Assignment_8
Sai
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 = "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.
log_reg_fit = glm(y ~ x1 + x2, family = binomial)
summary(log_reg_fit)##
## Call:
## glm(formula = y ~ x1 + x2, family = binomial)
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -0.08536 0.09009 -0.947 0.343
## x1 0.47083 0.31875 1.477 0.140
## x2 -0.20178 0.31656 -0.637 0.524
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 692.18 on 499 degrees of freedom
## Residual deviance: 689.53 on 497 degrees of freedom
## AIC: 695.53
##
## 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.
log_trng_data = data.frame(x1 = x1, x2 = x2, y = y)
log_reg_pred = predict(log_reg_fit, log_trng_data, type = "response")
log_reg_pred_prob = ifelse(log_reg_pred > 0.52, 1, 0)
log_reg_pred_prob_pos = log_trng_data[log_reg_pred_prob == 1, ]
log_reg_pred_prob_neg = log_trng_data[log_reg_pred_prob == 0, ]
plot(log_reg_pred_prob_pos$x1, log_reg_pred_prob_pos$x2, col = "green", xlab = "X1", ylab = "X2", pch = "+")
points(log_reg_pred_prob_neg$x1, log_reg_pred_prob_neg$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. X21 , X1×X2, log(X2),and so
forth).
log_reg_nonlinear = glm(y ~ poly(x1, 2) + poly(x2, 2) + I(x1 * x2), data = log_trng_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_nonlinear_pred = predict(log_reg_nonlinear, log_trng_data, type = "response")
log_reg_nonlinear_pred_prob = ifelse(log_reg_nonlinear_pred > 0.5, 1, 0)
log_reg_nonlinear_pred_prob_pos = log_trng_data[log_reg_nonlinear_pred_prob == 1, ]
log_reg_nonlinear_pred_prob_neg = log_trng_data[log_reg_nonlinear_pred_prob == 0, ]
plot(log_reg_nonlinear_pred_prob_pos$x1, log_reg_nonlinear_pred_prob_pos$x2, col = "red", xlab = "X1", ylab = "X2", pch = "+")
points(log_reg_nonlinear_pred_prob_neg$x1, log_reg_nonlinear_pred_prob_neg$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, log_trng_data, kernel = "linear", cost = 0.1)
svm.pred = predict(svm.fit, log_trng_data)
data.pos = log_trng_data[svm.pred == 1, ]
data.neg = log_trng_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_fit_radial= svm(as.factor(y) ~ x1 + x2, log_trng_data, kernel = "radial", gamma=1)
svm_fit_pred_radial = predict(svm_fit_radial, log_trng_data)
svm_fit_pred_pos_radial = log_trng_data[svm_fit_pred_radial == 1, ]
svm_fit_pred_neg_radial = log_trng_data[svm_fit_pred_radial == 0, ]
plot(svm_fit_pred_pos_radial$x1, svm_fit_pred_pos_radial$x2, col = "red", xlab = "X1", ylab = "X2", pch = "+")
points(svm_fit_pred_neg_radial$x1, svm_fit_pred_neg_radial$x2, col = "blue", pch = 4)
## (i) Comment on your results.
Support vector machines with radial kernels show a far more distinct non-linear border in Question (h) than other approaches. By comparison, the other approaches either show less clear-cut linear boundaries or fail to detect them at all.
