Assignment 7

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:

> x1 <- runif (500) - 0.5
> x2 <- runif (500) - 0.5
> y <- 1 x (x1^2 - x2^2 > 0)

set.seed(421)
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", pch = "+")
points(x1[y == 1], x2[y == 1], col = "blue", pch = 4)

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

lm.fit = glm(y ~ x1 + x2, family = binomial)
summary(lm.fit)

## 
## Call:
## glm(formula = y ~ x1 + x2, family = binomial)
## 
## Deviance Residuals: 
##    Min      1Q  Median      3Q     Max  
## -1.278  -1.227   1.089   1.135   1.175  
## 
## Coefficients:
##             Estimate Std. Error z value Pr(>|z|)
## (Intercept)  0.11999    0.08971   1.338    0.181
## x1          -0.16881    0.30854  -0.547    0.584
## x2          -0.08198    0.31476  -0.260    0.795
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 691.35  on 499  degrees of freedom
## Residual deviance: 690.99  on 497  degrees of freedom
## AIC: 696.99
## 
## 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(lm.fit, data, type = "response")
lm.pred = ifelse(lm.prob > 0.52, 1, 0)
data.pos = data[lm.pred == 1, ]
data.neg = data[lm.pred == 0, ]
plot(data.pos$x1, data.pos$x2, col = "blue", xlab = "X1", ylab = "X2", pch = "+")
points(data.neg$x1, data.neg$x2, col = "red", pch = 4)

(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).

lm.fit = glm(y ~ poly(x1, 2) + poly(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.

lm.prob = predict(lm.fit, 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 = "blue", xlab = "X1", ylab = "X2", pch = "+")
points(data.neg$x1, data.neg$x2, col = "red", pch = 4)

(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.fit = svm(as.factor(y) ~ x1 + x2, data, kernel = "linear", cost = 0.1)
svm.pred = predict(svm.fit, data)
data.pos = data[svm.pred == 1, ]
data.neg = data[svm.pred == 0, ]
plot(data.pos$x1, data.pos$x2, col = "blue", xlab = "X1", ylab = "X2", pch = "+")
points(data.neg$x1, data.neg$x2, col = "red", pch = 4)

(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.fit = svm(as.factor(y) ~ x1 + x2, data, gamma = 1)
svm.pred = predict(svm.fit, data)
data.pos = data[svm.pred == 1, ]
data.neg = data[svm.pred == 0, ]
plot(data.pos$x1, data.pos$x2, col = "blue", xlab = "X1", ylab = "X2", pch = "+")
points(data.neg$x1, data.neg$x2, col = "red", pch = 4)

(i) Comment on your results.

SVM with non-linear kernel and logistic regression with interaction terms are important for finding non-linear decision boundaries.

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.

library(ISLR2)
var <- ifelse(Auto$mpg > median(Auto$mpg), 1, 0)

I got an error here, “replacement has 0 rows”?

Auto$mpglevel <- as.factor(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.

This is the code I used, it runs in R studio just fine and gives me cost summary, but it knits with an error:
“Algorithm did not converge”
“Fitted probabilities numerically 0 or 1 occurred”
“Execution halted”

set.seed(1)
tune.out <- tune(svm, mpglevel ~ ., data = Auto, kernel = “linear”, ranges = list(cost = c(0.01, 0.1, 1, 5, 10, 100, 1000)))
summary(tune.out)

The best cost is cost of .01.

(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.

This is the code I used, it runs in R studio just fine and gives me cost summary, but it knits with the same error:
“Algorithm did not converge”
“Fitted probabilities numerically 0 or 1 occurred”
“Execution halted”

set.seed(1)
tune.out <- tune(svm, mpglevel ~ ., data = Auto, kernel = “polynomial”, ranges = list(cost = c(0.01, 0.1, 1, 5, 10, 100), degree = c(2, 3, 4)))
summary(tune.out)

For polynomial kernel, the lowest cross-validation error is a degree of 3 and cost of 0.1.

set.seed(1)
tune.out <- tune(svm, mpglevel ~ ., data = Auto, kernel = “radial”, ranges = list(cost = c(0.01, 0.1, 1, 5, 10, 100), gamma = c(0.01, 0.1, 1, 5, 10, 100)))
summary(tune.out)

For radial kernel, the lowest cross-validation error is a degree of 1 and cost of 0.01.

