Assignment_8
Karthik Munnam
2024-04-26
##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 * (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 = "green", 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.
logr_model = glm(y ~ x1 + x2, family = binomial)
summary(logr_model)##
## Call:
## glm(formula = y ~ x1 + x2, family = binomial)
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 0.03729 0.08970 0.416 0.678
## x1 0.36158 0.31741 1.139 0.255
## x2 0.15267 0.31217 0.489 0.625
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 692.95 on 499 degrees of freedom
## Residual deviance: 691.40 on 497 degrees of freedom
## AIC: 697.4
##
## 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.
logr_model_data = data.frame(x1 = x1, x2 = x2, y = y)
logr_model_data_pred = predict(logr_model, logr_model_data, type = "response")
logr_model_data_pred_prob = ifelse(logr_model_data_pred > 0.5, 1, 0)
logr_model_data_pred_prob_pos = logr_model_data[logr_model_data_pred_prob == 1, ]
logr_model_data_pred_prob_neg = logr_model_data[logr_model_data_pred_prob == 0, ]
plot(logr_model_data_pred_prob_pos$x1, logr_model_data_pred_prob_pos$x2, col = "green", xlab = "X1", ylab = "X2", pch = "+")
points(logr_model_data_pred_prob_neg$x1, logr_model_data_pred_prob_neg$x2, col = "blue", pch = 4)
## (e) 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).
log_reg_NLR = glm(y ~ poly(x1, 2) + poly(x2, 2) + I(x1 * x2), data = logr_model_data, family = binomial)## Warning: glm.fit: algorithm did not converge## Warning: glm.fit: fitted probabilities numerically 0 or 1 occurredf. 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.
log_reg_NLR_pred = predict(log_reg_NLR, logr_model_data, type = "response")
log_reg_NLR_pred_prob = ifelse(log_reg_NLR_pred > 0.5, 1, 0)
log_reg_NLR_pred_prob_pos = logr_model_data[log_reg_NLR_pred_prob == 1, ]
log_reg_NLR_pred_prob_neg = logr_model_data[log_reg_NLR_pred_prob == 0, ]
plot(log_reg_NLR_pred_prob_pos$x1, log_reg_NLR_pred_prob_pos$x2, col = "red", xlab = "X1", ylab = "X2", pch = "+")
points(log_reg_NLR_pred_prob_neg$x1, log_reg_NLR_pred_prob_neg$x2, col = "blue", 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, logr_model_data, kernel = "linear",cost=1)
svm_pred = predict(svm_fit, logr_model_data)
data.pos = logr_model_data[svm_pred == 1, ]
data.neg = logr_model_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.
library(e1071)
svm_radial= svm(as.factor(y) ~ x1 + x2, logr_model_data, kernel = "radial", gamma=1)
svm_pred_radial = predict(svm_radial, logr_model_data)
svm_pred_pos_radial = logr_model_data[svm_pred_radial == 1, ]
svm_pred_neg_radial = logr_model_data[svm_pred_radial == 0, ]
plot(svm_pred_pos_radial$x1, svm_pred_pos_radial$x2, col = "red", xlab = "X1", ylab = "X2", pch = "+")
points(svm_pred_neg_radial$x1, svm_pred_neg_radial$x2, col = "blue", pch = 4)
## (i) Comment on your results.
Compared to other methods, support vector machines with radial kernels exhibit a much more pronounced non-linear boundary in Question (h). In contrast, the other methods either fail to detect linear boundaries at all or display less distinct boundaries.
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(ISLR)## Warning: package 'ISLR' was built under R version 4.3.2library(e1071)
attach(Auto)median_mpg = median(Auto$mpg)
binary_var = ifelse(Auto$mpg > median_mpg, 1, 0)
Auto$mpg_binary = as.factor(binary_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.
set.seed(1)
svm_tune_lin= tune(svm, mpg~., data = Auto, kernel = "linear", ranges = list(cost = c(0.01, 0.1, 1, 10, 100)))
summary(svm_tune_lin)##
## Parameter tuning of 'svm':
##
## - sampling method: 10-fold cross validation
##
## - best parameters:
## cost
## 0.1
##
## - best performance: 8.981009
##
## - Detailed performance results:
## cost error dispersion
## 1 1e-02 10.305990 5.295587
## 2 1e-01 8.981009 4.750742
## 3 1e+00 9.647184 4.313908
## 4 1e+01 10.306219 4.953047
## 5 1e+02 10.684083 5.080506Interpretation:
The kernel that is “linear” has the least error, as we can see for cost = 0.1.
