Assignment 8

Problems 5, 7 and 8 from Chapter 9 of ISLR2

library(ISLR2)
library(tidyverse)

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library(caret)

## Loading required package: lattice
## 
## Attaching package: 'caret'
## 
## The following object is masked from 'package:purrr':
## 
##     lift

library(e1071)

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:

set.seed(23)
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.

data.frame(x1, x2) %>% 
  ggplot(aes(x1, x2)) +
  geom_point(aes(colour = as.factor(y))) +
  theme_minimal() +
  theme(legend.position = "none")

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

logit_model = glm(y ~ x1 + x2, family = "binomial")
logit_model

## 
## Call:  glm(formula = y ~ x1 + x2, family = "binomial")
## 
## Coefficients:
## (Intercept)           x1           x2  
##    -0.05418     -0.52477      0.02243  
## 
## Degrees of Freedom: 499 Total (i.e. Null);  497 Residual
## Null Deviance:       692.8 
## Residual Deviance: 689.9     AIC: 695.9

(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.

df = data.frame(x1, x2, y = as.factor(y))
df$preds = ifelse(predict(logit_model, type = "response") > 0.5, 1, 0)

df %>% 
  ggplot(aes(x1, x2)) +
  geom_point(aes(colour = as.factor(preds))) +
  theme_minimal() +
  theme(legend.position = "none")

(e) Now fit a logistic regression model to the data using non-linear functions of X1 and X2 as predictors (e.g. X12, X1 ×X2, log(X2), and so forth).

logit_model2 = glm(y ~ x1 + x2 + I(x1^2) + I(x2^2), family = "binomial")

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

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

logit_model2

## 
## Call:  glm(formula = y ~ x1 + x2 + I(x1^2) + I(x2^2), family = "binomial")
## 
## Coefficients:
## (Intercept)           x1           x2      I(x1^2)      I(x2^2)  
##       8.072      120.777     -121.585    35721.515   -35675.872  
## 
## Degrees of Freedom: 499 Total (i.e. Null);  495 Residual
## Null Deviance:       692.8 
## Residual Deviance: 1.303e-05     AIC: 10

(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.

df$preds2 = ifelse(predict(logit_model2, type = "response") > 0.5, 1, 0)

df %>% 
  ggplot(aes(x1, x2)) +
  geom_point(aes(colour = as.factor(preds2))) +
  theme_minimal() +
  theme(legend.position = "none")

(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.

svm_model = svm(y ~ x1 + x2, data = df, kernel = "linear", scale = F)
summary(svm_model)

## 
## Call:
## svm(formula = y ~ x1 + x2, data = df, kernel = "linear", scale = F)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  linear 
##        cost:  1 
## 
## Number of Support Vectors:  481
## 
##  ( 240 241 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  0 1

set.seed(23)
tune_svm = tune(svm, y ~ x1 + x2, data = df, kernal = "linear", 
                ranges = list(cost = c(10, 100, 500, 1000)))
summary(tune_svm)

## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##   500
## 
## - best performance: 0.02 
## 
## - Detailed performance results:
##   cost error dispersion
## 1   10 0.024 0.02458545
## 2  100 0.022 0.02201010
## 3  500 0.020 0.02108185
## 4 1000 0.028 0.02699794

best_model = tune_svm$best.model
df$preds3 = predict(best_model, df[ ,1:3])

df %>% 
  ggplot(aes(x1, x2)) +
  geom_point(aes(colour = as.factor(preds3))) +
  theme_minimal() +
  theme(legend.position = "none")

(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_model2 = svm(y ~ x1 + x2, data = df, kernel = "radial", gamma = 1)
summary(svm_model2)

## 
## Call:
## svm(formula = y ~ x1 + x2, data = df, kernel = "radial", gamma = 1)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  radial 
##        cost:  1 
## 
## Number of Support Vectors:  161
## 
##  ( 79 82 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  0 1

set.seed(23)
tune_svm2 = tune(svm, y ~ x1 + x2, data = df, kernal = "radial", 
                ranges = list(cost = c(10, 100, 500, 1000), gamma = c(0.1, 0.5, 1, 2, 3)))
summary(tune_svm2)

## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost gamma
##  1000   0.1
## 
## - best performance: 0.018 
## 
## - Detailed performance results:
##    cost gamma error dispersion
## 1    10   0.1 0.034 0.02674987
## 2   100   0.1 0.026 0.02503331
## 3   500   0.1 0.020 0.02309401
## 4  1000   0.1 0.018 0.02201010
## 5    10   0.5 0.024 0.02458545
## 6   100   0.5 0.022 0.02201010
## 7   500   0.5 0.020 0.02108185
## 8  1000   0.5 0.028 0.02699794
## 9    10   1.0 0.028 0.03155243
## 10  100   1.0 0.028 0.02347576
## 11  500   1.0 0.028 0.02149935
## 12 1000   1.0 0.032 0.02529822
## 13   10   2.0 0.030 0.03299832
## 14  100   2.0 0.026 0.02319004
## 15  500   2.0 0.034 0.03405877
## 16 1000   2.0 0.030 0.03018462
## 17   10   3.0 0.034 0.03272783
## 18  100   3.0 0.038 0.03326660
## 19  500   3.0 0.030 0.03162278
## 20 1000   3.0 0.030 0.03299832

best_model2 = tune_svm2$best.model
df$preds4 = predict(best_model, df[ ,1:3])

df %>% 
  ggplot(aes(x1, x2)) +
  geom_point(aes(colour = as.factor(preds4))) +
  theme_minimal() +
  theme(legend.position = "none")

(i) Comment on your results.

The main take away is that the SVM impressively modeled the non-linear relationship with very few parameters. This is going to be a lot more efficient that manually trying to calculate the relationship in a logit model.

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.

auto = Auto

(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.

auto$mpg_class = factor(as.numeric(auto$mpg > median(auto$mpg)))

(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.

svm_model = svm(mpg_class ~ . -mpg, data = auto, kernel = "linear", scale = F)

set.seed(23)
tune_svm = tune(svm, mpg_class ~ . -mpg, data = auto, kernal = "linear", 
                ranges = list(cost = c(0.001, 0.01, 0.1, 1, 10, 100, 1000)))
summary(tune_svm)

## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##   100
## 
## - best performance: 0.08692308 
## 
## - Detailed performance results:
##    cost      error dispersion
## 1 1e-03 0.56903846 0.05422978
## 2 1e-02 0.56903846 0.05422978
## 3 1e-01 0.15596154 0.07332520
## 4 1e+00 0.09205128 0.04410227
## 5 1e+01 0.09205128 0.04729922
## 6 1e+02 0.08692308 0.05148748
## 7 1e+03 0.10455128 0.05021545

df = data.frame(tune_svm$performances$cost, tune_svm$performances$error) 
df %>% 
  ggplot(aes(tune_svm.performances.cost, tune_svm.performances.error)) +
  geom_point()

(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(23)
tune_svm2 = tune(svm, mpg_class ~ . -mpg, data = auto, kernal = "radial", 
                ranges = list(cost = c(0.1, 1, 10, 100, 500, 1000), gamma = c(0.1, 0.5, 1, 2, 3)))
summary(tune_svm2)

## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost gamma
##    10   0.1
## 
## - best performance: 0.07647436 
## 
## - Detailed performance results:
##     cost gamma      error dispersion
## 1  1e-01   0.1 0.08948718 0.03889147
## 2  1e+00   0.1 0.08692308 0.04569673
## 3  1e+01   0.1 0.07647436 0.03785376
## 4  1e+02   0.1 0.09705128 0.02668837
## 5  5e+02   0.1 0.09974359 0.04923644
## 6  1e+03   0.1 0.09974359 0.04923644
## 7  1e-01   0.5 0.08698718 0.04068822
## 8  1e+00   0.5 0.08435897 0.04520148
## 9  1e+01   0.5 0.08423077 0.03606392
## 10 1e+02   0.5 0.08435897 0.04032997
## 11 5e+02   0.5 0.08435897 0.04032997
## 12 1e+03   0.5 0.08435897 0.04032997
## 13 1e-01   1.0 0.56903846 0.05422978
## 14 1e+00   1.0 0.07916667 0.03506781
## 15 1e+01   1.0 0.08423077 0.03429497
## 16 1e+02   1.0 0.08423077 0.03429497
## 17 5e+02   1.0 0.08423077 0.03429497
## 18 1e+03   1.0 0.08423077 0.03429497
## 19 1e-01   2.0 0.56903846 0.05422978
## 20 1e+00   2.0 0.11487179 0.06033198
## 21 1e+01   2.0 0.11480769 0.05532699
## 22 1e+02   2.0 0.11480769 0.05532699
## 23 5e+02   2.0 0.11480769 0.05532699
## 24 1e+03   2.0 0.11480769 0.05532699
## 25 1e-01   3.0 0.56903846 0.05422978
## 26 1e+00   3.0 0.40365385 0.15419418
## 27 1e+01   3.0 0.38596154 0.15519190
## 28 1e+02   3.0 0.38596154 0.15519190
## 29 5e+02   3.0 0.38596154 0.15519190
## 30 1e+03   3.0 0.38596154 0.15519190

