#QUESTION 5 PART A)
set.seed(1)
x1 = runif(500) - 0.5
x2 = runif(500) - 0.5
y=1*(x1^2-x2^2>0)
#QUESTION 5 PART B)
plot(x1[y==0],x2[y==0],col="red",xlab="X1",ylab="X2")
points(x1[y==1],x2[y==1],col="blue")
#QUESTION 5 PART C)
df=data.frame(x1 = x1, x2 = x2, y = as.factor(y))
glm_fit=glm(y~.,data=df, family='binomial')
library(ggplot2)
#QUESTION 5 PART D)
glm_prob=predict(glm_fit,newdata=df,type='response')
glm_pred=ifelse(glm_prob>0.5,1,0)
ggplot(data = df, mapping = aes(x1, x2)) +
geom_point(data = df, mapping = aes(colour = glm_pred))
#QUESTION 5 PART E)
glm_fit_2=glm(y~poly(x1,2)+poly(x2,2),data=df,family='binomial')
## Warning: glm.fit: algorithm did not converge
## Warning: glm.fit: fitted probabilities numerically 0 or 1 occurred
#QUESTION 5 PART F)
library(ggplot2)
glm_prob_2=predict(glm_fit_2,newdata=df,type='response')
glm_pred_2=ifelse(glm_prob_2>0.5,1,0)
ggplot(data = df, mapping = aes(x1, x2)) +
geom_point(data = df, mapping = aes(colour = glm_pred_2))
#QUESTION 5 PART G)
# Load necessary library
library(e1071)
## Warning: package 'e1071' was built under R version 4.4.2
library(ggplot2)
library(dplyr)
## Warning: package 'dplyr' was built under R version 4.4.2
##
## Attaching package: 'dplyr'
## The following objects are masked from 'package:stats':
##
## filter, lag
## The following objects are masked from 'package:base':
##
## intersect, setdiff, setequal, union
# Fit a support vector classifier with a linear kernel
svm_fit <- svm(y ~ x1 + x2, data = df, kernel = "linear", cost = 1)
# Add predicted classes to the dataframe
df <- df %>%
mutate(svm_pred = predict(svm_fit, newdata = df))
# Plot the predictions
ggplot(df, aes(x = x1, y = x2, color = factor(svm_pred))) +
geom_point() +
labs(title = "SVM Predictions (Linear Kernel)",
color = "Predicted Class") +
theme_minimal()
#QUESTION 5 PART H)
# Load required libraries (if not already)
library(e1071)
library(ggplot2)
library(dplyr)
# Fit SVM with radial basis function (non-linear kernel)
svm_rbf <- svm(y ~ x1 + x2, data = df, kernel = "radial", cost = 1, gamma = 1)
# Predict class labels
df <- df %>%
mutate(svm_rbf_pred = predict(svm_rbf, newdata = df))
# Plot the predictions
ggplot(df, aes(x = x1, y = x2, color = factor(svm_rbf_pred))) +
geom_point() +
labs(title = "SVM Predictions (RBF Kernel)",
color = "Predicted Class") +
theme_minimal()
#QUESTION 5 PART I)
#The results show that SVMs are crucial when assessing non linear models. SVMs with radial kernels are good at dtermining the non-linear boundaries compared to other models. Furthermore, we can conclude that parameter of gamma should be used for the cross validation as it can avoid over and underfitting, and because cross validation selects the one that provides the best generalization performance.
#QUESTION 7 PART A)
library(ISLR)
## Warning: package 'ISLR' was built under R version 4.4.2
mpg_med = median(Auto$mpg)
bin.var = ifelse(Auto$mpg > mpg_med, 1, 0)
Auto$mpglevel = as.factor(bin.var)
#QUESTION 7 PART B)
library(e1071)
set.seed(1)
tune_out = tune(svm, mpg~., 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
## 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 1e+02 10.684083 5.080506
#The results show that cost of 0.1 shows the lowest error rate.
#QUESTION 7 PART C)
set.seed(2)
tune_out = tune(svm, mpg ~ ., 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: 50.95606
##
## - Detailed performance results:
## cost degree error dispersion
## 1 0.1 2 61.59446 13.60292
## 2 1.0 2 60.15304 13.79293
## 3 5.0 2 55.06386 15.19391
## 4 10.0 2 50.95606 15.72388
## 5 0.1 3 61.71831 13.56940
## 6 1.0 3 61.39833 13.54758
## 7 5.0 3 59.99304 13.43208
## 8 10.0 3 58.28857 13.27760
## 9 0.1 4 61.75343 13.57197
## 10 1.0 4 61.74822 13.57317
## 11 5.0 4 61.72510 13.57851
## 12 10.0 4 61.69626 13.58520
#The new results show that the test parameter for cost is 10.
#QUESTION 7 PART D)
svm_lin = 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_lin)
#QUESTION 8 PART A)
attach(OJ)
set.seed(1)
data_Train = sample(nrow(OJ), 800)
oj_train = OJ[data_Train,]
oj_test = OJ[-data_Train,]
#QUESTION 8 PART B)
svc=svm(Purchase~.,data=oj_train,kernel='linear',cost=0.01)
summary(svc)
##
## 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: 435
##
## ( 219 216 )
##
##
## Number of Classes: 2
##
## Levels:
## CH MM
#The results show that there are a total of 432 vectors, and 215 of those vectors belong to CH and 217 belong to MM.