(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(2)
svm_poly_tune_fit = tune(svm, mpg ~ ., data = Auto, kernel = "polynomial", ranges = list(cost = c(0.01, 0.1, 1, 10, 100), degree = c(2, 3, 4)))
summary(svm_poly_tune_fit)##
## Parameter tuning of 'svm':
##
## - sampling method: 10-fold cross validation
##
## - best parameters:
## cost degree
## 100 2
##
## - best performance: 17.65639
##
## - Detailed performance results:
## cost degree error dispersion
## 1 1e-02 2 61.73797 13.574948
## 2 1e-01 2 61.59446 13.602921
## 3 1e+00 2 60.15304 13.792932
## 4 1e+01 2 50.95606 15.723882
## 5 1e+02 2 17.65639 6.118173
## 6 1e-02 3 61.75044 13.571592
## 7 1e-01 3 61.71831 13.569399
## 8 1e+00 3 61.39833 13.547581
## 9 1e+01 3 58.28857 13.277604
## 10 1e+02 3 41.11944 10.878007
## 11 1e-02 4 61.75395 13.571849
## 12 1e-01 4 61.75343 13.571969
## 13 1e+00 4 61.74822 13.573169
## 14 1e+01 4 61.69626 13.585205
## 15 1e+02 4 61.18770 13.708674set.seed(2)
svm_radial_tune_fit = tune(svm, mpg ~ ., data = Auto, kernel = "radial", ranges = list(gamma = c(0.01, 0.1, 1, 10, 100), cost = c(0.01, 0.1, 1, 10, 100), degree = c(2, 3, 4)))
summary(svm_radial_tune_fit)##
## Parameter tuning of 'svm':
##
## - sampling method: 10-fold cross validation
##
## - best parameters:
## gamma cost degree
## 0.1 10 2
##
## - best performance: 6.568396
##
## - Detailed performance results:
## gamma cost degree error dispersion
## 1 1e-02 1e-02 2 43.589029 12.099454
## 2 1e-01 1e-02 2 30.860013 10.718178
## 3 1e+00 1e-02 2 60.022394 13.337004
## 4 1e+01 1e-02 2 61.700791 13.484368
## 5 1e+02 1e-02 2 61.723405 13.484402
## 6 1e-02 1e-01 2 13.785012 5.925585
## 7 1e-01 1e-01 2 9.994363 4.477300
## 8 1e+00 1e-01 2 46.903846 11.752198
## 9 1e+01 1e-01 2 61.386759 13.250273
## 10 1e+02 1e-01 2 61.609286 13.250083
## 11 1e-02 1e+00 2 7.748636 3.531173
## 12 1e-01 1e+00 2 6.691829 2.643121
## 13 1e+00 1e+00 2 21.684672 6.769173
## 14 1e+01 1e+00 2 59.571108 12.871977
## 15 1e+02 1e+00 2 61.303858 12.780671
## 16 1e-02 1e+01 2 6.933472 2.745711
## 17 1e-01 1e+01 2 6.568396 1.851267
## 18 1e+00 1e+01 2 20.535666 5.787001
## 19 1e+01 1e+01 2 59.223154 11.676800
## 20 1e+02 1e+01 2 61.037610 11.751463
## 21 1e-02 1e+02 2 7.243347 2.386674
## 22 1e-01 1e+02 2 6.851118 1.622121
## 23 1e+00 1e+02 2 20.535666 5.787001
## 24 1e+01 1e+02 2 59.223154 11.676800
## 25 1e+02 1e+02 2 61.037610 11.751463
## 26 1e-02 1e-02 3 43.589029 12.099454
## 27 1e-01 1e-02 3 30.860013 10.718178
## 28 1e+00 1e-02 3 60.022394 13.337004
## 29 1e+01 1e-02 3 61.700791 13.484368
## 30 1e+02 1e-02 3 61.723405 13.484402
## 31 1e-02 1e-01 3 13.785012 5.925585
## 32 1e-01 1e-01 3 9.994363 4.477300
## 33 1e+00 1e-01 3 46.903846 11.752198
## 34 1e+01 1e-01 3 61.386759 13.250273
## 35 1e+02 1e-01 3 61.609286 13.250083
## 36 1e-02 1e+00 3 7.748636 3.531173
## 37 1e-01 1e+00 3 6.691829 