set.seed(23)
tune_svm3 = tune(svm, mpg_class ~ . -mpg, data = auto, kernal = "polynomial", 
                ranges = list(cost = c(0.1, 1, 10, 100, 500, 1000), gamma = c(0.1, 0.5, 1, 2, 3)))
summary(tune_svm3)

## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost gamma
##    10   0.1
## 
## - best performance: 0.07647436 
## 
## - Detailed performance results:
##     cost gamma      error dispersion
## 1  1e-01   0.1 0.08948718 0.03889147
## 2  1e+00   0.1 0.08692308 0.04569673
## 3  1e+01   0.1 0.07647436 0.03785376
## 4  1e+02   0.1 0.09705128 0.02668837
## 5  5e+02   0.1 0.09974359 0.04923644
## 6  1e+03   0.1 0.09974359 0.04923644
## 7  1e-01   0.5 0.08698718 0.04068822
## 8  1e+00   0.5 0.08435897 0.04520148
## 9  1e+01   0.5 0.08423077 0.03606392
## 10 1e+02   0.5 0.08435897 0.04032997
## 11 5e+02   0.5 0.08435897 0.04032997
## 12 1e+03   0.5 0.08435897 0.04032997
## 13 1e-01   1.0 0.56903846 0.05422978
## 14 1e+00   1.0 0.07916667 0.03506781
## 15 1e+01   1.0 0.08423077 0.03429497
## 16 1e+02   1.0 0.08423077 0.03429497
## 17 5e+02   1.0 0.08423077 0.03429497
## 18 1e+03   1.0 0.08423077 0.03429497
## 19 1e-01   2.0 0.56903846 0.05422978
## 20 1e+00   2.0 0.11487179 0.06033198
## 21 1e+01   2.0 0.11480769 0.05532699
## 22 1e+02   2.0 0.11480769 0.05532699
## 23 5e+02   2.0 0.11480769 0.05532699
## 24 1e+03   2.0 0.11480769 0.05532699
## 25 1e-01   3.0 0.56903846 0.05422978
## 26 1e+00   3.0 0.40365385 0.15419418
## 27 1e+01   3.0 0.38596154 0.15519190
## 28 1e+02   3.0 0.38596154 0.15519190
## 29 5e+02   3.0 0.38596154 0.15519190
## 30 1e+03   3.0 0.38596154 0.15519190

The radial and polynomial outperformed the linear kernal.

(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.

auto = auto %>% 
  dplyr::select(-mpg)

svm_fit = svm(mpg_class ~ ., data = auto, kernal = "radial", scale = T, cost = 10, gamma = 0.1)

plot(svm_fit, auto, weight ~ horsepower)

Mode <- function(x) {
  ux <- unique(x)
  ux[which.max(tabulate(match(x, ux)))]
}

plot(svm_fit, 
     auto, 
     weight ~ horsepower, 
     slice = list(cylinders = median(auto$cylinders), 
                  displacement = median(auto$displacement),
                  acceleration = median(auto$acceleration),
                  year = median(auto$year), 
                  origin = Mode(auto$origin), 
                  name = Mode(auto$name)))

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.

oj = OJ

set.seed(23)

train_index = sample(1:nrow(oj), 800)
train = oj[train_index, ]
test = oj[-train_index, ]

(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_fit = svm(Purchase ~ ., data = train, kernel = "linear", scale = T, cost = 0.01)
summary(svm_fit)

## 
## Call:
## svm(formula = Purchase ~ ., data = train, kernel = "linear", cost = 0.01, 
##     scale = T)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  linear 
##        cost:  0.01 
## 
## Number of Support Vectors:  441
## 
##  ( 220 221 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM

There are two classes CH and MM. This is a large number of support vectors compared to the 800 training size.