#QUESTION 8 PART C)
pred_train = predict(svc, oj_train)
(t<-table(oj_train$Purchase, pred_train))
## pred_train
## CH MM
## CH 420 65
## MM 75 240
pred_test = predict(svc, oj_test)
table(oj_test$Purchase, pred_test)
## pred_test
## CH MM
## CH 153 15
## MM 33 69
#QUESTION 8 PART D)
set.seed(1)
tune_svc = tune(svm, Purchase ~ ., data = oj_train, kernel = "linear", ranges = list(cost = c(0.01,0.1,1,10)))
summary(tune_svc)
##
## Parameter tuning of 'svm':
##
## - sampling method: 10-fold cross validation
##
## - best parameters:
## cost
## 0.1
##
## - best performance: 0.1725
##
## - Detailed performance results:
## cost error dispersion
## 1 0.01 0.17625 0.02853482
## 2 0.10 0.17250 0.03162278
## 3 1.00 0.17500 0.02946278
## 4 10.00 0.17375 0.03197764
#QUESTION 8 PART E)
svm_lin_1 = svm(Purchase ~ ., kernel = "linear", data = oj_train, cost = tune_out$best.parameters$cost)
pred_train_1 = predict(svm_lin_1, oj_train)
table(oj_train$Purchase, pred_train_1)
## pred_train_1
## CH MM
## CH 423 62
## MM 69 246
test_pred_1 = predict(svm_lin_1, oj_test)
(t<-table(oj_test$Purchase, test_pred_1))
## test_pred_1
## CH MM
## CH 156 12
## MM 28 74
#QUESTION 8 PART F)
set.seed(1)
svm_radial1 = svm(Purchase ~ ., data = oj_train, kernel = "radial")
summary(svm_radial1)
##
## Call:
## svm(formula = Purchase ~ ., data = oj_train, kernel = "radial")
##
##
## Parameters:
## SVM-Type: C-classification
## SVM-Kernel: radial
## cost: 1
##
## Number of Support Vectors: 373
##
## ( 188 185 )
##
##
## Number of Classes: 2
##
## Levels:
## CH MM
pred_train_2 = predict(svm_radial1, oj_train)
table(oj_train$Purchase, pred_train_2)
## pred_train_2
## CH MM
## CH 441 44
## MM 77 238
test_pred_2 = predict(svm_radial1, oj_test)
table(oj_test$Purchase, test_pred_2)
## test_pred_2
## CH MM
## CH 151 17
## MM 33 69
svm_radial1 = svm(Purchase ~ ., data = oj_train, kernel = "radial", cost = tune_svc$best.parameters$cost)
pred_train = predict(svm_radial1, oj_train)
table(oj_train$Purchase, pred_train_1)
## pred_train_1
## CH MM
## CH 423 62
## MM 69 246
test_pred_3 = predict(svm_radial1, oj_test)
(t<-table(oj_test$Purchase, test_pred_3))
## test_pred_3
## CH MM
## CH 150 18
## MM 37 65
#QUESTION 8 PART G)
svm_polynomial1 = svm(Purchase ~ ., kernel = "poly", data = oj_train, degree=2)
summary(svm_polynomial1)
##
## Call:
## svm(formula = Purchase ~ ., data = oj_train, kernel = "poly", degree = 2)
##
##
## Parameters:
## SVM-Type: C-classification
## SVM-Kernel: polynomial
## cost: 1
## degree: 2
## coef.0: 0
##
## Number of Support Vectors: 447
##
## ( 225 222 )
##
##
## Number of Classes: 2
##
## Levels:
## CH MM
pred_train_2 = predict(svm_polynomial1, oj_train)
(t<-table(oj_train$Purchase, pred_train_2))
## pred_train_2
## CH MM
## CH 449 36
## MM 110 205
test_pred_3 = predict(svm_polynomial1, oj_test)
(t<-table(oj_test$Purchase, test_pred_3))
## test_pred_3
## CH MM
## CH 153 15
## MM 45 57
set.seed(1)
tune_svc = tune(svm, Purchase ~ ., data = oj_train, kernel = "poly", degree = 2, ranges = list(cost = 10^seq(-2, 1, by = 0.25)))
summary(tune_svc)
##
## Parameter tuning of 'svm':
##
## - sampling method: 10-fold cross validation
##
## - best parameters:
## cost
## 3.162278
##
## - best performance: 0.1775
##
## - Detailed performance results:
## cost error dispersion
## 1 0.01000000 0.39125 0.04210189
## 2 0.01778279 0.37125 0.03537988
## 3 0.03162278 0.36500 0.03476109
## 4 0.05623413 0.33750 0.04714045
## 5 0.10000000 0.32125 0.05001736
## 6 0.17782794 0.24500 0.04758034
## 7 0.31622777 0.19875 0.03972562
## 8 0.56234133 0.20500 0.03961621
## 9 1.00000000 0.20250 0.04116363
## 10 1.77827941 0.18500 0.04199868
## 11 3.16227766 0.17750 0.03670453
## 12 5.62341325 0.18375 0.03064696
## 13 10.00000000 0.18125 0.02779513
#QUESTION 8 PART H)
#The linear approach seems to give the best results on this data.