2.643121
## 38 1e+00 1e+00 3 21.684672 6.769173
## 39 1e+01 1e+00 3 59.571108 12.871977
## 40 1e+02 1e+00 3 61.303858 12.780671
## 41 1e-02 1e+01 3 6.933472 2.745711
## 42 1e-01 1e+01 3 6.568396 1.851267
## 43 1e+00 1e+01 3 20.535666 5.787001
## 44 1e+01 1e+01 3 59.223154 11.676800
## 45 1e+02 1e+01 3 61.037610 11.751463
## 46 1e-02 1e+02 3 7.243347 2.386674
## 47 1e-01 1e+02 3 6.851118 1.622121
## 48 1e+00 1e+02 3 20.535666 5.787001
## 49 1e+01 1e+02 3 59.223154 11.676800
## 50 1e+02 1e+02 3 61.037610 11.751463
## 51 1e-02 1e-02 4 43.589029 12.099454
## 52 1e-01 1e-02 4 30.860013 10.718178
## 53 1e+00 1e-02 4 60.022394 13.337004
## 54 1e+01 1e-02 4 61.700791 13.484368
## 55 1e+02 1e-02 4 61.723405 13.484402
## 56 1e-02 1e-01 4 13.785012 5.925585
## 57 1e-01 1e-01 4 9.994363 4.477300
## 58 1e+00 1e-01 4 46.903846 11.752198
## 59 1e+01 1e-01 4 61.386759 13.250273
## 60 1e+02 1e-01 4 61.609286 13.250083
## 61 1e-02 1e+00 4 7.748636 3.531173
## 62 1e-01 1e+00 4 6.691829 2.643121
## 63 1e+00 1e+00 4 21.684672 6.769173
## 64 1e+01 1e+00 4 59.571108 12.871977
## 65 1e+02 1e+00 4 61.303858 12.780671
## 66 1e-02 1e+01 4 6.933472 2.745711
## 67 1e-01 1e+01 4 6.568396 1.851267
## 68 1e+00 1e+01 4 20.535666 5.787001
## 69 1e+01 1e+01 4 59.223154 11.676800
## 70 1e+02 1e+01 4 61.037610 11.751463
## 71 1e-02 1e+02 4 7.243347 2.386674
## 72 1e-01 1e+02 4 6.851118 1.622121
## 73 1e+00 1e+02 4 20.535666 5.787001
## 74 1e+01 1e+02 4 59.223154 11.676800
## 75 1e+02 1e+02 4 61.037610 11.751463Interpretation:
The lowest error for a polynomial kernel is at degree = 2 and cost = 100, while the lowest error for a radial kernel is at gamma = 0.1, degree = 2, and cost = 10.
(d) Make some plots to back up your assertions in (b) and (c).
svm_linear= svm(mpg_binary ~ ., data = Auto, kernel = "linear", cost = 0.1)
svm_poly = svm(mpg_binary ~ ., data = Auto, kernel = "polynomial", cost = 100,
degree = 2)
svm_radial = svm(mpg_binary ~ ., data = Auto, kernel = "radial", cost = 10, gamma = 0.1)
plotpairs <- function(fit) {
for (name in names(Auto)[!(names(Auto) %in% c("mpg", "binary_var", "name"))]) {
if (is.numeric(Auto[[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 ISLR2
package. ## (a) Create a training set containing a random sample of 800
observations, and a test set containing the remaining observations.
attach(OJ)
set.seed(1)
OJ_Split_data = sample(nrow(OJ), 800)
OJ_train_data = OJ[OJ_Split_data,]
OJ_test_data = OJ[-OJ_Split_data,](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.
linear_SVM1 = svm(Purchase ~ ., kernel = "linear", data = OJ_train_data, cost = 0.01)
summary(linear_SVM1)##
## Call:
## svm(formula = Purchase ~ ., data = OJ_train_data, 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 MMInterpretation:
219 support vectors are at level CH and 216 are at level MM out of a total of 435 support vectors.
