Assignment 8, Chapter 9, Q5,7,8

Q5

library(caret)

## Loading required package: ggplot2

## Loading required package: lattice

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)`

set.seed(1)
x1 <- runif (500) - 0.5
x2 <- runif (500) - 0.5
#Note: Y is the class!
y <- 1 * (x1 ^2 - x2 ^2 > 0)

(b) z

plot(x=x1[y==0], y=x2[y==0], xlab="X1",ylab="X2", col="black")
points(x1[y==1],x2[y==1],col="lightblue")

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

logRegMod.fit <- glm(y~x1+x2, family = 'binomial')
summary(logRegMod.fit)

## 
## Call:
## glm(formula = y ~ x1 + x2, family = "binomial")
## 
## Deviance Residuals: 
##    Min      1Q  Median      3Q     Max  
## -1.179  -1.139  -1.112   1.206   1.257  
## 
## Coefficients:
##              Estimate Std. Error z value Pr(>|z|)
## (Intercept) -0.087260   0.089579  -0.974    0.330
## x1           0.196199   0.316864   0.619    0.536
## x2          -0.002854   0.305712  -0.009    0.993
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 692.18  on 499  degrees of freedom
## Residual deviance: 691.79  on 497  degrees of freedom
## AIC: 697.79
## 
## 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.

df <-  data.frame(x1 = x1, x2 = x2, y = y)
lm.prob <-  predict(logRegMod.fit, newdata=df, type = "response")
lm.pred  <-  ifelse(lm.prob > 0.50, 1, 0)
ggplot(data = df, mapping = aes(x1, x2)) +
    geom_point(data = df, mapping = aes(colour = lm.pred))

(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).

logRegMod.fit2 <-  glm(y ~ poly(x1, 3) + poly(x2, 3), data = df, family = binomial)

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

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

summary(logRegMod.fit2)

## 
## Call:
## glm(formula = y ~ poly(x1, 3) + poly(x2, 3), family = binomial, 
##     data = df)
## 
## Deviance Residuals: 
##        Min          1Q      Median          3Q         Max  
## -8.116e-04  -2.000e-08  -2.000e-08   2.000e-08   1.098e-03  
## 
## Coefficients:
##              Estimate Std. Error z value Pr(>|z|)
## (Intercept)    -112.8     7700.8  -0.015    0.988
## poly(x1, 3)1   2320.9   202067.7   0.011    0.991
## poly(x1, 3)2  26012.9   817440.0   0.032    0.975
## poly(x1, 3)3   -238.4   100966.6  -0.002    0.998
## poly(x2, 3)1    200.2    85132.7   0.002    0.998
## poly(x2, 3)2 -27719.8   872877.7  -0.032    0.975
## poly(x2, 3)3    387.8    62341.3   0.006    0.995
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 6.9218e+02  on 499  degrees of freedom
## Residual deviance: 3.3141e-06  on 493  degrees of freedom
## AIC: 14
## 
## Number of Fisher Scoring iterations: 25

(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.

lm.prob2 <-  predict(logRegMod.fit2, newdata=df, type = "response")
lm.pred2  <-  ifelse(lm.prob2 > 0.50, 1, 0)
ggplot(data = df, mapping = aes(x1, x2)) +
    geom_point(data = df, mapping = aes(colour = lm.pred2))

(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_lin <- svm(y~.,data=df,kernel='linear',cost=0.01)
plot(svm_lin,df)

(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_lin_2=svm(y~.,data=df,kernel='radial',gamma=1)
plot(svm_lin_2,data=df)

Q7

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.

library(ISLR)
attach(Auto)

## The following object is masked from package:ggplot2:
## 
##     mpg

(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$MPG_Abv_Med <- as.factor(ifelse(Auto$mpg > median(Auto$mpg),1,0))

(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)
library(e1071)
tune.7b = tune(svm, MPG_Abv_Med ~ ., data = Auto, kernel = "linear", ranges = list(cost = c(0.01, 0.1, 1, 5, 10, 100)))
summary(tune.7b)

