Homework 8

Results

Q5

library(titanic)
library(caret)
library(lattice)
library(ggplot2)
library(gam)
library(car)
library(ROCR)
library(ISLR)
library(e1071)
library(data.table)
library(ggmosaic)

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.

set.seed(1)
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.

dt <- data.table(y = as.factor(y), x1=x1, x2=x2)
ggplot(dt, aes(x=x1, y=x2, col=y)) + geom_point()

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

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

prob <- predict(log.fit, dt, type="response")
log.pred <- ifelse(prob > 0.5, 1, 0)
ggplot(data.table(x1=dt$x1, x2=dt$x2, pred = as.factor(log.pred)), aes(x=x1,y=x2,col=pred)) + geom_point()

e.Now fit a logistic regression model to the data using non-linear functions of X1 and X2 as predictors.

log.poly <- glm(y ~ poly(x1, 2) + poly(x2, 2), family = binomial)

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

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

summary(log.poly)

## 
## Call:
## glm(formula = y ~ poly(x1, 2) + poly(x2, 2), family = binomial)
## 
## Deviance Residuals: 
##        Min          1Q      Median          3Q         Max  
## -1.079e-03  -2.000e-08  -2.000e-08   2.000e-08   1.297e-03  
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)
## (Intercept)     -94.48    2963.78  -0.032    0.975
## poly(x1, 2)1   3442.52  104411.28   0.033    0.974
## poly(x1, 2)2  30110.74  858421.66   0.035    0.972
## poly(x2, 2)1    162.82   26961.99   0.006    0.995
## poly(x2, 2)2 -31383.76  895267.48  -0.035    0.972
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 6.9218e+02  on 499  degrees of freedom
## Residual deviance: 4.2881e-06  on 495  degrees of freedom
## AIC: 10
## 
## 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.

prob <- predict(log.poly, dt, type="response")
log.pred <- ifelse(prob > 0.5, 1, 0)
ggplot(data.table(x1=dt$x1, x2=dt$x2, pred = as.factor(log.pred)), aes(x=x1,y=x2,col=pred)) + geom_point()

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.

svm.fit <- svm(as.factor(y) ~ x1 + x2, dt, kernel = "linear", cost = 1)
svm.pred <- predict(svm.fit, dt)
ggplot(data.table(x1=dt$x1, x2=dt$x2, pred = as.factor(svm.pred)), aes(x=x1,y=x2,col=pred)) + geom_point()

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.fit <- svm(as.factor(y) ~ x1 + x2, dt, gamma = 1)
svm.pred <- predict(svm.fit, dt)
ggplot(data.table(x1=dt$x1, x2=dt$x2, pred = as.factor(svm.pred)), aes(x=x1,y=x2,col=pred)) + geom_point()

i. Comment on your findings

We can see from these models that SVM with a non-linear kernel are great at finding non-linear boundaries.

Q7

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.

gas.med = median(Auto$mpg)
new.var = ifelse(Auto$mpg > gas.med, 1, 0)
Auto$mpglevel = as.factor(new.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.

set.seed(1)
tune.out = tune(svm, mpglevel ~ ., 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
##     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.

set.seed(1)
tune.out = tune(svm, mpglevel ~ ., 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: 0.5130128 
## 
## - Detailed performance results:
##    cost degree     error dispersion
## 1   0.1      2 0.5511538 0.04366593
## 2   1.0      2 0.5511538 0.04366593
## 3   5.0      2 0.5511538 0.04366593
## 4  10.0      2 0.5130128 0.08963366
## 5   0.1      3 0.5511538 0.04366593
## 6   1.0      3 0.5511538 0.04366593
## 7   5.0      3 0.5511538 0.04366593
## 8  10.0      3 0.5511538 0.04366593
## 9   0.1      4 0.5511538 0.04366593
## 10  1.0      4 0.5511538 0.04366593
## 11  5.0      4 0.5511538 0.04366593
## 12 10.0      4 0.5511538 0.04366593

set.seed(1)
tune.out = tune(svm, mpglevel ~ ., 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.out)

## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost gamma
##    10  0.01
## 
## - best performance: 0.02557692 
## 
## - Detailed performance results:
##    cost gamma      error dispersion
## 1   0.1 1e-02 0.08929487 0.04382379
## 2   1.0 1e-02 0.07403846 0.03522110
## 3   5.0 1e-02 0.04852564 0.03303346
## 4  10.0 1e-02 0.02557692 0.02093679
## 5   0.1 1e-01 0.07903846 0.03874545
## 6   1.0 1e-01 0.05371795 0.03525162
## 7   5.0 1e-01 0.02820513 0.03299190
## 8  10.0 1e-01 0.03076923 0.03375798
## 9   0.1 1e+00 0.55115385 0.04366593
## 10  1.0 1e+00 0.06384615 0.04375618
## 11  5.0 1e+00 0.05884615 0.04020934
## 12 10.0 1e+00 0.05884615 0.04020934
## 13  0.1 5e+00 0.55115385 0.04366593
## 14  1.0 5e+00 0.49493590 0.04724924
## 15  5.0 5e+00 0.48217949 0.05470903
## 16 10.0 5e+00 0.48217949 0.05470903
## 17  0.1 1e+01 0.55115385 0.04366593
## 18  1.0 1e+01 0.51794872 0.05063697
## 19  5.0 1e+01 0.51794872 0.04917316
## 20 10.0 1e+01 0.51794872 0.04917316
## 21  0.1 1e+02 0.55115385 0.04366593
## 22  1.0 1e+02 0.55115385 0.04366593
## 23  5.0 1e+02 0.55115385 0.04366593
## 24 10.0 1e+02 0.55115385 0.04366593

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

svm.linear = 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.linear)

plotpairs(svm.radial)

plotpairs(svm.poly)

Q8

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

set.seed(1)
train = sample(dim(OJ)[1], 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.linear = svm(Purchase ~ ., kernel = "linear", data = OJ.train, cost = 0.01)
summary(svm.linear)

## 
## 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

C.What are the training and test error rates?

train.pred = predict(svm.linear, OJ.train)
table(OJ.train$Purchase, train.pred)

##     train.pred
##       CH  MM
##   CH 420  65
##   MM  75 240

test.pred = predict(svm.linear, OJ.test)
table(OJ.test$Purchase, test.pred)

##     test.pred
##       CH  MM
##   CH 153  15
##   MM  33  69

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

set.seed(1554)
tune.out = tune(svm, Purchase ~ ., data = OJ.train, kernel = "linear", ranges = list(cost = 10^seq(-2, 
    1, by = 0.25)))
summary(tune.out)

## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##       cost
##  0.3162278
## 
## - best performance: 0.17125 
## 
## - Detailed performance results:
##           cost   error dispersion
## 1   0.01000000 0.17750 0.06635343
## 2   0.01778279 0.17750 0.05916080
## 3   0.03162278 0.17500 0.06095308
## 4   0.05623413 0.17375 0.06755913
## 5   0.10000000 0.17625 0.06755913
## 6   0.17782794 0.17625 0.06573569
## 7   0.31622777 0.17125 0.06483151
## 8   0.56234133 0.17375 0.06573569
## 9   1.00000000 0.17250 0.06258328
## 10  1.77827941 0.17500 0.06997023
## 11  3.16227766 0.17250 0.06661456
## 12  5.62341325 0.17625 0.07155272
## 13 10.00000000 0.17875 0.07072295

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

svm.linear = svm(Purchase ~ ., kernel = "linear", data = OJ.train, cost = tune.out$best.parameters$cost)
train.pred = predict(svm.linear, OJ.train)
table(OJ.train$Purchase, train.pred)

##     train.pred
##       CH  MM
##   CH 423  62
##   MM  71 244

test.pred = predict(svm.linear, OJ.test)
table(OJ.test$Purchase, test.pred)

##     test.pred
##       CH  MM
##   CH 155  13
##   MM  29  73

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.radial = svm(Purchase ~ ., data = OJ.train, kernel = "radial")
summary(svm.radial)

## 
## 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

train.pred = predict(svm.radial, OJ.train)
table(OJ.train$Purchase, train.pred)

##     train.pred
##       CH  MM
##   CH 441  44
##   MM  77 238

test.pred = predict(svm.radial, OJ.test)
table(OJ.test$Purchase, test.pred)

##     test.pred
##       CH  MM
##   CH 151  17
##   MM  33  69

set.seed(1)
tune.out = tune(svm, Purchase ~ ., data = OJ.train, kernel = "radial", ranges = list(cost = 10^seq(-2, 
    1, by = 0.25)))
summary(tune.out)

## 
## 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.radial = svm(Purchase ~ ., data = OJ.train, kernel = "radial", cost = tune.out$best.parameters$cost)
train.pred = predict(svm.radial, OJ.train)
table(OJ.train$Purchase, train.pred)

##     train.pred
##       CH  MM
##   CH 437  48
##   MM  71 244

h.Overall, which approach seems to give the best results on this data? there appears to be minimun misclaffidication on both the train and testing data.

Overall the radial based kernel seems to be giving the best results within the training and testing data.