Assignment 8

Chapter 09 Chapter 09 (page 398): 5, 7, 8

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.

1. 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:

set.seed(421)
x1 = runif(500) - 0.5 
x2 = runif(500) - 0.5
y=1*(x1^2-x2^2>0)

2. 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 = "red", xlab = "X1", ylab = "X2", pch = "+")
points(x1[y == 1], x2[y == 1], col = "blue", pch = 4)

3. 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)
## 
## 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

4. 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)

5. 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).

lm.fit = glm(y ~ poly(x1, 2) + poly(x2, 2) + I(x1 * x2), data = data, family = binomial)

## Warning: glm.fit: algorithm did not converge

## Warning: glm.fit: fitted probabilities numerically 0 or 1 occurred

6. 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)

7. 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)

## Warning: package 'e1071' was built under R version 4.3.3

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)

8. 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)

1. Comment on your results.

The experiment perfromed covers the idea of SVMS are important to use for finding non linear models using cross validation would be easier with the parameter of gamma.

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.

1. 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)
gas.med = median(Auto$mpg)
new.var = ifelse(Auto$mpg > gas.med, 1, 0)
Auto$mpglevel = as.factor(new.var)

2. 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.

library(e1071)
set.seed(3255)
tune.out = tune(svm, mpglevel ~ ., data = Auto, kernel = "linear", ranges = list(cost = c(0.01, 
    0.1, 1, 5, 10, 100)))
summary(tune.out)

## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##     1
## 
## - best performance: 0.01269231 
## 
## - Detailed performance results:
##    cost      error dispersion
## 1 1e-02 0.07397436 0.06863413
## 2 1e-01 0.05102564 0.06923024
## 3 1e+00 0.01269231 0.02154160
## 4 5e+00 0.01519231 0.01760469
## 5 1e+01 0.02025641 0.02303772
## 6 1e+02 0.03294872 0.02898463

3. 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(21)
tune.out = tune(svm, mpglevel ~ ., data = Auto, kernel = "polynomial", ranges = list(cost = c(0.1, 
    1, 5, 10), degree = c(2, 3, 4)))
summary(tune.out)

## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost degree
##    10      2
## 
## - best performance: 0.5435897 
## 
## - Detailed performance results:
##    cost degree     error dispersion
## 1   0.1      2 0.5587821 0.04538579
## 2   1.0      2 0.5587821 0.04538579
## 3   5.0      2 0.5587821 0.04538579
## 4  10.0      2 0.5435897 0.05611162
## 5   0.1      3 0.5587821 0.04538579
## 6   1.0      3 0.5587821 0.04538579
## 7   5.0      3 0.5587821 0.04538579
## 8  10.0      3 0.5587821 0.04538579
## 9   0.1      4 0.5587821 0.04538579
## 10  1.0      4 0.5587821 0.04538579
## 11  5.0      4 0.5587821 0.04538579
## 12 10.0      4 0.5587821 0.04538579

set.seed(463)
tune.out = tune(svm, mpglevel ~ ., data = Auto, kernel = "radial", ranges = list(cost = c(0.1, 
    1, 5, 10), gamma = c(0.01, 0.1, 1, 5, 10, 100)))
summary(tune.out)

## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost gamma
##    10  0.01
## 
## - best performance: 0.02551282 
## 
## - Detailed performance results:
##    cost gamma      error dispersion
## 1   0.1 1e-02 0.09429487 0.04814900
## 2   1.0 1e-02 0.07897436 0.03875105
## 3   5.0 1e-02 0.05352564 0.02532795
## 4  10.0 1e-02 0.02551282 0.02417610
## 5   0.1 1e-01 0.07891026 0.03847631
## 6   1.0 1e-01 0.05602564 0.02881876
## 7   5.0 1e-01 0.03826923 0.03252085
## 8  10.0 1e-01 0.03320513 0.02964746
## 9   0.1 1e+00 0.57660256 0.05479863
## 10  1.0 1e+00 0.06628205 0.02996211
## 11  5.0 1e+00 0.06115385 0.02733573
## 12 10.0 1e+00 0.06115385 0.02733573
## 13  0.1 5e+00 0.57660256 0.05479863
## 14  1.0 5e+00 0.51538462 0.06642516
## 15  5.0 5e+00 0.50775641 0.07152757
## 16 10.0 5e+00 0.50775641 0.07152757
## 17  0.1 1e+01 0.57660256 0.05479863
## 18  1.0 1e+01 0.53833333 0.05640443
## 19  5.0 1e+01 0.53070513 0.05708644
## 20 10.0 1e+01 0.53070513 0.05708644
## 21  0.1 1e+02 0.57660256 0.05479863
## 22  1.0 1e+02 0.57660256 0.05479863
## 23  5.0 1e+02 0.57660256 0.05479863
## 24 10.0 1e+02 0.57660256 0.05479863

4. 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.

svm.linear = svm(mpglevel ~ ., data = Auto, kernel = "linear", cost = 1)
svm.poly = svm(mpglevel ~ ., data = Auto, kernel = "polynomial", cost = 10, 
    degree = 2)
svm.radial = svm(mpglevel ~ ., data = Auto, kernel = "radial", cost = 10, 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 = "")))
    }
}
plotpairs(svm.linear)

plotpairs(svm.poly)

plotpairs(svm.radial)

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

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

set.seed(1)
train = sample(1:nrow(OJ), 800)
OJtrain = OJ[train, ]
OJtest = OJ[-train, ]

2. 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.

svmlinOJ = svm(Purchase ~ ., data = OJtrain, kernel = "linear", cost = 0.01)
summary(svmlinOJ)

## 
## Call:
## svm(formula = Purchase ~ ., data = OJtrain, 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

The svm created 435 support vectors, with two classes (CH and MM). 219 are CH and 216 are MM.

