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Question 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.

  1. 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(12345)
x1 = runif(500) - 0.5
x2 = runif(500) - 0.5
y = 1 * (x1^2 - x2^2 > 0)
  1. 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, x2, xlab = "X1", ylab = "X2", col = ifelse(y, 'purple', 'black'))

  1. Fit a logistic regression model to the data, using X1 and X2 as predictors.
data5 =data.frame(x1 = x1, x2 = x2, y=as.factor(y))
glm.fit5 = glm(y~., data=data5, family = 'binomial')
glm.fit5
## 
## Call:  glm(formula = y ~ ., family = "binomial", data = data5)
## 
## Coefficients:
## (Intercept)           x1           x2  
##     0.07563      0.46494      0.41377  
## 
## Degrees of Freedom: 499 Total (i.e. Null);  497 Residual
## Null Deviance:       692.2 
## Residual Deviance: 688   AIC: 694
  1. 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.
set.seed(12345)
train5 <- sample(250, 250)
data.train5 <- data5[train5, ]
data.test5 <- data5[-train5, ]
glm.probs5 = predict(glm.fit5, newdata=data5, type = "response")
glm.preds5 = rep(0,500)
glm.preds5[ glm.probs5 > 0.5 ] = 1
plot(x1,x2,col=ifelse(glm.preds5 > 0, 'purple', 'black'), pch=ifelse(as.integer(glm.preds5>0)== y,3,1))

  1. 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).
set.seed(12345)
glm.fite = glm(y ~ poly(x1, 2) + poly(x2, 2) + I(x1 * x2), data = data5, family = binomial)
## Warning: glm.fit: algorithm did not converge
## Warning: glm.fit: fitted probabilities numerically 0 or 1 occurred
  1. 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.
set.seed(12345)
glm.poly = glm(y ~ poly(x1, 2) + poly(x2, 2) + I(x1 * x2), data = data.train5, family = binomial)
## Warning: glm.fit: algorithm did not converge
## Warning: glm.fit: fitted probabilities numerically 0 or 1 occurred
lm.probe = predict(glm.poly, data5, type = "response")
lm.prede = ifelse(lm.probe > 0.5, 1, 0)
data.pos5f = data5[lm.prede == 1, ]
data.neg5f = data5[lm.prede == 0, ]
plot(data.pos5f$x1, data.pos5f$x2, col = "purple", xlab = "X1", ylab = "X2", pch = 3)
points(data.neg5f$x1, data.neg5f$x2, col = "black", pch = 1)

  1. 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.fitg5 <- svm(as.factor(y) ~ x1 + x2, data5, kernel = "linear", cost = 0.01)
svm.predsg5 <- predict(svm.fitg5, data5)

data.pos5g <- data5[svm.predsg5 == 1, ]
data.neg5g <- data5[svm.predsg5 == 0, ]
plot(data.pos5g$x1, data.pos5g$x2, col = "purple", xlab = "X1", ylab = "X2", pch = 1)
points(data.neg5g$x1, data.neg5g$x2, col = "black", pch = 4)

  1. 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.fit5h <- svm(as.factor(y) ~ x1 + x2, data5, gamma = 1)
svm.preds5h <- predict(svm.fit5h, data5)

data.pos5h <- data5[svm.preds5h == 1, ]
data.neg5h <- data5[svm.preds5h == 0, ]
plot(data.pos5h$x1, data.pos5h$x2, col = "black", xlab = "X1", ylab = "X2", pch = 1)
points(data.neg5h$x1, data.neg5h$x2, col = "purple", pch = 4)

  1. Comment on your results.

Question 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.