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 4.3.2attach(Auto)
library(e1071)med_auto_mpg = median(Auto$mpg)
vari_bin = ifelse(Auto$mpg > med_auto_mpg, 1, 0)
Auto$mpg_binary_var = as.factor(vari_bin)(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 = tune(svm, mpg~., data = Auto, kernel = "linear", ranges = list(cost = c(0.01, 0.1, 1, 5, 10, 50, 100)))
summary(svm_tune)##
## 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 5e+00 10.149220 4.755080
## 5 1e+01 10.306219 4.953047
## 6 5e+01 10.631566 5.129439
## 7 1e+02 10.684083 5.080506Interpretation:
We notice the lowest error for kernel- “linear” at 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(svm, mpg ~ ., data = Auto, kernel = "polynomial", ranges = list(cost = c(0.01, 0.1, 1, 5, 10, 50, 100), degree = c(2, 3, 4, 5)))
summary(svm_poly)##
## 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 5e+00 2 55.06386 15.193907
## 5 1e+01 2 50.95606 15.723882
## 6 5e+01 2 28.51889 9.993313
## 7 1e+02 2 17.65639 6.118173
## 8 1e-02 3 61.75044 13.571592
## 9 1e-01 3 61.71831 13.569399
## 10 1e+00 3 61.39833 13.547581
## 11 5e+00 3 59.99304 13.432082
## 12 1e+01 3 58.28857 13.277604
## 13 5e+01 3 47.62254 12.244469
## 14 1e+02 3 41.11944 10.878007
## 15 1e-02 4 61.75395 13.571849
## 16 1e-01 4 61.75343 13.571969
## 17 1e+00 4 61.74822 13.573169
## 18 5e+00 4 61.72510 13.578511
## 19 1e+01 4 61.69626 13.585205
## 20 5e+01 4 61.46773 13.639401
## 21 1e+02 4 61.18770 13.708674
## 22 1e-02 5 61.75400 13.571835
## 23 1e-01 5 61.75395 13.571833
## 24 1e+00 5 61.75336 13.571812
## 25 5e+00 5 61.75076 13.571718
## 26 1e+01 5 61.74752 13.571601
## 27 5e+01 5 61.72156 13.570669
## 28 1e+02 5 61.68916 13.569517set.seed(2)
svm_radial = 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, 5, 10, 50, 100), degree = c(2, 3, 4, 5)))
summary(svm_radial)##
## Parameter tuning of 'svm':
##
## - sampling method: 10-fold cross validation
##
## - best parameters:
## gamma cost degree
## 0.1 5 2
##
## - best performance: 6.498043
##
## - 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 5e+00 2 6.942622 2.873995
## 17 1e-01 5e+00 2 6.498043 1.915277
## 18 1e+00 5e+00 2 20.556407 5.817490
## 19 1e+01 5e+00 2 59.223154 11.676800
## 20 1e+02 5e+00 2 61.037610 11.751463
## 21 1e-02 1e+01 2 6.933472 2.745711
## 22 1e-01 1e+01 2 6.568396 1.851267
## 23 1e+00 1e+01 2 20.535666 5.787001
## 24 1e+01 1e+01 2 59.223154 11.676800
## 25 1e+02 1e+01 2 61.037610 11.751463
## 26 1e-02 5e+01 2 7.190176 2.258457
## 27 1e-01 5e+01 2 6.693108 1.681847
## 28 1e+00 5e+01 2 20.535666 5.787001
## 29 1e+01 5e+01 2 59.223154 11.676800
## 30 1e+02 5e+01 2 61.037610 11.751463
## 31 1e-02 1e+02 2 7.243347 2.386674
## 32 1e-01 1e+02 2 6.851118 1.622121
## 33 1e+00 1e+02 2 20.535666 5.787001
## 34 1e+01 1e+02 2 59.223154 11.676800
## 35 1e+02 1e+02 2 61.037610 11.751463
## 36 1e-02 1e-02 3 43.589029 12.099454
## 37 1e-01 1e-02 3 30.860013 10.718178
## 38 1e+00 1e-02 3 60.022394 13.337004
## 39 1e+01 1e-02 3 61.700791 13.484368
## 40 1e+02 1e-02 3 