Assignment 8

Exercise 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.

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*(x1^2-x22 > 0)

set.seed(100)
x1 = runif(500) - 0.5
x2 = runif(500) - 0.5
y = 1 * (x1^2 - x2^2 > 0)

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")
points(x1[y == 1], x2[y == 1], col = "blue", pch = 4)

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

data <- data.frame(x1 = x1, x2 = x2, y = as.factor(y))
lm_fit <- glm(y ~ ., data = data, family = 'binomial')

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.

lm_probs <- predict(lm_fit, newdata = data, type = 'response')
lm_preds <- ifelse(lm_probs > 0.5, 1, 0)
plot(data$x1, data$x2, col = lm_preds + 2)

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_nl = 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

summary(lm_fit_nl)

## 
## Call:
## glm(formula = y ~ poly(x1, 2) + poly(x2, 2) + I(x1 * x2), family = binomial, 
##     data = data)
## 
## Deviance Residuals: 
##        Min          1Q      Median          3Q         Max  
## -1.564e-03  -2.000e-08  -2.000e-08   2.000e-08   1.502e-03  
## 
## Coefficients:
##                Estimate Std. Error z value Pr(>|z|)
## (Intercept)     -222.21   13258.36  -0.017    0.987
## poly(x1, 2)1    3902.33  364800.33   0.011    0.991
## poly(x1, 2)2   33807.47  937118.99   0.036    0.971
## poly(x2, 2)1    -656.84  387630.34  -0.002    0.999
## poly(x2, 2)2  -35213.02 1032138.51  -0.034    0.973
## I(x1 * x2)        64.72  130327.27   0.000    1.000
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 6.9276e+02  on 499  degrees of freedom
## Residual deviance: 5.1269e-06  on 494  degrees of freedom
## AIC: 12
## 
## Number of Fisher Scoring iterations: 25

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_probs = predict(lm_fit, data, type = "response")
lm_preds = ifelse(lm_probs > 0.5, 1, 0)
data_pos = data[lm_preds == 1, ]
data_neg = data[lm_preds == 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)

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(y ~ ., data = data, kernel = 'linear', cost = 0.01)
plot(svm_fit, data)

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_preds = predict(svm_fit, data)
data_pos = data[svm_preds == 1, ]
data_neg = data[svm_preds == 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)

Comment on your results.

Using SVMs are best to use for non-linear models.

Exercise 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.

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)
gas_med = median(Auto$mpg)
new_gas_var = ifelse(Auto$mpg > gas_med, 1, 0)
Auto$mpglevel = as.factor(new_gas_var)

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(100)
horl_gas = tune(svm, mpglevel ~ ., data = Auto, kernel = "linear", ranges = list(cost = c(0.01, 
    0.1, 1, 5, 10, 100)))
summary(horl_gas)

## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##     1
## 
## - best performance: 0.01512821 
## 
## - Detailed performance results:
##    cost      error dispersion
## 1 1e-02 0.07391026 0.04398186
## 2 1e-01 0.05102564 0.03408666
## 3 1e+00 0.01512821 0.02421271
## 4 5e+00 0.01775641 0.01700310
## 5 1e+01 0.02538462 0.02372507
## 6 1e+02 0.03564103 0.02125655

The best performance is 0.01512821 for the best cost of 1.

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(100)
horl_gas2 = tune(svm, mpglevel ~ ., data = Auto, kernel = "polynomial", ranges = list(cost = c(0.1, 
    1, 5, 10), degree = c(2, 3, 4)))
summary(horl_gas2)

## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost degree
##   0.1      3
## 
## - best performance: 0.5435897 
## 
## - Detailed performance results:
##    cost degree     error dispersion
## 1   0.1      2 0.5485897 0.06532230
## 2   1.0      2 0.5485897 0.06532230
## 3   5.0      2 0.5485897 0.06532230
## 4  10.0      2 0.5539744 0.07981094
## 5   0.1      3 0.5435897 0.07674778
## 6   1.0      3 0.5435897 0.07674778
## 7   5.0      3 0.5435897 0.07674778
## 8  10.0      3 0.5435897 0.07674778
## 9   0.1      4 0.5485897 0.06532230
## 10  1.0      4 0.5485897 0.06532230
## 11  5.0      4 0.5485897 0.06532230
## 12 10.0      4 0.5485897 0.06532230

The best performance is 0.5435897 for the best cost of 1 with a degree of 3.