  1. 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 4.0.5
data(Auto)
attach(Auto)
str(Auto)
## 'data.frame':    392 obs. of  9 variables:
##  $ mpg         : num  18 15 18 16 17 15 14 14 14 15 ...
##  $ cylinders   : num  8 8 8 8 8 8 8 8 8 8 ...
##  $ displacement: num  307 350 318 304 302 429 454 440 455 390 ...
##  $ horsepower  : num  130 165 150 150 140 198 220 215 225 190 ...
##  $ weight      : num  3504 3693 3436 3433 3449 ...
##  $ acceleration: num  12 11.5 11 12 10.5 10 9 8.5 10 8.5 ...
##  $ year        : num  70 70 70 70 70 70 70 70 70 70 ...
##  $ origin      : num  1 1 1 1 1 1 1 1 1 1 ...
##  $ name        : Factor w/ 304 levels "amc ambassador brougham",..: 49 36 231 14 161 141 54 223 241 2 ...
mpg.auto.med = median(Auto$mpg)
mpg.new = ifelse(Auto$mpg > mpg.auto.med, 1, 0)
Auto$mpg.new = as.factor(mpg.new)
  1. 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(1)
linearsvm7b =tune(svm,mpg.new~.,data=Auto,kernel ="linear",ranges=list(cost=c(0.001, 0.01, 0.1, 1,5,10,100)))
summary(linearsvm7b)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##     1
## 
## - best performance: 0.01025641 
## 
## - Detailed performance results:
##    cost      error dispersion
## 1 1e-03 0.09442308 0.04519425
## 2 1e-02 0.07653846 0.03617137
## 3 1e-01 0.04596154 0.03378238
## 4 1e+00 0.01025641 0.01792836
## 5 5e+00 0.02051282 0.02648194
## 6 1e+01 0.02051282 0.02648194
## 7 1e+02 0.03076923 0.03151981
  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(5)
polysvm7c = tune(svm, mpg.new ~ ., data = Auto, kernel = "polynomial", ranges = list(cost = c(0.1, 1, 5, 10), degree = c(2, 3, 4)))
summary(polysvm7c)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost degree
##    10      2
## 
## - best performance: 0.5738462 
## 
## - Detailed performance results:
##    cost degree     error dispersion
## 1   0.1      2 0.5841026 0.04075521
## 2   1.0      2 0.5841026 0.04075521
## 3   5.0      2 0.5841026 0.04075521
## 4  10.0      2 0.5738462 0.03762359
## 5   0.1      3 0.5841026 0.04075521
## 6   1.0      3 0.5841026 0.04075521
## 7   5.0      3 0.5841026 0.04075521
## 8  10.0      3 0.5841026 0.04075521
## 9   0.1      4 0.5841026 0.04075521
## 10  1.0      4 0.5841026 0.04075521
## 11  5.0      4 0.5841026 0.04075521
## 12 10.0      4 0.5841026 0.04075521
  1. Make some plots to back up your assertions in (b) and (c).
svm.linear = svm(mpg.new ~ ., data = Auto, kernel = "linear", cost = 1)
svm.poly = svm(mpg.new ~ ., data = Auto, kernel = "polynomial", cost = 10, 
    degree = 2)
svm.radial = svm(mpg.new ~ ., data = Auto, kernel = "radial", cost = 10, gamma = 0.01)
plotpairs = function(fit) {
    for (name in names(Auto)[!(names(Auto) %in% c("mpg", "mpg.new", "name"))]) {
        plot(fit, Auto, as.formula(paste("mpg~", name, sep = "")))
    }
}
plotpairs(svm.poly)

plotpairs(svm.radial)

Question 8. This problem involves the OJ data set which is part of the ISLR package.