61.723405 13.484402
## 41 1e-02 1e-01 3 13.785012 5.925585
## 42 1e-01 1e-01 3 9.994363 4.477300
## 43 1e+00 1e-01 3 46.903846 11.752198
## 44 1e+01 1e-01 3 61.386759 13.250273
## 45 1e+02 1e-01 3 61.609286 13.250083
## 46 1e-02 1e+00 3 7.748636 3.531173
## 47 1e-01 1e+00 3 6.691829 2.643121
## 48 1e+00 1e+00 3 21.684672 6.769173
## 49 1e+01 1e+00 3 59.571108 12.871977
## 50 1e+02 1e+00 3 61.303858 12.780671
## 51 1e-02 5e+00 3 6.942622 2.873995
## 52 1e-01 5e+00 3 6.498043 1.915277
## 53 1e+00 5e+00 3 20.556407 5.817490
## 54 1e+01 5e+00 3 59.223154 11.676800
## 55 1e+02 5e+00 3 61.037610 11.751463
## 56 1e-02 1e+01 3 6.933472 2.745711
## 57 1e-01 1e+01 3 6.568396 1.851267
## 58 1e+00 1e+01 3 20.535666 5.787001
## 59 1e+01 1e+01 3 59.223154 11.676800
## 60 1e+02 1e+01 3 61.037610 11.751463
## 61 1e-02 5e+01 3 7.190176 2.258457
## 62 1e-01 5e+01 3 6.693108 1.681847
## 63 1e+00 5e+01 3 20.535666 5.787001
## 64 1e+01 5e+01 3 59.223154 11.676800
## 65 1e+02 5e+01 3 61.037610 11.751463
## 66 1e-02 1e+02 3 7.243347 2.386674
## 67 1e-01 1e+02 3 6.851118 1.622121
## 68 1e+00 1e+02 3 20.535666 5.787001
## 69 1e+01 1e+02 3 59.223154 11.676800
## 70 1e+02 1e+02 3 61.037610 11.751463
## 71 1e-02 1e-02 4 43.589029 12.099454
## 72 1e-01 1e-02 4 30.860013 10.718178
## 73 1e+00 1e-02 4 60.022394 13.337004
## 74 1e+01 1e-02 4 61.700791 13.484368
## 75 1e+02 1e-02 4 61.723405 13.484402
## 76 1e-02 1e-01 4 13.785012 5.925585
## 77 1e-01 1e-01 4 9.994363 4.477300
## 78 1e+00 1e-01 4 46.903846 11.752198
## 79 1e+01 1e-01 4 61.386759 13.250273
## 80 1e+02 1e-01 4 61.609286 13.250083
## 81 1e-02 1e+00 4 7.748636 3.531173
## 82 1e-01 1e+00 4 6.691829 2.643121
## 83 1e+00 1e+00 4 21.684672 6.769173
## 84 1e+01 1e+00 4 59.571108 12.871977
## 85 1e+02 1e+00 4 61.303858 12.780671
## 86 1e-02 5e+00 4 6.942622 2.873995
## 87 1e-01 5e+00 4 6.498043 1.915277
## 88 1e+00 5e+00 4 20.556407 5.817490
## 89 1e+01 5e+00 4 59.223154 11.676800
## 90 1e+02 5e+00 4 61.037610 11.751463
## 91 1e-02 1e+01 4 6.933472 2.745711
## 92 1e-01 1e+01 4 6.568396 1.851267
## 93 1e+00 1e+01 4 20.535666 5.787001
## 94 1e+01 1e+01 4 59.223154 11.676800
## 95 1e+02 1e+01 4 61.037610 11.751463
## 96 1e-02 5e+01 4 7.190176 2.258457
## 97 1e-01 5e+01 4 6.693108 1.681847
## 98 1e+00 5e+01 4 20.535666 5.787001
## 99 1e+01 5e+01 4 59.223154 11.676800
## 100 1e+02 5e+01 4 61.037610 11.751463
## 101 1e-02 1e+02 4 7.243347 2.386674
## 102 1e-01 1e+02 4 6.851118 1.622121
## 103 1e+00 1e+02 4 20.535666 5.787001
## 104 1e+01 1e+02 4 59.223154 11.676800
## 105 1e+02 1e+02 4 61.037610 11.751463
## 106 1e-02 1e-02 5 43.589029 12.099454
## 107 1e-01 1e-02 5 30.860013 10.718178
## 108 1e+00 1e-02 5 60.022394 13.337004
## 109 1e+01 1e-02 5 61.700791 13.484368
## 110 1e+02 1e-02 5 61.723405 13.484402
## 111 1e-02 1e-01 5 13.785012 5.925585
## 112 1e-01 1e-01 5 9.994363 4.477300
## 113 1e+00 1e-01 5 46.903846 11.752198
## 114 1e+01 1e-01 5 61.386759 13.250273
## 115 1e+02 1e-01 5 61.609286 13.250083
## 116 1e-02 1e+00 5 7.748636 3.531173
## 117 