(c) What are the training and test error rates?
pred_OJ_train_data = predict(linear_SVM1, OJ_train_data)
(t<-table(OJ_train_data$Purchase, pred_OJ_train_data))## pred_OJ_train_data
## CH MM
## CH 420 65
## MM 75 240misclassified_svm1 <- sum(OJ_train_data$Purchase != pred_OJ_train_data)
training_error_rate_svm1 <- misclassified_svm1 / nrow(OJ_train_data)
print(paste("Training Error Rate:", training_error_rate_svm1))## [1] "Training Error Rate: 0.175"pred_OJ_test_data= predict(linear_SVM1, OJ_test_data)
(tt<-table(OJ_test_data$Purchase, pred_OJ_test_data))## pred_OJ_test_data
## CH MM
## CH 153 15
## MM 33 69misclassified_Test_svm1 <- sum(OJ_test_data$Purchase != pred_OJ_test_data)
test_error_rate_svm1 <- misclassified_Test_svm1 / nrow(OJ_test_data)
print(paste("Test Error Rate:", test_error_rate_svm1))## [1] "Test Error Rate: 0.177777777777778"(d) Use the tune() function to select an optimal cost. Consider values in the range 0.01 to 10.
set.seed(1)
svm_optimal_tune1 = tune(svm, Purchase ~ ., data = OJ_train_data, kernel = "linear", ranges = list(cost = c(0.01,0.1,1,10)))
summary(svm_optimal_tune1)##
## Parameter tuning of 'svm':
##
## - sampling method: 10-fold cross validation
##
## - best parameters:
## cost
## 0.1
##
## - best performance: 0.1725
##
## - Detailed performance results:
## cost error dispersion
## 1 0.01 0.17625 0.02853482
## 2 0.10 0.17250 0.03162278
## 3 1.00 0.17500 0.02946278
## 4 10.00 0.17375 0.03197764Interpretation:
With a dispersion (ideal cost) of 0.03162278, the lowest inaccuracy occurs at 0.1 cost.
(e) Compute the training and test error rates using this new value for cost.
svm_new_cost_fit1 = svm(Purchase ~ ., kernel = "linear", data = OJ_train_data, cost = svm_optimal_tune1$best.parameters$cost)
pred_svm_new_cost_fit1 = predict(svm_new_cost_fit1, OJ_train_data)
table(OJ_train_data$Purchase, pred_svm_new_cost_fit1)## pred_svm_new_cost_fit1
## CH MM
## CH 422 63
## MM 69 246misclassified_new_train1 <- sum(OJ_train_data$Purchase != pred_svm_new_cost_fit1)
training_error_rate_new1 <- misclassified_new_train1 / nrow(OJ_train_data)
print(paste("Training Error Rate:", training_error_rate_new1))## [1] "Training Error Rate: 0.165"pred_svm_new_cost_fit_test1 = predict(svm_new_cost_fit1, OJ_test_data)
table(OJ_test_data$Purchase, pred_svm_new_cost_fit_test1)## pred_svm_new_cost_fit_test1
## CH MM
## CH 155 13
## MM 31 71misclassified_new_test1 <- sum(OJ_test_data$Purchase != pred_svm_new_cost_fit_test1)
test_error_rate_new1 <- misclassified_new_test1 / nrow(OJ_test_data)
print(paste("Test Error Rate:", test_error_rate_new1))## [1] "Test Error Rate: 0.162962962962963"(f) Repeat parts (b) through (e) using a support vector machine with a radial kernel. Use the default value for gamma.
set.seed(1)
svm_rad_fit2 = svm(Purchase ~ ., data = OJ_train_data, kernel = "radial")
summary(svm_rad_fit2)##
## Call:
## svm(formula = Purchase ~ ., data = OJ_train_data, kernel = "radial")
##
##
## Parameters:
## SVM-Type: C-classification
## SVM-Kernel: radial
## cost: 1
##
## Number of Support Vectors: 373
##
## ( 188 185 )
##
##
## Number of Classes: 2
##
## Levels:
## CH MMInterpretation:
There are total of 373 support vectors out Of which, 188 vectos are at level CH, and 185 vectors are at level MM.