  1. Create a training set containing a random sample of 800 observations, and a test set containing the remaining observations.
library(ISLR)
attach(OJ)
str(OJ)
## 'data.frame':    1070 obs. of  18 variables:
##  $ Purchase      : Factor w/ 2 levels "CH","MM": 1 1 1 2 1 1 1 1 1 1 ...
##  $ WeekofPurchase: num  237 239 245 227 228 230 232 234 235 238 ...
##  $ StoreID       : num  1 1 1 1 7 7 7 7 7 7 ...
##  $ PriceCH       : num  1.75 1.75 1.86 1.69 1.69 1.69 1.69 1.75 1.75 1.75 ...
##  $ PriceMM       : num  1.99 1.99 2.09 1.69 1.69 1.99 1.99 1.99 1.99 1.99 ...
##  $ DiscCH        : num  0 0 0.17 0 0 0 0 0 0 0 ...
##  $ DiscMM        : num  0 0.3 0 0 0 0 0.4 0.4 0.4 0.4 ...
##  $ SpecialCH     : num  0 0 0 0 0 0 1 1 0 0 ...
##  $ SpecialMM     : num  0 1 0 0 0 1 1 0 0 0 ...
##  $ LoyalCH       : num  0.5 0.6 0.68 0.4 0.957 ...
##  $ SalePriceMM   : num  1.99 1.69 2.09 1.69 1.69 1.99 1.59 1.59 1.59 1.59 ...
##  $ SalePriceCH   : num  1.75 1.75 1.69 1.69 1.69 1.69 1.69 1.75 1.75 1.75 ...
##  $ PriceDiff     : num  0.24 -0.06 0.4 0 0 0.3 -0.1 -0.16 -0.16 -0.16 ...
##  $ Store7        : Factor w/ 2 levels "No","Yes": 1 1 1 1 2 2 2 2 2 2 ...
##  $ PctDiscMM     : num  0 0.151 0 0 0 ...
##  $ PctDiscCH     : num  0 0 0.0914 0 0 ...
##  $ ListPriceDiff : num  0.24 0.24 0.23 0 0 0.3 0.3 0.24 0.24 0.24 ...
##  $ STORE         : num  1 1 1 1 0 0 0 0 0 0 ...
set.seed(1)
intrain <- sample(nrow(OJ), 800)
train8 = OJ[intrain, ]
test8 = OJ[-intrain, ]
  1. 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.linear8 <- svm(Purchase ~ ., data = train8, kernel = "linear", cost = 0.01)
summary(svm.linear8)
## 
## Call:
## svm(formula = Purchase ~ ., data = train8, 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
  1. What are the training and test error rates?
train.pred8 <- predict(svm.linear8, train8)
table(train8$Purchase, train.pred8)
##     train.pred8
##       CH  MM
##   CH 420  65
##   MM  75 240
(75 + 65) / (420 + 240 + 75 + 65)
## [1] 0.175
test_pred <- predict(svm.linear8, test8)
table(test8$Purchase, test_pred)
##     test_pred
##       CH  MM
##   CH 153  15
##   MM  33  69
(33 + 15) / (153 + 69 + 33 + 15)
## [1] 0.1777778
  1. Use the tune() function to select an optimal cost. Consider values in the range 0.01 to 10.
set.seed(1)
tune.op <- tune(svm, Purchase ~ ., data = train8, kernel = "linear", ranges = list(cost = 10^seq(-2, 1, by = 0.25)))
summary(tune.op)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##      cost
##  3.162278
## 
## - best performance: 0.16875 
## 
## - Detailed performance results:
##           cost   error dispersion
## 1   0.01000000 0.17625 0.02853482
## 2   0.01778279 0.17625 0.03143004
## 3   0.03162278 0.17125 0.02829041
## 4   0.05623413 0.17625 0.02853482
## 5   0.10000000 0.17250 0.03162278
## 6   0.17782794 0.17125 0.02829041
## 7   0.31622777 0.17125 0.02889757
## 8   0.56234133 0.17125 0.02703521
## 9   1.00000000 0.17500 0.02946278
## 10  1.77827941 0.17375 0.02729087
## 11  3.16227766 0.16875 0.03019037
## 12  5.62341325 0.17375 0.03304563
## 13 10.00000000 0.17375 0.03197764