## 
## 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-02 0.07653846 0.03617137
## 2 1e-01 0.04596154 0.03378238
## 3 1e+00 0.01025641 0.01792836
## 4 5e+00 0.02051282 0.02648194
## 5 1e+01 0.02051282 0.02648194
## 6 1e+02 0.03076923 0.03151981

(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.

tune.7c1 = tune(svm, MPG_Abv_Med ~ ., data = Auto, kernel = "polynomial", ranges = list(cost = c(0.1, 1, 5, 10), degree = c(2, 3, 4)))
summary(tune.7c1)

## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost degree
##    10      2
## 
## - best performance: 0.5841667 
## 
## - Detailed performance results:
##    cost degree     error dispersion
## 1   0.1      2 0.6019231 0.06346118
## 2   1.0      2 0.6019231 0.06346118
## 3   5.0      2 0.6019231 0.06346118
## 4  10.0      2 0.5841667 0.07806609
## 5   0.1      3 0.6019231 0.06346118
## 6   1.0      3 0.6019231 0.06346118
## 7   5.0      3 0.6019231 0.06346118
## 8  10.0      3 0.6019231 0.06346118
## 9   0.1      4 0.6019231 0.06346118
## 10  1.0      4 0.6019231 0.06346118
## 11  5.0      4 0.6019231 0.06346118
## 12 10.0      4 0.6019231 0.06346118

tune.7c2 = tune(svm, MPG_Abv_Med ~ ., data = Auto, kernel = "radial", ranges = list(cost = c(0.1, 1, 5, 10), gamma = c(0.01, 0.1, 1, 5, 10, 100)))
summary(tune.7c2)

## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost gamma
##    10  0.01
## 
## - best performance: 0.02044872 
## 
## - Detailed performance results:
##    cost gamma      error dispersion
## 1   0.1 1e-02 0.08698718 0.04560056
## 2   1.0 1e-02 0.07160256 0.03373099
## 3   5.0 1e-02 0.05121795 0.02967002
## 4  10.0 1e-02 0.02044872 0.01077927
## 5   0.1 1e-01 0.07673077 0.03419344
## 6   1.0 1e-01 0.05121795 0.03203768
## 7   5.0 1e-01 0.02557692 0.01709522
## 8  10.0 1e-01 0.02301282 0.02244393
## 9   0.1 1e+00 0.59461538 0.08083319
## 10  1.0 1e+00 0.06141026 0.03026776
## 11  5.0 1e+00 0.06397436 0.02789391
## 12 10.0 1e+00 0.06397436 0.02789391
## 13  0.1 5e+00 0.59461538 0.08083319
## 14  1.0 5e+00 0.52051282 0.09421163
## 15  5.0 5e+00 0.51538462 0.10051415
## 16 10.0 5e+00 0.51538462 0.10051415
## 17  0.1 1e+01 0.59461538 0.08083319
## 18  1.0 1e+01 0.55384615 0.09432787
## 19  5.0 1e+01 0.54358974 0.09085645
## 20 10.0 1e+01 0.54358974 0.09085645
## 21  0.1 1e+02 0.59461538 0.08083319
## 22  1.0 1e+02 0.59461538 0.08083319
## 23  5.0 1e+02 0.59461538 0.08083319
## 24 10.0 1e+02 0.59461538 0.08083319

(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`.

svm.linear  <-  svm(MPG_Abv_Med ~ ., data = Auto, kernel = "linear", cost = 1)
svm.poly <-  svm(MPG_Abv_Med ~ ., data = Auto, kernel = "polynomial", cost = 10, degree = 2)
svm.radial <-  svm(MPG_Abv_Med ~ ., data = Auto, kernel = "radial", cost = 10, gamma = 0.01)
plotpairs <-  function(fit) {
    for (name in names(Auto)[!(names(Auto) %in% c("mpg", "MPG_Abv_Med", "name"))]) {
        plot(fit, Auto, as.formula(paste("mpg~", name, sep = "")))
    }
}
plotpairs(svm.linear)

Q8

This problem involves the `OJ` data set which is part of the `ISLR2` package.

library(ISLR2)

## 
## Attaching package: 'ISLR2'

## The following object is masked _by_ '.GlobalEnv':
## 
##     Auto

## The following objects are masked from 'package:ISLR':
## 
##     Auto, Credit

attach(OJ)
set.seed(1)