pred_svm_rad_fit2 = predict(svm_rad_fit2, OJ_train_data)
table(OJ_train_data$Purchase, pred_svm_rad_fit2)## pred_svm_rad_fit2
## CH MM
## CH 441 44
## MM 77 238misclassified_svm_rad_fit2 <- sum(OJ_train_data$Purchase != pred_svm_rad_fit2)
training_error_misclassified_svm_rad_fit2 <- misclassified_svm_rad_fit2 / nrow(OJ_train_data)
print(paste("Training Error Rate:", training_error_misclassified_svm_rad_fit2))## [1] "Training Error Rate: 0.15125"pred_svm_rad_fit1_test1 = predict(svm_rad_fit2, OJ_test_data)
table(OJ_test_data$Purchase, pred_svm_rad_fit1_test1)## pred_svm_rad_fit1_test1
## CH MM
## CH 151 17
## MM 33 69misclassified_svm_rad_fit1_test1 <- sum(OJ_test_data$Purchase != pred_svm_rad_fit1_test1)
test_error_misclassified_svm_rad_fit2 <- misclassified_svm_rad_fit1_test1 / nrow(OJ_test_data)
print(paste("Test Error Rate:", test_error_misclassified_svm_rad_fit2))## [1] "Test Error Rate: 0.185185185185185"svm_rad_1_tune2 = svm(Purchase ~ ., data = OJ_train_data, kernel = "radial", cost = 1)
pred_train_tune_svm2 = predict(svm_rad_1_tune2, OJ_train_data)
table(OJ_train_data$Purchase, pred_train_tune_svm2)## pred_train_tune_svm2
## CH MM
## CH 441 44
## MM 77 238misclassified2_train_tune_svm2<- sum(OJ_train_data$Purchase != pred_train_tune_svm2)
training_error_misclassified_svm_rad <- misclassified2_train_tune_svm2 / nrow(OJ_train_data)
print(paste("Training Error Rate:", training_error_misclassified_svm_rad))## [1] "Training Error Rate: 0.15125"pred_svm_rad_fit1_test_tune1 = predict(svm_rad_1_tune2, OJ_test_data)
table(OJ_test_data$Purchase, pred_svm_rad_fit1_test_tune1)## pred_svm_rad_fit1_test_tune1
## CH MM
## CH 151 17
## MM 33 69misclassified_svm_rad_tune_test2 <- sum(OJ_test_data$Purchase != pred_svm_rad_fit1_test_tune1)
test_error_misclassified_svm_rad1 <- misclassified_svm_rad_tune_test2 / nrow(OJ_test_data)
print(paste("Test Error Rate:", test_error_misclassified_svm_rad1))## [1] "Test Error Rate: 0.185185185185185"svm_rad_1_tune_linear1 = svm(Purchase ~ ., data = OJ_train_data, kernel = "linear", cost = 1)
pred_train_tune_svm_linear1 = predict(svm_rad_1_tune_linear1, OJ_train_data)
table(OJ_train_data$Purchase, pred_train_tune_svm_linear1)## pred_train_tune_svm_linear1
## CH MM
## CH 424 61
## MM 70 245misclassified2_train_tune_svm_linear2<- sum(OJ_train_data$Purchase != pred_train_tune_svm_linear1)
training_error_misclassified_svm_rad2_lin2 <- misclassified2_train_tune_svm_linear2 / nrow(OJ_train_data)
print(paste("Training Error Rate:", training_error_misclassified_svm_rad2_lin2))## [1] "Training Error Rate: 0.16375"pred_svm_rad_fit1_test_tune_linear1 = predict(svm_rad_1_tune_linear1, OJ_test_data)
table(OJ_test_data$Purchase, pred_svm_rad_fit1_test_tune_linear1)## pred_svm_rad_fit1_test_tune_linear1
## CH MM
## CH 155 13
## MM 29 73misclassified_svm_rad_tune_test_lin2 <- sum(OJ_test_data$Purchase != pred_svm_rad_fit1_test_tune_linear1)
test_error_misclassified_svm_rad2_lin2 <- misclassified_svm_rad_tune_test_lin2 / nrow(OJ_test_data)
print(paste("Test Error Rate:", test_error_misclassified_svm_rad2_lin2))## [1] "Test Error Rate: 0.155555555555556"(g) Repeat parts (b) through (e) using a support vector machine with a polynomial kernel. Set degree = 2.
svm_pol_degree21 = svm(Purchase ~ ., kernel = "poly", data = OJ_train_data, degree=2)
summary(svm_pol_degree21)##
## Call:
## svm(formula = Purchase ~ ., data = OJ_train_data, kernel = "poly",
## degree = 2)
##
##
## Parameters:
## SVM-Type: C-classification
## SVM-Kernel: polynomial
## cost: 1
## degree: 2
## coef.0: 0
##
## Number of Support Vectors: 447
##
## ( 225 222 )
##
##
## Number of Classes: 2
##
## Levels:
## CH MMInterpretation:
A total of 447 support vectors with a polynominal kernel, degree of 2, and cost of 1 are present; of these, 255 vectors are at level CH and 222 at level MM.