Optimal Cost = 3.16

  1. Compute the training and test error rates using this new value for cost.
svm.line <- svm(Purchase ~ ., kernel = "linear", data = train8, cost = tune.op$best.parameter$cost)
train.pred8e <- predict(svm.line, train8)
table(train8$Purchase, train.pred8e)
##     train.pred8e
##       CH  MM
##   CH 423  62
##   MM  70 245
(70 + 62) / (423 + 245 + 70 + 62)
## [1] 0.165
test.pred8e <- predict(svm.line, test8)
table(test8$Purchase, test.pred8e)
##     test.pred8e
##       CH  MM
##   CH 156  12
##   MM  29  73
(29 + 12) / (156 + 73 + 29 + 12)
## [1] 0.1518519
  1. Repeat parts (b) through (e) using a support vector machine with a radial kernel. Use the default value for gamma.
svm.radial8f <- svm(Purchase ~ ., kernel = "radial", data = train8)
summary(svm.radial8f)
## 
## Call:
## svm(formula = Purchase ~ ., data = train8, 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
train.pred8f <- predict(svm.radial8f, train8)
table(train8$Purchase, train.pred8f)
##     train.pred8f
##       CH  MM
##   CH 441  44
##   MM  77 238
(77 + 44) / (441 + 238 + 77 + 44)
## [1] 0.15125
test.pred8f <- predict(svm.radial8f, test8)
table(test8$Purchase, test.pred8f)
##     test.pred8f
##       CH  MM
##   CH 151  17
##   MM  33  69
(33 + 17) / (151 + 69 + 33 + 17)
## [1] 0.1851852
set.seed(1)
tune.op8f <- tune(svm, Purchase ~ ., data = train8, kernel = "radial", ranges = list(cost = 10^seq(-2, 
    1, by = 0.25)))
summary(tune.op8f)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##       cost
##  0.5623413
## 
## - best performance: 0.16875 
## 
## - Detailed performance results:
##           cost   error dispersion
## 1   0.01000000 0.39375 0.04007372
## 2   0.01778279 0.39375 0.04007372
## 3   0.03162278 0.35750 0.05927806
## 4   0.05623413 0.19500 0.02443813
## 5   0.10000000 0.18625 0.02853482
## 6   0.17782794 0.18250 0.03291403
## 7   0.31622777 0.17875 0.03230175
## 8   0.56234133 0.16875 0.02651650
## 9   1.00000000 0.17125 0.02128673
## 10  1.77827941 0.17625 0.02079162
## 11  3.16227766 0.17750 0.02266912
## 12  5.62341325 0.18000 0.02220485
## 13 10.00000000 0.18625 0.02853482
svm.radial8f <- svm(Purchase ~ ., kernel = "radial", data = train8, cost = tune.op8f$best.parameter$cost)
summary(svm.radial8f)
## 
## Call:
## svm(formula = Purchase ~ ., data = train8, kernel = "radial", cost = tune.op8f$best.parameter$cost)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  radial 
##        cost:  0.5623413 
## 
## Number of Support Vectors:  397
## 
##  ( 200 197 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM
train.pred8ff <- predict(svm.radial8f, train8)
table(train8$Purchase, train.pred8ff)
##     train.pred8ff
##       CH  MM
##   CH 437  48
##   MM  71 244
(71 + 48) / (437 + 244 + 48 + 71)
## [1] 0.14875
test.pred8ff <- predict(svm.radial8f, test8)
table(test8$Purchase, test.pred8ff)
##     test.pred8ff
##       CH  MM
##   CH 150  18
##   MM  30  72
(30 + 18) / (150 + 72 + 30 + 18)
## [1] 0.1777778
  1. Repeat parts (b) through (e) using a support vector machine with a polynomial kernel. Set degree=2.
svm.poly8g <- svm(Purchase ~ ., kernel = "polynomial", data = train8, degree = 2)
summary(svm.poly8g)
## 
## Call:
## svm(formula = Purchase ~ ., data = train8, kernel = "polynomial", 
##     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
train.pred8g <- predict(svm.poly8g, train8)
table(train8$Purchase, train.pred8g)
##     train.pred8g
##       CH  MM
##   CH 449  36
##   MM 110 205
(110 + 36) / (449 + 205 + 110 + 36)
## [1] 0.1825
test.pred8gg <- predict(svm.poly8g, test8)
table(test8$Purchase, test.pred8gg)
##     test.pred8gg
##       CH  MM
##   CH 153  15
##   MM  45  57
(45 + 15) / (153 + 57 + 45 +15)
## [1] 0.2222222
set.seed(2)
tune.op8g <- tune(svm, Purchase ~ ., data = train8, kernel = "polynomial", degree = 2, ranges = list(cost = 10^seq(-2, 
    1, by = 0.25)))
summary(tune.op8g)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##      cost
##  3.162278
## 
## - best performance: 0.18 
## 
## - Detailed performance results:
##           cost   error dispersion
## 1   0.01000000 0.39000 0.03670453
## 2   0.01778279 0.37000 0.03395258
## 3   0.03162278 0.36375 0.03197764
## 4   0.05623413 0.34500 0.03291403
## 5   0.10000000 0.32125 0.03866254
## 6   0.17782794 0.24750 0.03322900
## 7   0.31622777 0.20250 0.04073969
## 8   0.56234133 0.20250 0.03670453
## 9   1.00000000 0.19625 0.03910900
## 10  1.77827941 0.19125 0.03586723
## 11  3.16227766 0.18000 0.04005205
## 12  5.62341325 0.18000 0.04133199
## 13 10.00000000 0.18125 0.03830162
svm.poly8g <- svm(Purchase ~ ., kernel = "polynomial", degree = 2, data = train8, cost = tune.op8g$best.parameter$cost)
summary(svm.poly8g)
## 
## Call:
## svm(formula = Purchase ~ ., data = train8, kernel = "polynomial", 
##     degree = 2, cost = tune.op8g$best.parameter$cost)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  polynomial 
##        cost:  3.162278 
##      degree:  2 
##      coef.0:  0 
## 
## Number of Support Vectors:  385
## 
##  ( 197 188 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM
train.pred8t <- predict(svm.poly8g, train8)
table(train8$Purchase, train.pred8t)
##     train.pred8t
##       CH  MM
##   CH 451  34
##   MM  90 225
(90 + 34) / (451 + 225 + 90 + 34)
## [1] 0.155
test.pred8t <- predict(svm.poly8g, test8)
table(test8$Purchase, test.pred8t)
##     test.pred8t
##       CH  MM
##   CH 154  14
##   MM  41  61
(41 + 14)/(154 + 61 + 41 + 14)
## [1] 0.2037037
  1. Overall, which approach seems to give the best results on this data? Radial.