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

svm.linear <- svm(mpglevel ~ ., data = Auto, kernel = “linear”, cost = 1)
svm.poly <- svm(mpglevel ~ ., data = Auto, kernel = “polynomial”, cost = 100, degree = 2)
svm.radial <- svm(mpglevel ~ ., data = Auto, kernel = “radial”, cost = 100, 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 = ““)))
}
}

I got an error here, “object ‘mpg’ not found”:

plotpairs(svm.linear)
plotpairs(svm.poly)
plotpairs(svm.radial)

Problem 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(1)
train <- sample(nrow(OJ), 800)
OJ.train <- OJ[train, ]
OJ.test <- OJ[-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.

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

## 
## 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

SVC creates 435 support vectors out of 800 training data points. 219 of these belowng to level CH and 216 belong to MM.

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

train.pred <- predict(svm.linear, OJ.train)
table(OJ.train$Purchase, train.pred)

##     train.pred
##       CH  MM
##   CH 420  65
##   MM  75 240

(78 + 55) / (439 + 228 + 78 + 55)

## [1] 0.16625

test.pred <- predict(svm.linear, OJ.test)
table(OJ.test$Purchase, test.pred)

##     test.pred
##       CH  MM
##   CH 153  15
##   MM  33  69

(31 + 18) / (141 + 80 + 31 + 18)

## [1] 0.1814815

The training error rate is 16.6% and the test error rate is 18.1%.

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

set.seed(2)
tune.out <- tune(svm, Purchase ~ ., data = OJ.train, kernel = "linear", ranges = list(cost = 10^seq(-2, 1, by = 0.25)))
summary(tune.out)

## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##      cost
##  1.778279
## 
## - best performance: 0.1675 
## 
## - Detailed performance results:
##           cost   error dispersion
## 1   0.01000000 0.17625 0.04059026
## 2   0.01778279 0.17625 0.04348132
## 3   0.03162278 0.17125 0.04604120
## 4   0.05623413 0.17000 0.04005205
## 5   0.10000000 0.17125 0.04168749
## 6   0.17782794 0.17000 0.04090979
## 7   0.31622777 0.17125 0.04411554
## 8   0.56234133 0.17125 0.04084609
## 9   1.00000000 0.17000 0.04090979
## 10  1.77827941 0.16750 0.03782269
## 11  3.16227766 0.16750 0.03782269
## 12  5.62341325 0.16750 0.03545341
## 13 10.00000000 0.17000 0.03736085

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

svm.linear <- svm(Purchase ~ ., kernel = "linear", data = OJ.train, cost = tune.out$best.parameter$cost)
train.pred <- predict(svm.linear, OJ.train)
table(OJ.train$Purchase, train.pred)

##     train.pred
##       CH  MM
##   CH 423  62
##   MM  69 246

(71 + 56) / (438 + 235 + 71 + 56)

## [1] 0.15875

test.pred <- predict(svm.linear, OJ.test)
table(OJ.test$Purchase, test.pred)

##     test.pred
##       CH  MM
##   CH 156  12
##   MM  29  73

(32 + 19) / (140 + 79 + 32 + 19)

## [1] 0.1888889

Using the new cost, the training error rate is 15.9% and the test error rate is 18.9%.

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

svm.radial <- svm(Purchase ~ ., kernel = "radial", data = OJ.train)
summary(svm.radial)

## 
## 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

train.pred <- predict(svm.radial, OJ.train)
table(OJ.train$Purchase, train.pred)

##     train.pred
##       CH  MM
##   CH 441  44
##   MM  77 238

(77 + 39) / (455 + 229 + 77 + 39)

## [1] 0.145

test.pred <- predict(svm.radial, OJ.test)
table(OJ.test$Purchase, test.pred)

##     test.pred
##       CH  MM
##   CH 151  17
##   MM  33  69

(28 + 18) / (141 + 83 + 28 + 18)

## [1] 0.1703704

Radial kernal creates 373 support vectors, 188 at level CH and 185 at level MM. The training error is 14.5% and the test error is 17.0%.

set.seed(2)
tune.out <- tune(svm, Purchase ~ ., data = OJ.train, kernel = "radial", ranges = list(cost = 10^seq(-2, 
    1, by = 0.25)))
summary(tune.out)

## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##     1
## 
## - best performance: 0.1725 
## 
## - Detailed performance results:
##           cost   error dispersion
## 1   0.01000000 0.39375 0.03240906
## 2   0.01778279 0.39375 0.03240906
## 3   0.03162278 0.34750 0.05552777
## 4   0.05623413 0.19250 0.03016160
## 5   0.10000000 0.19500 0.03782269
## 6   0.17782794 0.18000 0.04048319
## 7   0.31622777 0.17250 0.03809710
## 8   0.56234133 0.17500 0.04124790
## 9   1.00000000 0.17250 0.03162278
## 10  1.77827941 0.17750 0.03717451
## 11  3.16227766 0.18375 0.03438447
## 12  5.62341325 0.18500 0.03717451
## 13 10.00000000 0.18750 0.03173239

svm.radial <- svm(Purchase ~ ., kernel = "radial", data = OJ.train, cost = tune.out$best.parameter$cost)
summary(svm.radial)