1e-01 1e+00 5 6.691829 2.643121
## 118 1e+00 1e+00 5 21.684672 6.769173
## 119 1e+01 1e+00 5 59.571108 12.871977
## 120 1e+02 1e+00 5 61.303858 12.780671
## 121 1e-02 5e+00 5 6.942622 2.873995
## 122 1e-01 5e+00 5 6.498043 1.915277
## 123 1e+00 5e+00 5 20.556407 5.817490
## 124 1e+01 5e+00 5 59.223154 11.676800
## 125 1e+02 5e+00 5 61.037610 11.751463
## 126 1e-02 1e+01 5 6.933472 2.745711
## 127 1e-01 1e+01 5 6.568396 1.851267
## 128 1e+00 1e+01 5 20.535666 5.787001
## 129 1e+01 1e+01 5 59.223154 11.676800
## 130 1e+02 1e+01 5 61.037610 11.751463
## 131 1e-02 5e+01 5 7.190176 2.258457
## 132 1e-01 5e+01 5 6.693108 1.681847
## 133 1e+00 5e+01 5 20.535666 5.787001
## 134 1e+01 5e+01 5 59.223154 11.676800
## 135 1e+02 5e+01 5 61.037610 11.751463
## 136 1e-02 1e+02 5 7.243347 2.386674
## 137 1e-01 1e+02 5 6.851118 1.622121
## 138 1e+00 1e+02 5 20.535666 5.787001
## 139 1e+01 1e+02 5 59.223154 11.676800
## 140 1e+02 1e+02 5 61.037610 11.751463Interpretation:
For a radial kernel, we see a lowest of the error at gamma = 0.1, degree=2, and cost = 5 and with polynomial kernel the lowest error is at degree = 2 and cost = 100.
(d) Make some plots to back up your assertions in (b) and (c).
svm_linear = svm(mpg_binary_var ~ ., data = Auto, kernel = "linear", cost = 0.1)
svm_poly = svm(mpg_binary_var ~ ., data = Auto, kernel = "polynomial", cost = 100,
degree = 2)
svm_radial = svm(mpg_binary_var ~ ., data = Auto, kernel = "radial", cost = 5, gamma = 0.1)
plotpairs = function(fit) {
for (name in names(Auto)[!(names(Auto) %in% c("mpg", "mpg_binary_var", "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 = sample(nrow(OJ), 800)
OJ_Split_train = OJ[OJ_Split,]
OJ_Split_test = OJ[-OJ_Split,](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_SVM = svm(Purchase ~ ., kernel = "linear", data = OJ_Split_train, cost = 0.01)
summary(linear_SVM)##
## Call:
## svm(formula = Purchase ~ ., data = OJ_Split_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 MMInterpretation:
There are total of 435 support vectors out Of which, 219 vectos are at level CH, and 216 vectors are at level MM.
(c) What are the training and test error rates?
pred_OJ_Split_train = predict(linear_SVM, OJ_Split_train)
(t<-table(OJ_Split_train$Purchase, pred_OJ_Split_train))## pred_OJ_Split_train
## CH MM
## CH 420 65
## MM 75 240misclassified <- sum(OJ_Split_train$Purchase != pred_OJ_Split_train)
training_error_rate <- misclassified / nrow(OJ_Split_train)
print(paste("Training Error Rate:", training_error_rate))## [1] "Training Error Rate: 0.175"pred_OJ_Split_test= predict(linear_SVM, OJ_Split_test)
(tt<-table(OJ_Split_test$Purchase, pred_OJ_Split_test))## pred_OJ_Split_test
## CH MM
## CH 153 15
## MM 33 69misclassified_Test <- sum(OJ_Split_test$Purchase != pred_OJ_Split_test)
test_error_rate <- misclassified_Test / nrow(OJ_Split_test)
print(paste("Test Error Rate:", test_error_rate))## [1] "Test Error Rate: 0.177777777777778"(d) Use the tune() function to select an optimal cost. Consider values in the range 0.01 to 10.