(a) Create a training set containing a random sample of 800 observations, and a test set containing the remaining observations.

train  <-  sample(1:nrow(OJ), 800)
oj.train  <-  OJ[train,]
oj.test <-  OJ[-train,]

(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.8b  <-  svm(Purchase ~ ., data = oj.train, kernel = "linear", cost = 0.01)
summary(svm.8b)

## 
## 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

435 Support vectors were made, with CH and MM being the main variables to split.

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

train.pred<-predict(svm.8b, oj.train)
test.pred<-predict(svm.8b, oj.test)
mean(train.pred != oj.train$Purchase)

## [1] 0.175

mean(test.pred != oj.test$Purchase)

## [1] 0.1777778

Both error raters were just over 17%

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

tune.8d = tune(svm, Purchase ~ ., data = oj.train, kernel = "linear", ranges = list(cost = seq(0.01, 10, by = 0.1)))
summary(tune.8d)

## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##  6.51
## 
## - best performance: 0.17 
## 
## - Detailed performance results:
##     cost   error dispersion
## 1   0.01 0.17375 0.03884174
## 2   0.11 0.18000 0.03184162
## 3   0.21 0.18000 0.03291403
## 4   0.31 0.17875 0.03438447
## 5   0.41 0.17875 0.03438447
## 6   0.51 0.17625 0.03197764
## 7   0.61 0.17625 0.03197764
## 8   0.71 0.17625 0.03197764
## 9   0.81 0.17625 0.03197764
## 10  0.91 0.17625 0.03197764
## 11  1.01 0.17500 0.03061862
## 12  1.11 0.17500 0.03061862
## 13  1.21 0.17500 0.03061862
## 14  1.31 0.17500 0.03061862
## 15  1.41 0.17500 0.03061862
## 16  1.51 0.17500 0.03061862
## 17  1.61 0.17500 0.03061862
## 18  1.71 0.17500 0.03061862
## 19  1.81 0.17375 0.02972676
## 20  1.91 0.17375 0.02972676
## 21  2.01 0.17375 0.02972676
## 22  2.11 0.17375 0.03030516
## 23  2.21 0.17375 0.03030516
## 24  2.31 0.17375 0.03030516
## 25  2.41 0.17375 0.03304563
## 26  2.51 0.17250 0.03374743
## 27  2.61 0.17125 0.03230175
## 28  2.71 0.17125 0.03230175
## 29  2.81 0.17250 0.03106892
## 30  2.91 0.17125 0.03175973
## 31  3.01 0.17250 0.03270236
## 32  3.11 0.17125 0.03175973
## 33  3.21 0.17250 0.03270236
## 34  3.31 0.17250 0.03270236
## 35  3.41 0.17250 0.03270236
## 36  3.51 0.17375 0.03408018
## 37  3.61 0.17250 0.03322900
## 38  3.71 0.17250 0.03322900
## 39  3.81 0.17250 0.03322900
## 40  3.91 0.17250 0.03322900
## 41  4.01 0.17250 0.03322900
## 42  4.11 0.17250 0.03322900
## 43  4.21 0.17250 0.03322900
## 44  4.31 0.17250 0.03322900
## 45  4.41 0.17250 0.03322900
## 46  4.51 0.17250 0.03322900
## 47  4.61 0.17250 0.03322900
## 48  4.71 0.17250 0.03322900
## 49  4.81 0.17250 0.03322900
## 50  4.91 0.17250 0.03322900
## 51  5.01 0.17250 0.03322900
## 52  5.11 0.17250 0.03322900
## 53  5.21 0.17250 0.03322900
## 54  5.31 0.17250 0.03322900
## 55  5.41 0.17250 0.03322900
## 56  5.51 0.17250 0.03322900
## 57  5.61 0.17250 0.03322900
## 58  5.71 0.17250 0.03322900
## 59  5.81 0.17250 0.03322900
## 60  5.91 0.17250 0.03322900
## 61  6.01 0.17125 0.03438447
## 62  6.11 0.17125 0.03438447
## 63  6.21 0.17125 0.03438447
## 64  6.31 0.17125 0.03438447
## 65  6.41 0.17125 0.03438447
## 66  6.51 0.17000 0.03593976
## 67  6.61 0.17125 0.03438447
## 68  6.71 0.17125 0.03438447
## 69  6.81 0.17125 0.03438447
## 70  6.91 0.17125 0.03438447
## 71  7.01 0.17125 0.03438447
## 72  7.11 0.17000 0.03593976
## 73  7.21 0.17125 0.03438447
## 74  7.31 0.17125 0.03438447
## 75  7.41 0.17125 0.03438447
## 76  7.51 0.17125 0.03438447
## 77  7.61 0.17125 0.03438447
## 78  7.71 0.17125 0.03438447
## 79  7.81 0.17125 0.03438447
## 80  7.91 0.17125 0.03438447
## 81  8.01 0.17000 0.03593976
## 82  8.11 0.17000 0.03593976
## 83  8.21 0.17000 0.03593976
## 84  8.31 0.17000 0.03593976
## 85  8.41 0.17000 0.03593976
## 86  8.51 0.17000 0.03593976
## 87  8.61 0.17000 0.03593976
## 88  8.71 0.17000 0.03593976
## 89  8.81 0.17125 0.03488573
## 90  8.91 0.17125 0.03488573
## 91  9.01 0.17125 0.03488573
## 92  9.11 0.17125 0.03488573
## 93  9.21 0.17125 0.03488573
## 94  9.31 0.17125 0.03488573
## 95  9.41 0.17125 0.03488573
## 96  9.51 0.17125 0.03488573
## 97  9.61 0.17125 0.03488573
## 98  9.71 0.17125 0.03488573
## 99  9.81 0.17125 0.03488573
## 100 9.91 0.17125 0.03488573