## 
## Call:
## svm(formula = Purchase ~ ., data = OJ.train, kernel = "radial", cost = tune.out$best.parameter$cost)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  radial 
##        cost:  1 
## 
## Number of Support Vectors:  373
## 
##  ( 188 185 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM

train.pred <- predict(svm.radial, OJ.train)
table(OJ.train$Purchase, train.pred)

##     train.pred
##       CH  MM
##   CH 441  44
##   MM  77 238

(77 + 39) / (455 + 229 + 77 + 39)

## [1] 0.145

test.pred <- predict(svm.radial, OJ.test)
table(OJ.test$Purchase, test.pred)

##     test.pred
##       CH  MM
##   CH 151  17
##   MM  33  69

(28 + 18) / (141 + 83 + 28 + 18)

## [1] 0.1703704

Tuning does not reduce the error rates, they stay the same.

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

svm.poly <- svm(Purchase ~ ., kernel = "polynomial", data = OJ.train, degree = 2)
summary(svm.poly)

## 
## Call:
## svm(formula = Purchase ~ ., data = OJ.train, kernel = "polynomial", 
##     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

train.pred <- predict(svm.poly, OJ.train)
table(OJ.train$Purchase, train.pred)

##     train.pred
##       CH  MM
##   CH 449  36
##   MM 110 205

(105 + 33) / (461 + 201 + 105 + 33)

## [1] 0.1725

test.pred <- predict(svm.poly, OJ.test)
table(OJ.test$Purchase, test.pred)

##     test.pred
##       CH  MM
##   CH 153  15
##   MM  45  57

(41 + 10) / (149 + 70 + 41 + 10)

## [1] 0.1888889

Polynomial kernal creates 454 support vectors. 225 are level CH and 222 are level MM. The training error rate is 17.3% and the test error rate is 18.9%.

set.seed(2)
tune.out <- tune(svm, Purchase ~ ., data = OJ.train, kernel = "polynomial", degree = 2, ranges = list(cost = 10^seq(-2, 
    1, by = 0.25)))
summary(tune.out)

## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##      cost
##  3.162278
## 
## - best performance: 0.18 
## 
## - Detailed performance results:
##           cost   error dispersion
## 1   0.01000000 0.39000 0.03670453
## 2   0.01778279 0.37000 0.03395258
## 3   0.03162278 0.36375 0.03197764
## 4   0.05623413 0.34500 0.03291403
## 5   0.10000000 0.32125 0.03866254
## 6   0.17782794 0.24750 0.03322900
## 7   0.31622777 0.20250 0.04073969
## 8   0.56234133 0.20250 0.03670453
## 9   1.00000000 0.19625 0.03910900
## 10  1.77827941 0.19125 0.03586723
## 11  3.16227766 0.18000 0.04005205
## 12  5.62341325 0.18000 0.04133199
## 13 10.00000000 0.18125 0.03830162

svm.poly <- svm(Purchase ~ ., kernel = "polynomial", degree = 2, data = OJ.train, cost = tune.out$best.parameter$cost)
summary(svm.poly)

## 
## Call:
## svm(formula = Purchase ~ ., data = OJ.train, kernel = "polynomial", 
##     degree = 2, cost = tune.out$best.parameter$cost)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  polynomial 
##        cost:  3.162278 
##      degree:  2 
##      coef.0:  0 
## 
## Number of Support Vectors:  385
## 
##  ( 197 188 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM

train.pred <- predict(svm.poly, OJ.train)
table(OJ.train$Purchase, train.pred)

##     train.pred
##       CH  MM
##   CH 451  34
##   MM  90 225

(72 + 44) / (450 + 234 + 72 + 44)

## [1] 0.145

test.pred <- predict(svm.poly, OJ.test)
table(OJ.test$Purchase, test.pred)

##     test.pred
##       CH  MM
##   CH 154  14
##   MM  41  61

(31 + 19) / (140 + 80 + 31 + 19)

## [1] 0.1851852

Tuning decreases both the training and test error rates.

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

Radial kernel gives the minimum error on both train and test data.

Assignment 7

ikq392

2022-11-13

Problem 5.

Problem 7.

Problem 8.