set.seed(1)
svm_optimal_tune_fit = tune(svm, Purchase ~ ., data = OJ_Split_train, kernel = "linear", ranges = list(cost = c(0.01,0.1,1,10)))
summary(svm_optimal_tune_fit)##
## Parameter tuning of 'svm':
##
## - sampling method: 10-fold cross validation
##
## - best parameters:
## cost
## 0.1
##
## - best performance: 0.1725
##
## - Detailed performance results:
## cost error dispersion
## 1 0.01 0.17625 0.02853482
## 2 0.10 0.17250 0.03162278
## 3 1.00 0.17500 0.02946278
## 4 10.00 0.17375 0.03197764Interpretation:
The lowest error is at 0.1 cost, with dispersion(optimal cost) is 0.03162278.
(e) Compute the training and test error rates using this new value for cost.
svm_new_cost_fit = svm(Purchase ~ ., kernel = "linear", data = OJ_Split_train, cost = svm_optimal_tune_fit$best.parameters$cost)
pred_svm_new_cost_fit = predict(svm_new_cost_fit, OJ_Split_train)
table(OJ_Split_train$Purchase, pred_svm_new_cost_fit)## pred_svm_new_cost_fit
## CH MM
## CH 422 63
## MM 69 246misclassified_new_train <- sum(OJ_Split_train$Purchase != pred_svm_new_cost_fit)
training_error_rate_new <- misclassified_new_train / nrow(OJ_Split_train)
print(paste("Training Error Rate:", training_error_rate_new))## [1] "Training Error Rate: 0.165"pred_svm_new_cost_fit_test = predict(svm_new_cost_fit, OJ_Split_test)
table(OJ_Split_test$Purchase, pred_svm_new_cost_fit_test)## pred_svm_new_cost_fit_test
## CH MM
## CH 155 13
## MM 31 71misclassified_new_test <- sum(OJ_Split_test$Purchase != pred_svm_new_cost_fit_test)
test_error_rate_new <- misclassified_new_test / nrow(OJ_Split_test)
print(paste("Test Error Rate:", test_error_rate_new))## [1] "Test Error Rate: 0.162962962962963"(f) Repeat parts (b) through (e) using a support vector machine with a radial kernel. Use the default value for gamma.
set.seed(1)
svm_rad_fit1 = svm(Purchase ~ ., data = OJ_Split_train, kernel = "radial")
summary(svm_rad_fit1)##
## Call:
## svm(formula = Purchase ~ ., data = OJ_Split_train, kernel = "radial")
##
##
## Parameters:
## SVM-Type: C-classification
## SVM-Kernel: radial
## cost: 1
##
## Number of Support Vectors: 373
##
## ( 188 185 )
##
##
## Number of Classes: 2
##
## Levels:
## CH MMInterpretation:
There are total of 373 support vectors out Of which, 188 vectos are at level CH, and 185 vectors are at level MM.