tune.8d$best.parameters$cost

## [1] 6.51

The best performance is at an error rate of 0.1675, which occurss at multiple cost values between 3.81 and 4.41. Using the code tune.8d$best.parameters$cost, we get a cost value of 3.81, which I will use going forward.

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

svm.8e<-svm(Purchase ~ ., kernel='linear', data=oj.train, cost=tune.8d$best.parameters$cost)
train.pred<-predict(svm.8e, oj.train)
test.pred<-predict(svm.8e, oj.test)
mean(train.pred != oj.train$Purchase)

## [1] 0.1625

mean(test.pred != oj.test$Purchase)

## [1] 0.1518519

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

#8b repeat
svm.8b  <-  svm(Purchase ~ ., data = oj.train, kernel = "radial", cost = 0.01)
summary(svm.8b)

## 
## Call:
## svm(formula = Purchase ~ ., data = oj.train, kernel = "radial", cost = 0.01)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  radial 
##        cost:  0.01 
## 
## Number of Support Vectors:  634
## 
##  ( 319 315 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM

#8c repeat
train.pred<-predict(svm.8b, oj.train)
test.pred<-predict(svm.8b, oj.test)
mean(train.pred != oj.train$Purchase)

## [1] 0.39375

mean(test.pred != oj.test$Purchase)

## [1] 0.3777778

#8d repeat
tune.8d = tune(svm, Purchase ~ ., data = oj.train, kernel = "radial", ranges = list(cost = seq(0.01, 10, by = 0.1)))
summary(tune.8d)

## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##  0.71
## 
## - best performance: 0.1675 
## 
## - Detailed performance results:
##     cost   error dispersion
## 1   0.01 0.39375 0.06568284
## 2   0.11 0.18125 0.04759858
## 3   0.21 0.17500 0.04487637
## 4   0.31 0.17375 0.04505013
## 5   0.41 0.17375 0.04543387
## 6   0.51 0.17375 0.04767147
## 7   0.61 0.16750 0.04495368
## 8   0.71 0.16750 0.04005205
## 9   0.81 0.17125 0.03821086
## 10  0.91 0.17250 0.04031129
## 11  1.01 0.17625 0.03793727
## 12  1.11 0.17625 0.03793727
## 13  1.21 0.17375 0.04016027
## 14  1.31 0.17500 0.03864008
## 15  1.41 0.17875 0.03682259
## 16  1.51 0.17750 0.03717451
## 17  1.61 0.18000 0.03641962
## 18  1.71 0.17750 0.03622844
## 19  1.81 0.17750 0.03622844
## 20  1.91 0.17625 0.03606033
## 21  2.01 0.17625 0.03606033
## 22  2.11 0.17625 0.03606033
## 23  2.21 0.17875 0.03910900
## 24  2.31 0.17875 0.04084609
## 25  2.41 0.17875 0.04084609
## 26  2.51 0.18000 0.04048319
## 27  2.61 0.18125 0.03919768
## 28  2.71 0.18125 0.03919768
## 29  2.81 0.18125 0.03919768
## 30  2.91 0.18125 0.03919768
## 31  3.01 0.18500 0.03987829
## 32  3.11 0.18500 0.03987829
## 33  3.21 0.18250 0.03917553
## 34  3.31 0.18250 0.03917553
## 35  3.41 0.18375 0.03998698
## 36  3.51 0.18125 0.04379958
## 37  3.61 0.18125 0.04379958
## 38  3.71 0.18125 0.04379958
## 39  3.81 0.18000 0.04297932
## 40  3.91 0.18000 0.04297932
## 41  4.01 0.18000 0.04297932
## 42  4.11 0.18125 0.04299952
## 43  4.21 0.18000 0.04377975
## 44  4.31 0.18000 0.04377975
## 45  4.41 0.18000 0.04377975
## 46  4.51 0.18125 0.04497299
## 47  4.61 0.18250 0.04417453
## 48  4.71 0.18375 0.04168749
## 49  4.81 0.18250 0.04133199
## 50  4.91 0.18250 0.04133199
## 51  5.01 0.18125 0.04299952
## 52  5.11 0.18250 0.04090979
## 53  5.21 0.18250 0.04090979
## 54  5.31 0.18250 0.04090979
## 55  5.41 0.18250 0.04090979
## 56  5.51 0.18250 0.04090979
## 57  5.61 0.18250 0.04090979
## 58  5.71 0.18250 0.04090979
## 59  5.81 0.18125 0.04093101
## 60  5.91 0.18125 0.04093101
## 61  6.01 0.18125 0.04093101
## 62  6.11 0.18125 0.04093101
## 63  6.21 0.18000 0.04297932
## 64  6.31 0.18000 0.04297932
## 65  6.41 0.18000 0.04297932
## 66  6.51 0.18000 0.04297932
## 67  6.61 0.18125 0.04135299
## 68  6.71 0.18125 0.04135299
## 69  6.81 0.18125 0.04135299
## 70  6.91 0.18125 0.04135299
## 71  7.01 0.18125 0.04135299
## 72  7.11 0.18125 0.04135299
## 73  7.21 0.18125 0.04135299
## 74  7.31 0.18125 0.04007372
## 75  7.41 0.18125 0.04007372
## 76  7.51 0.18125 0.04007372
## 77  7.61 0.18125 0.04007372
## 78  7.71 0.18000 0.04005205
## 79  7.81 0.18000 0.04005205
## 80  7.91 0.18000 0.04005205
## 81  8.01 0.18000 0.04005205
## 82  8.11 0.18000 0.04005205
## 83  8.21 0.17875 0.04126894
## 84  8.31 0.17875 0.04126894
## 85  8.41 0.17875 0.04126894
## 86  8.51 0.17875 0.04126894
## 87  8.61 0.17875 0.04126894
## 88  8.71 0.17875 0.04126894
## 89  8.81 0.17875 0.04126894
## 90  8.91 0.17875 0.04126894
## 91  9.01 0.17875 0.04126894
## 92  9.11 0.17875 0.04126894
## 93  9.21 0.17875 0.04126894
## 94  9.31 0.17875 0.04126894
## 95  9.41 0.18000 0.04216370
## 96  9.51 0.18000 0.04216370
## 97  9.61 0.18125 0.04340139
## 98  9.71 0.18125 0.04340139
## 99  9.81 0.18125 0.04340139
## 100 9.91 0.18125 0.04340139

tune.8d$best.parameters$cost

## [1] 0.71

#8e repeat
svm.8e<-svm(Purchase ~ ., kernel='radial', data=oj.train, cost=tune.8d$best.parameters$cost)
train.pred<-predict(svm.8e, oj.train)
test.pred<-predict(svm.8e, oj.test)
mean(train.pred != oj.train$Purchase)

## [1] 0.15125

mean(test.pred != oj.test$Purchase)

## [1] 0.1851852

8b repeat: 634 Support vectors were made, with CH and MM being the main variables to split.