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

data.frame(
  training_error = mean(predict(svm_fit, train) != train$Purchase),
  testing_error = mean(predict(svm_fit, test) != test$Purchase))

##   training_error testing_error
## 1          0.165      0.162963

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

set.seed(23)
tune_svm = tune(svm, Purchase ~ ., data = train, kernel = "linear", 
                ranges = list(cost = c(0.01, 0.1, 1, 10)))
summary(tune_svm)

## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##    10
## 
## - best performance: 0.16625 
## 
## - Detailed performance results:
##    cost   error dispersion
## 1  0.01 0.17125 0.05434266
## 2  0.10 0.17250 0.04779877
## 3  1.00 0.17000 0.04901814
## 4 10.00 0.16625 0.06039511

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

svm_fit = svm(Purchase ~ ., data = train, kernel = "linear", scale = T, cost = 10)

linear_df = data.frame(
  training_error = mean(predict(svm_fit, train) != train$Purchase),
  testing_error = mean(predict(svm_fit, test) != test$Purchase))
linear_df

##   training_error testing_error
## 1        0.16125     0.1703704

Much lower error.

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

svm_fit2 = svm(Purchase ~ ., data = train, kernel = 'radial', decision.values = T, cost = 0.01)
summary(svm_fit2)

## 
## Call:
## svm(formula = Purchase ~ ., data = train, kernel = "radial", decision.values = T, 
##     cost = 0.01)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  radial 
##        cost:  0.01 
## 
## Number of Support Vectors:  616
## 
##  ( 306 310 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM

data.frame(
  training_error = mean(predict(svm_fit2, train) != train$Purchase),
  testing_error = mean(predict(svm_fit2, test) != test$Purchase))

##   training_error testing_error
## 1         0.3825     0.4111111

set.seed(23)
tune_svm = tune(svm, Purchase ~ ., data = train, kernel = "radial", decision.values = T,
                ranges = list(cost = c(0.01, 0.1, 1, 10)))
summary(tune_svm)

## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##     1
## 
## - best performance: 0.165 
## 
## - Detailed performance results:
##    cost   error dispersion
## 1  0.01 0.38250 0.05277047
## 2  0.10 0.19500 0.06072479
## 3  1.00 0.16500 0.04401704
## 4 10.00 0.18375 0.05036326

svm_fit = svm(Purchase ~ ., data = train, kernal = "radial", cost = 1, decision.values = T)

radial_df = data.frame(
  training_error = mean(predict(svm_fit, train) != train$Purchase),
  testing_error = mean(predict(svm_fit, test) != test$Purchase))
radial_df

##   training_error testing_error
## 1        0.14375      0.162963

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

svm_fit = svm(Purchase ~ ., data = train, kernel = "polynomial", cost = 0.01, degree = 2)
summary(svm_fit)

## 
## Call:
## svm(formula = Purchase ~ ., data = train, kernel = "polynomial", 
##     cost = 0.01, degree = 2)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  polynomial 
##        cost:  0.01 
##      degree:  2 
##      coef.0:  0 
## 
## Number of Support Vectors:  618
## 
##  ( 306 312 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM

data.frame(
  training_error = mean(predict(svm_fit, train) != train$Purchase),
  testing_error = mean(predict(svm_fit, test) != test$Purchase))

##   training_error testing_error
## 1         0.3825     0.4111111

set.seed(23)
tune_svm = tune(svm, Purchase ~ ., data = train, kernel = "polynomial", degree = 2,
                ranges = list(cost = c(0.01, 0.1, 1, 10)))
summary(tune_svm)

## 
## 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 0.38250 0.05277047
## 2  0.10 0.31625 0.07289881
## 3  1.00 0.19375 0.05179085
## 4 10.00 0.18125 0.05179085

svm_fit = svm(Purchase ~ ., data = train, kernel = "polynomial", degree = 2, cost = 10)

poly_df = data.frame(
  training_error = mean(predict(svm_fit, train) != train$Purchase),
  testing_error = mean(predict(svm_fit, test) != test$Purchase))
poly_df

##   training_error testing_error
## 1        0.14375     0.1666667

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

res = cbind("Kernel" = c("Poly", "Radial", "Linerar"), rbind(poly_df, radial_df, linear_df))
res

##    Kernel training_error testing_error
## 1    Poly        0.14375     0.1666667
## 2  Radial        0.14375     0.1629630
## 3 Linerar        0.16125     0.1703704

The radial model seemed to do the best but they all had very similar results.