pred_svm_rad_fit1 = predict(svm_rad_fit1, OJ_Split_train)
table(OJ_Split_train$Purchase, pred_svm_rad_fit1)## pred_svm_rad_fit1
## CH MM
## CH 441 44
## MM 77 238misclassified_svm_rad_fit1 <- sum(OJ_Split_train$Purchase != pred_svm_rad_fit1)
training_error_misclassified_svm_rad_fit1 <- misclassified_svm_rad_fit1 / nrow(OJ_Split_train)
print(paste("Training Error Rate:", training_error_misclassified_svm_rad_fit1))## [1] "Training Error Rate: 0.15125"pred_svm_rad_fit1_test = predict(svm_rad_fit1, OJ_Split_test)
table(OJ_Split_test$Purchase, pred_svm_rad_fit1_test)## pred_svm_rad_fit1_test
## CH MM
## CH 151 17
## MM 33 69misclassified_svm_rad_fit1_test <- sum(OJ_Split_test$Purchase != pred_svm_rad_fit1_test)
test_error_misclassified_svm_rad_fit1 <- misclassified_svm_rad_fit1_test / nrow(OJ_Split_test)
print(paste("Test Error Rate:", test_error_misclassified_svm_rad_fit1))## [1] "Test Error Rate: 0.185185185185185"svm_rad_1_tune = svm(Purchase ~ ., data = OJ_Split_train, kernel = "radial", cost = 1)
pred_train_tune_svm = predict(svm_rad_1_tune, OJ_Split_train)
table(OJ_Split_train$Purchase, pred_train_tune_svm)## pred_train_tune_svm
## CH MM
## CH 441 44
## MM 77 238misclassified2_train_tune_svm<- sum(OJ_Split_train$Purchase != pred_train_tune_svm)
training_error_misclassified_svm_rad2 <- misclassified2_train_tune_svm / nrow(OJ_Split_train)
print(paste("Training Error Rate:", training_error_misclassified_svm_rad2))## [1] "Training Error Rate: 0.15125"pred_svm_rad_fit1_test_tune = predict(svm_rad_1_tune, OJ_Split_test)
table(OJ_Split_test$Purchase, pred_svm_rad_fit1_test_tune)## pred_svm_rad_fit1_test_tune
## CH MM
## CH 151 17
## MM 33 69misclassified_svm_rad_tune_test <- sum(OJ_Split_test$Purchase != pred_svm_rad_fit1_test_tune)
test_error_misclassified_svm_rad2 <- misclassified_svm_rad_tune_test / nrow(OJ_Split_test)
print(paste("Test Error Rate:", test_error_misclassified_svm_rad2))## [1] "Test Error Rate: 0.185185185185185"svm_rad_1_tune_linear = svm(Purchase ~ ., data = OJ_Split_train, kernel = "linear", cost = 1)
pred_train_tune_svm_linear = predict(svm_rad_1_tune_linear, OJ_Split_train)
table(OJ_Split_train$Purchase, pred_train_tune_svm_linear)## pred_train_tune_svm_linear
## CH MM
## CH 424 61
## MM 70 245misclassified2_train_tune_svm_linear<- sum(OJ_Split_train$Purchase != pred_train_tune_svm_linear)
training_error_misclassified_svm_rad2_lin <- misclassified2_train_tune_svm_linear / nrow(OJ_Split_train)
print(paste("Training Error Rate:", training_error_misclassified_svm_rad2_lin))## [1] "Training Error Rate: 0.16375"pred_svm_rad_fit1_test_tune_linear = predict(svm_rad_1_tune_linear, OJ_Split_test)
table(OJ_Split_test$Purchase, pred_svm_rad_fit1_test_tune_linear)## pred_svm_rad_fit1_test_tune_linear
## CH MM
## CH 155 13
## MM 29 73misclassified_svm_rad_tune_test_lin <- sum(OJ_Split_test$Purchase != pred_svm_rad_fit1_test_tune_linear)
test_error_misclassified_svm_rad2_lin <- misclassified_svm_rad_tune_test_lin / nrow(OJ_Split_test)
print(paste("Test Error Rate:", test_error_misclassified_svm_rad2_lin))## [1] "Test Error Rate: 0.155555555555556"(g) Repeat parts (b) through (e) using a support vector machine with a polynomial kernel. Set degree = 2.
svm_pol_degree2 = svm(Purchase ~ ., kernel = "poly", data = OJ_Split_train, degree=2)
summary(svm_pol_degree2)##
## Call:
## svm(formula = Purchase ~ ., data = OJ_Split_train, kernel = "poly",
## degree = 2)
##
##
## Parameters:
## SVM-Type: C-classification
## SVM-Kernel: polynomial
## cost: 1
## degree: 2
## coef.0: 0
##
## Number of Support Vectors: 447
##
## ( 225 222 )
##
##
## Number of Classes: 2
##
## Levels:
## CH MMInterpretation:
There are total of 447 support vectors out Of which, 255 vectos are at level CH, and 222 vectors are at level MM with kernel as polynominal, degree is 2 and cost as 1.
