  polynomial 
##        cost:  0.01 
##      degree:  3 
##      coef.0:  0 
## 
## Number of Support Vectors:  614
## 
##  ( 309 305 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM

Polynomial Training Error

train.pred.oj.poly <- predict(svm.oj.poly, training.oj)
mean(train.pred.oj.poly != training.oj$Purchase)

## [1] 0.3675

Polynomial Test Error

test.pred.oj.poly <- predict(svm.oj.poly, test.oj)
mean(test.pred.oj.poly != test.oj$Purchase)

## [1] 0.3740741

Tune Polynomial

set.seed(123)
tune.oj.poly<- tune(svm, Purchase ~ ., data = OJ, kernel = "polynomial", ranges = list(cost = c(0.01, 0.1, 1, 5, 10)))

tune.oj.poly

## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##     5
## 
## - best performance: 0.1841121

New Polynomial Training Error

train.cost.oj.poly <- svm(Purchase ~ ., data = training.oj,  kernel = "polynomial", cost = tune.oj.poly$best.parameter$cost)
pred.train.cost.poly <- predict(train.cost.oj.poly, training.oj)
mean(pred.train.cost.poly != training.oj$Purchase)

## [1] 0.15125

New Polynomial Test Error

test.cost.oj.poly <- svm(Purchase ~ ., data = test.oj,  kernel = "polynomial", cost = tune.oj.poly$best.parameter$cost)
pred.test.cost.poly <- predict(test.cost.oj.poly, test.oj)
mean(pred.test.cost.poly != test.oj$Purchase)

## [1] 0.1

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

Linear Test: 0.1703704
Linear Tuned Test: 0.1296

Radial Test: 0.3925926
Radial Tuned Test: 0.1296296

Polynomial Test: 0.3740741
Polynomial Tuned Test: 0.1

The Polynomial Tuned Test gives the best results.

detach(OJ)

ISLR Assignment 8

Chapter 9 (page 368) Problems 5, 7, 8

Misty Stultz

4/26/22

Problem 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. For instance, you can do this as follows:

(b) Plot the observations, colored according to their class labels. Your plot should display \(X_1\) on the x-axis, and \(X_2\) on the y-axis.

(c) Fit a logistic regression model to the data, using \(X_1\) and \(X_2\) as predictors.

(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.

(e) Now fit a logistic regression model to the data using non-linear functions of \(X_1\) and \(X_2\) as predictors (e.g. \({X_1}^2\), \(X_1\) × \(X_2\), \(log(X_2)\), and so forth).

(g) Fit a support vector classifier to the data with \(X_1\) and \(X_2\) as predictors. Obtain a class prediction for each training observation. Plot the observations, colored according to the predicted class labels.

(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.

(i) Comment on your results.

Problem 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.

(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.

(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.

(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. Essentially, instead of typing

where svmfit contains your fitted model and dat is a data frame containing your data, you can type

in order to plot just the first and fourth variables. However, you must replace x1 and x4 with the correct variable names. To find out more, type ?plot.svm.

Problem 8

This problem involves the OJ data set which is part of the ISLR package.

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

(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.

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

Training Error

Test Error

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

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

Training Error

Test Error

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

Radial Kernel

Summary Radial

Radial Training Error

Radial Test Error

Tune Radial

New Radial Training Error

New Radial Test Error

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

Polynomial Kernel

Summary Polynomial

Polynomial Training Error

Polynomial Test Error

Tune Polynomial

New Polynomial Training Error

New Polynomial Test Error

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

**(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.**

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.

(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.

**(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.**

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. Essentially, instead of typing

where `svmfit` contains your fitted model and `dat` is a data frame containing your data, you can type

in order to plot just the first and fourth variables. However, you must replace `x1` and `x4` with the correct variable names. To find out more, type `?plot.svm`.

This problem involves the `OJ` data set which is part of the `ISLR` package.

(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.

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

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

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

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