8c repeat: Both error rates were just at 39% and 37%

8d repeat: Using the code tune.8d$best.parameters$cost, we get a best cost value of 0.61 with an error rate of .18375 which I will use going forward.

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

#8b repeat
svm.8b  <-  svm(Purchase ~ ., data = oj.train, kernel = "polynomial", cost = 0.01, degree=2)
summary(svm.8b)

## 
## Call:
## svm(formula = Purchase ~ ., data = oj.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:  636
## 
##  ( 321 315 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM

#8c repeat
train.pred<-predict(svm.8b, oj.train)
test.pred<-predict(svm.8b, oj.test)
mean(train.pred != oj.train$Purchase)

## [1] 0.3725

mean(test.pred != oj.test$Purchase)

## [1] 0.3666667

#8d repeat
tune.8d = tune(svm, Purchase ~ ., data = oj.train, kernel = "polynomial", degree=2, ranges = list(cost = seq(0.01, 10, by = 0.1)))
summary(tune.8d)

## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##  5.11
## 
## - best performance: 0.1825 
## 
## - Detailed performance results:
##     cost   error dispersion
## 1   0.01 0.39000 0.08287373
## 2   0.11 0.31125 0.06958458
## 3   0.21 0.23000 0.06826216
## 4   0.31 0.20875 0.06209592
## 5   0.41 0.20500 0.05470883
## 6   0.51 0.20750 0.05809475
## 7   0.61 0.20375 0.05466120
## 8   0.71 0.21000 0.05916080
## 9   0.81 0.20625 0.05628857
## 10  0.91 0.20500 0.05627314
## 11  1.01 0.20375 0.05653477
## 12  1.11 0.20375 0.05923412
## 13  1.21 0.20125 0.05726704
## 14  1.31 0.19500 0.05749396
## 15  1.41 0.19375 0.05311479
## 16  1.51 0.19125 0.05529278
## 17  1.61 0.19250 0.05898446
## 18  1.71 0.19375 0.05899918
## 19  1.81 0.19500 0.05809475
## 20  1.91 0.19375 0.05781015
## 21  2.01 0.19250 0.06043821
## 22  2.11 0.19125 0.05804991
## 23  2.21 0.18750 0.06123724
## 24  2.31 0.18625 0.06248611
## 25  2.41 0.18750 0.06236096
## 26  2.51 0.18750 0.06152010
## 27  2.61 0.18625 0.05964304
## 28  2.71 0.18625 0.05964304
## 29  2.81 0.18750 0.05833333
## 30  2.91 0.18625 0.05816941
## 31  3.01 0.18875 0.05756940
## 32  3.11 0.19000 0.06061032
## 33  3.21 0.18750 0.05833333
## 34  3.31 0.18625 0.05905800
## 35  3.41 0.18750 0.05803495
## 36  3.51 0.18625 0.05816941
## 37  3.61 0.18625 0.05816941
## 38  3.71 0.18625 0.05816941
## 39  3.81 0.18750 0.05621141
## 40  3.91 0.18750 0.05621141
## 41  4.01 0.18500 0.06202598
## 42  4.11 0.18500 0.06202598
## 43  4.21 0.18750 0.06038074
## 44  4.31 0.18750 0.06038074
## 45  4.41 0.18750 0.06152010
## 46  4.51 0.18500 0.05676462
## 47  4.61 0.18500 0.05676462
## 48  4.71 0.18625 0.05510407
## 49  4.81 0.18500 0.05329426
## 50  4.91 0.18375 0.05104804
## 51  5.01 0.18375 0.05104804
## 52  5.11 0.18250 0.04901814
## 53  5.21 0.18250 0.04901814
## 54  5.31 0.18250 0.04901814
## 55  5.41 0.18250 0.04901814
## 56  5.51 0.18250 0.04901814
## 57  5.61 0.18375 0.04825065
## 58  5.71 0.18375 0.04825065
## 59  5.81 0.18375 0.04825065
## 60  5.91 0.18500 0.04706674
## 61  6.01 0.18375 0.04788949
## 62  6.11 0.18375 0.04788949
## 63  6.21 0.18375 0.04788949
## 64  6.31 0.18375 0.04788949
## 65  6.41 0.18375 0.04788949
## 66  6.51 0.18375 0.04788949
## 67  6.61 0.18375 0.04788949
## 68  6.71 0.18375 0.04788949
## 69  6.81 0.18375 0.04788949
## 70  6.91 0.18375 0.04788949
## 71  7.01 0.18250 0.05041494
## 72  7.11 0.18250 0.05041494
## 73  7.21 0.18375 0.05239076
## 74  7.31 0.18375 0.05239076
## 75  7.41 0.18375 0.05239076
## 76  7.51 0.18375 0.05239076
## 77  7.61 0.18375 0.05239076
## 78  7.71 0.18375 0.05239076
## 79  7.81 0.18375 0.05239076
## 80  7.91 0.18375 0.05239076
## 81  8.01 0.18500 0.05230785
## 82  8.11 0.18375 0.05239076
## 83  8.21 0.18500 0.04993051
## 84  8.31 0.18500 0.04993051
## 85  8.41 0.18375 0.05036326
## 86  8.51 0.18375 0.05036326
## 87  8.61 0.18375 0.05205833
## 88  8.71 0.18375 0.05205833
## 89  8.81 0.18375 0.05205833
## 90  8.91 0.18375 0.05205833
## 91  9.01 0.18375 0.05205833
## 92  9.11 0.18375 0.05205833
## 93  9.21 0.18500 0.05130248
## 94  9.31 0.18500 0.05130248
## 95  9.41 0.18500 0.05130248
## 96  9.51 0.18500 0.05130248
## 97  9.61 0.18625 0.05084358
## 98  9.71 0.18625 0.05084358
## 99  9.81 0.18500 0.04958158
## 100 9.91 0.18500 0.04958158