pred_train_tune_svm_poly = predict(svm_pol_degree2, OJ_Split_train)
table(OJ_Split_train$Purchase, pred_train_tune_svm_poly)## pred_train_tune_svm_poly
## CH MM
## CH 449 36
## MM 110 205misclassified2_train_tune_svm_poly<- sum(OJ_Split_train$Purchase != pred_train_tune_svm_poly)
training_error_misclassified_svm_poly <- misclassified2_train_tune_svm_poly / nrow(OJ_Split_train)
print(paste("Training Error Rate:", training_error_misclassified_svm_poly))## [1] "Training Error Rate: 0.1825"pred_svm_rad_fit1_test_tune_poly = predict(svm_pol_degree2, OJ_Split_test)
table(OJ_Split_test$Purchase, pred_svm_rad_fit1_test_tune_poly)## pred_svm_rad_fit1_test_tune_poly
## CH MM
## CH 153 15
## MM 45 57misclassified_svm_rad_tune_test_poly <- sum(OJ_Split_test$Purchase != pred_svm_rad_fit1_test_tune_poly)
test_error_misclassified_svm_poly <- misclassified_svm_rad_tune_test_poly / nrow(OJ_Split_test)
print(paste("Test Error Rate:", test_error_misclassified_svm_poly))## [1] "Test Error Rate: 0.222222222222222"set.seed(1)
tune_svm_poly = tune(svm, Purchase ~ ., data = OJ_Split_train, kernel = "poly", degree = 2, ranges = list(cost = c(0.01,0.1,1,10)))
summary(tune_svm_poly)##
## Parameter tuning of 'svm':
##
## - sampling method: 10-fold cross validation
##
## - best parameters:
## cost
## 10
##
## - best performance: 0.18125
##
## - Detailed performance results:
## cost error dispersion
## 1 0.01 0.39125 0.04210189
## 2 0.10 0.32125 0.05001736
## 3 1.00 0.20250 0.04116363
## 4 10.00 0.18125 0.02779513Interpretation:
The lowest error is at 10 cost, with dispersion(optimal cost) is 0.02779513.
svm_poly_tune_cost_optim = svm(Purchase ~ ., data = OJ_Split_train, kernel = "poly", degree = 2, cost = tune_svm_poly$best.parameters$cost)
train_pred_optim_cost_svm_poly = predict(svm_poly_tune_cost_optim, OJ_Split_train)
table(OJ_Split_train$Purchase, train_pred_optim_cost_svm_poly)## train_pred_optim_cost_svm_poly
## CH MM
## CH 447 38
## MM 82 233misclassified2_train_tune_svm_poly_optim<- sum(OJ_Split_train$Purchase != train_pred_optim_cost_svm_poly)
training_error_misclassified_svm_poly_optim <- misclassified2_train_tune_svm_poly_optim / nrow(OJ_Split_train)
print(paste("Training Error Rate:", training_error_misclassified_svm_poly_optim))## [1] "Training Error Rate: 0.15"pred_svm_fit1_test_tune_poly_optim = predict(svm_poly_tune_cost_optim, OJ_Split_test)
table(OJ_Split_test$Purchase, pred_svm_fit1_test_tune_poly_optim)## pred_svm_fit1_test_tune_poly_optim
## CH MM
## CH 154 14
## MM 37 65misclassified_svm_tune_test_poly_optim <- sum(OJ_Split_test$Purchase != pred_svm_fit1_test_tune_poly_optim)
test_error_misclassified_svm_poly_optim <- misclassified_svm_tune_test_poly_optim / nrow(OJ_Split_test)
print(paste("Test Error Rate:", test_error_misclassified_svm_poly_optim))## [1] "Test Error Rate: 0.188888888888889"(h) Overall, which approach seems to give the best results on this data?
With this dataset, the tuned SVM-linear model provides the best results.