3. What are the training and test error rates?

train.pred = predict(svmlinOJ, OJtrain)
table(OJtrain$Purchase, train.pred)

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

1 - (420 + 240) / (420 + 240 + 65 + 75)

## [1] 0.175

test.pred = predict(svmlinOJ, OJtest)
table(OJtest$Purchase, test.pred)

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

1 - (153 + 69) / (153 + 69 + 15 + 33)

## [1] 0.1777778

The Train error is 17.50%, and the Test error is 17.78%

4. Use the tune() function to select an optimal cost. Consider values in the range 0.01 to 10. The lowest error is with a cost of 10

set.seed(1)
tunesvm = tune(svm, Purchase ~ ., data = OJtrain, kernel = "linear", ranges = list(cost = c(0.01, 1, 10)))
summary(tunesvm)

## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##    10
## 
## - best performance: 0.17375 
## 
## - Detailed performance results:
##    cost   error dispersion
## 1  0.01 0.17625 0.02853482
## 2  1.00 0.17500 0.02946278
## 3 10.00 0.17375 0.03197764

5. Compute the training and test error rates using this new value for cost.

svm10OJ = svm(Purchase ~ ., data = OJtrain, kernel = "linear", cost = 10)
train10pred = predict(svm10OJ, OJtrain)
table(OJtrain$Purchase, train10pred)

##     train10pred
##       CH  MM
##   CH 423  62
##   MM  69 246

1 - (423 + 246) / 800

## [1] 0.16375

svm10OJ = svm(Purchase ~ ., data = OJtest, kernel = "linear", cost = 10)
test10pred = predict(svm10OJ, OJtest)
table(OJtest$Purchase, test10pred)

##     test10pred
##       CH  MM
##   CH 154  14
##   MM  31  71

1 - (154 + 71) / 270

## [1] 0.1666667

The training error for the SVM with a cost of 10 is 16.375%. The test error for the SVM with a cost of 10 is 16.667%

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

svmradOJ = svm(Purchase ~ ., data = OJtrain, kernel = "radial")
trainpred = predict(svmradOJ, OJtrain)
table(OJtrain$Purchase, trainpred)

##     trainpred
##       CH  MM
##   CH 441  44
##   MM  77 238

1 - (441 + 238) / 800

## [1] 0.15125

testpred = predict(svmradOJ, OJtest)
table(OJtest$Purchase, testpred)

##     testpred
##       CH  MM
##   CH 151  17
##   MM  33  69

1 - (151 + 69) / 270

## [1] 0.1851852

set.seed(1)
tunesvm = tune(svm, Purchase ~ ., data = OJtrain, kernal = "radial", ranges = list(cost = c(0.01, 1, 10)))
summary(tunesvm)

## 
## 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  1.00 0.17125 0.02128673
## 3 10.00 0.18625 0.02853482

svmradOJ = svm(Purchase ~ ., data = OJtrain, kernel = "radial", cost = 1)
trainpred = predict(svmradOJ, OJtrain)
table(OJtrain$Purchase, trainpred)

##     trainpred
##       CH  MM
##   CH 441  44
##   MM  77 238

1 - (441 + 238) / 800

## [1] 0.15125

testpred = predict(svmradOJ, OJtest)
table(OJtest$Purchase, testpred)

##     testpred
##       CH  MM
##   CH 151  17
##   MM  33  69

1 - (151 + 69) / 270

## [1] 0.1851852

Train error for default radial svm is 15.125%. Test error for default radial svm is 18.519%

Tuned training error for radial svm is 15.125%. Tuned test error for radial svm is 18.519%

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

svmpolyOJ = svm(Purchase ~ ., data = OJtrain, kernel = "polynomial", degree = 2)
trainpred = predict(svmpolyOJ, OJtrain)
table(OJtrain$Purchase, trainpred)

##     trainpred
##       CH  MM
##   CH 449  36
##   MM 110 205

1 - (449 + 205) / 800

## [1] 0.1825

testpred = predict(svmpolyOJ, OJtest)
table(OJtest$Purchase, testpred)

##     testpred
##       CH  MM
##   CH 153  15
##   MM  45  57

1 - (154 + 65) / 270

## [1] 0.1888889

set.seed(1)
tunesvm = tune(svm, Purchase ~ ., data = OJtrain, kernel = "polynomial", degree = 2, ranges = list(cost = c(0.01, 1, 10)))
summary(tunesvm)

## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##    10
## 
## - best performance: 0.18125 
## 
## - Detailed performance results:
##    cost   error dispersion
## 1  0.01 0.39125 0.04210189
## 2  1.00 0.20250 0.04116363
## 3 10.00 0.18125 0.02779513

svmpolyOJ = svm(Purchase ~ ., data = OJtrain, kernel = "polynomial", degree = 2, cost = 10)
trainpred = predict(svmpolyOJ, OJtrain)
table(OJtrain$Purchase, trainpred)

##     trainpred
##       CH  MM
##   CH 447  38
##   MM  82 233

1 - (447 + 233) / 800

## [1] 0.15

Train error for tuned polynomial ia 15.00%

testpred = predict(svmpolyOJ, OJtest)
table(OJtest$Purchase, testpred)

##     testpred
##       CH  MM
##   CH 154  14
##   MM  37  65

1 - (154 + 65) / 270

## [1] 0.1888889

Test error for tuned polynomial 18.89%

8. Overall, which approach seems to give the best results on this data? The linear tuned model with a cost of 10 had the lowest test error with 16.667%. The radial and polynomial approach both had similar error rates of close to 19%