pred_train_tune_svm_poly1 = predict(svm_pol_degree21, OJ_train_data)
table(OJ_train_data$Purchase, pred_train_tune_svm_poly1)## pred_train_tune_svm_poly1
## CH MM
## CH 449 36
## MM 110 205misclassified2_train_tune_svm_poly1<- sum(OJ_train_data$Purchase != pred_train_tune_svm_poly1)
training_error_misclassified_svm_poly1 <- misclassified2_train_tune_svm_poly1 / nrow(OJ_train_data)
print(paste("Training Error Rate:", training_error_misclassified_svm_poly1))## [1] "Training Error Rate: 0.1825"pred_svm_rad_fit1_test_tune_poly1 = predict(svm_pol_degree21, OJ_test_data)
table(OJ_test_data$Purchase, pred_svm_rad_fit1_test_tune_poly1)## pred_svm_rad_fit1_test_tune_poly1
## CH MM
## CH 153 15
## MM 45 57misclassified_svm_rad_tune_test_poly1 <- sum(OJ_test_data$Purchase != pred_svm_rad_fit1_test_tune_poly1)
test_error_misclassified_svm_poly1 <- misclassified_svm_rad_tune_test_poly1 / nrow(OJ_test_data)
print(paste("Test Error Rate:", test_error_misclassified_svm_poly1))## [1] "Test Error Rate: 0.222222222222222"set.seed(1)
tune_svm_poly1 = tune(svm, Purchase ~ ., data = OJ_train_data, kernel = "poly", degree = 2, ranges = list(cost = c(0.01,0.1,1,10)))
summary(tune_svm_poly1)##
## Parameter tuning of 'svm':
##
## - sampling method: 10-fold cross validation
##
## - best parameters:
## cost
## 10
##
## - best performance: 0.18125
##
## - Detailed performance results:
## cost error dispersion
## 1 0.01 0.39125 0.04210189
## 2 0.10 0.32125 0.05001736
## 3 1.00 0.20250 0.04116363
## 4 10.00 0.18125 0.02779513Interpretation:
The lowest error is at 10 cost, with dispersion(optimal cost) is 0.02779513.
svm_poly_tune_cost_optim1 = svm(Purchase ~ ., data = OJ_train_data, kernel = "poly", degree = 2, cost = tune_svm_poly1$best.parameters$cost)
train_pred_optim_cost_svm_poly1 = predict(svm_poly_tune_cost_optim1, OJ_train_data)
table(OJ_train_data$Purchase, train_pred_optim_cost_svm_poly1)## train_pred_optim_cost_svm_poly1
## CH MM
## CH 447 38
## MM 82 233misclassified2_train_tune_svm_poly_optim1<- sum(OJ_train_data$Purchase != train_pred_optim_cost_svm_poly1)
training_error_misclassified_svm_poly_optim1 <- misclassified2_train_tune_svm_poly_optim1 / nrow(OJ_train_data)
print(paste("Training Error Rate:", training_error_misclassified_svm_poly_optim1))## [1] "Training Error Rate: 0.15"pred_svm_fit1_test_tunes1 = predict(svm_poly_tune_cost_optim1, OJ_test_data)
table(OJ_test_data$Purchase, pred_svm_fit1_test_tunes1)## pred_svm_fit1_test_tunes1
## CH MM
## CH 154 14
## MM 37 65misclassified_svm_tune_test1 <- sum(OJ_train_data$Purchase != pred_svm_fit1_test_tunes1)## Warning in `!=.default`(OJ_train_data$Purchase, pred_svm_fit1_test_tunes1):
## longer object length is not a multiple of shorter object length## Warning in is.na(e1) | is.na(e2): longer object length is not a multiple of
## shorter object lengthtest_error_misclassified_svm_poly1 <- misclassified_svm_tune_test1 / nrow(OJ_test_data)
print(paste("Test Error Rate:", test_error_misclassified_svm_poly1))## [1] "Test Error Rate: 1.3962962962963"(h) Overall, which approach seems to give the best results on this data?
The tuned SVM-linear model performs well with this dataset.