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 = 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)

Exercise 8

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

set.seed(100)
train = sample(dim(OJ)[1], 800)
oj_train = OJ[train, ]
oj_test = OJ[-train, ]

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 ~ ., kernel = "linear", data = oj_train, 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:  432
## 
##  ( 216 216 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM

What are the training and test error rates?

train_preds = predict(svm_linear, oj_train)
table(oj_train$Purchase, train_preds)

##     train_preds
##       CH  MM
##   CH 433  55
##   MM  78 234

test_preds = predict(svm_linear, oj_test)
table(oj_test$Purchase, test_preds)

##     test_preds
##       CH  MM
##   CH 147  18
##   MM  26  79

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

set.seed(100)
tune_out_op_cost = tune(svm, Purchase ~ ., data = oj_train, kernel = "linear", ranges = list(cost = 10^seq(-2, 
    1, by = 0.25)))
summary(tune_out_op_cost)

## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##        cost
##  0.05623413
## 
## - best performance: 0.16625 
## 
## - Detailed performance results:
##           cost   error dispersion
## 1   0.01000000 0.17500 0.04639804
## 2   0.01778279 0.17375 0.03972562
## 3   0.03162278 0.16875 0.04299952
## 4   0.05623413 0.16625 0.04411554
## 5   0.10000000 0.16875 0.03875224
## 6   0.17782794 0.16875 0.04007372
## 7   0.31622777 0.17375 0.03928617
## 8   0.56234133 0.17500 0.03908680
## 9   1.00000000 0.17375 0.04101575
## 10  1.77827941 0.17125 0.03230175
## 11  3.16227766 0.17375 0.03251602
## 12  5.62341325 0.17000 0.04005205
## 13 10.00000000 0.17000 0.03782269

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_op_cost$best.parameters$cost)
train_preds = predict(svm_linear, oj_train)
table(oj_train$Purchase, train_preds)

##     train_preds
##       CH  MM
##   CH 427  61
##   MM  67 245

test_preds = predict(svm_linear, oj_test)
table(oj_test$Purchase, test_preds)

##     test_preds
##       CH  MM
##   CH 141  24
##   MM  25  80

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

set.seed(100)
svm_radial = svm(Purchase ~ ., data = oj_train, kernel = "radial")
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:  368
## 
##  ( 187 181 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM

train_preds = predict(svm_radial, oj_train)
table(oj_train$Purchase, train_preds)

##     train_preds
##       CH  MM
##   CH 448  40
##   MM  69 243

test_preds = predict(svm_radial, oj_test)
table(oj_test$Purchase, test_preds)

##     test_preds
##       CH  MM
##   CH 147  18
##   MM  32  73

set.seed(755)
tune_out_rad = tune(svm, Purchase ~ ., data = oj_train, kernel = "radial", ranges = list(cost = 10^seq(-2, 
    1, by = 0.25)))
summary(tune_out_rad)

## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##      cost
##  1.778279
## 
## - best performance: 0.15875 
## 
## - Detailed performance results:
##           cost   error dispersion
## 1   0.01000000 0.39000 0.06529846
## 2   0.01778279 0.39000 0.06529846
## 3   0.03162278 0.34000 0.08203150
## 4   0.05623413 0.19375 0.05535554
## 5   0.10000000 0.18250 0.06015027
## 6   0.17782794 0.17125 0.06010696
## 7   0.31622777 0.16750 0.05109903
## 8   0.56234133 0.16625 0.04715886
## 9   1.00000000 0.16000 0.04241004
## 10  1.77827941 0.15875 0.04084609
## 11  3.16227766 0.17000 0.04377975
## 12  5.62341325 0.17000 0.04090979
## 13 10.00000000 0.17375 0.04143687

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

Overall, the models are very similar, but the radial kernel is a slightly better than the others.

Assignment 8

Alex De Leon

8/13/2021

Exercise 5

Exercise 7

Exercise 8