tune.8d$best.parameters$cost

## [1] 5.11

#8e repeat
svm.8e<-svm(Purchase ~ ., kernel='polynomial', degree=2, data=oj.train, cost=tune.8d$best.parameters$cost)
train.pred<-predict(svm.8e, oj.train)
test.pred<-predict(svm.8e, oj.test)
mean(train.pred != oj.train$Purchase)

## [1] 0.15625

mean(test.pred != oj.test$Purchase)

## [1] 0.1814815

8b repeat: 636 Support vectors were made, with CH and MM being the main variables to split.

8c repeat: Both error rates were just at 37% and 36%

8d repeat: Using the code tune.8d$best.parameters$cost, we get a best cost value of 2.11 with an error rate of .18 which I will use going forward.

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

The linear approach was best in my case as it had the lowest error rate for training and test with the smallest difference between them. Linear had 16% and 15%, Radial had 14% and 18%, and polynomial had 15% and 21%

detach(OJ)

Assignment 8, Chapter 9, Q5,7,8

Garrett Garceau

2023-03-31

Q5

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) z

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

(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.

(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).

(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.

(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.

Q7

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.

(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.

(d) Make some plots to back up your assertions in (b) and (c).

Q8

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.

(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.

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

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

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

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

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

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

Assignment 8, Chapter 9, Q5,7,8

Garrett Garceau

2023-03-31

Q5

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) z

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

(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.

(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).

(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.

(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.

Q7

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.

(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.

(d) Make some plots to back up your assertions in (b) and (c).

Q8

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.

(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.

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

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

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

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

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

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

`x1 <- runif (500) - 0.5`

`x2 <- runif (500) - 0.5`

`y <- 1 * (x1 ^2 - x2 ^2 > 0)`

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.

(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.

This problem involves the `OJ` data set which is part of the `ISLR2` package.

(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.

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

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

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

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