Data Mining Homework 8

Chapter 09 (page 368): 5, 7, 8

Problem 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.

Part 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*(x12-x22 > 0)

set.seed(421)
x1 = runif(500) - 0.5
x2 = runif(500) - 0.5
y = 1 * (x1^2 - x2^2 > 0)

Part B

Plot the observations, colored according to their class labels. Your plot should display \(X_1\) on the x-axis, and \(X_2\) 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)

The plot does appear to indicate a non-linear boundry decision.

Part C

Fit a logistic regression model to the data, using \(X_1\) and \(X_2\) as predictors.

lm.fit = glm(y ~ x1 + x2, family = binomial)
summary(lm.fit)
## 
## Call:
## glm(formula = y ~ x1 + x2, family = binomial)
## 
## Deviance Residuals: 
##    Min      1Q  Median      3Q     Max  
## -1.278  -1.227   1.089   1.135   1.175  
## 
## Coefficients:
##             Estimate Std. Error z value Pr(>|z|)
## (Intercept)  0.11999    0.08971   1.338    0.181
## x1          -0.16881    0.30854  -0.547    0.584
## x2          -0.08198    0.31476  -0.260    0.795
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 691.35  on 499  degrees of freedom
## Residual deviance: 690.99  on 497  degrees of freedom
## AIC: 696.99
## 
## Number of Fisher Scoring iterations: 3

Both variables are insignificant for predicting y.

Part 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.

data = data.frame(x1 = x1, x2 = x2, y = y)
lm.prob = predict(lm.fit, data, type = "response")
lm.pred = ifelse(lm.prob > 0.52, 1, 0)
data.pos = data[lm.pred == 1, ]
data.neg = data[lm.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)

With the given model and a probability threshold of 0.5, all points are classified to single class and no decision boundary can be shown. Hence we shift the probability threshold to 0.52 to show a meaningful decision boundary. This boundary is linear as seen in the figure.

Part E

Now fit a logistic regression model to the data using non-linear functions of \(X_1\) and \(X_2\) as predictors (e.g. \(X^2_1\) , \(X^1×X^2\), log(\(X_2\)), and so forth).

lm.fit = glm(y ~ poly(x1, 2) + poly(x2, 2) + I(x1 * x2), data = data, family = binomial)
## Warning: glm.fit: algorithm did not converge
## Warning: glm.fit: fitted probabilities numerically 0 or 1 occurred

Part 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.prob = predict(lm.fit, data, type = "response")
lm.pred = ifelse(lm.prob > 0.5, 1, 0)
data.pos = data[lm.pred == 1, ]
data.neg = data[lm.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)

This non-linear decision boundary closely resembles the true decision boundary.

Part G

Fit a support vector classifier to the data with \(X_1\) and \(X_2\) as predictors. Obtain a class prediction for each training observation. Plot the observations, colored according to the predicted class labels.

library(e1071)
## Warning: package 'e1071' was built under R version 3.5.3
svm.fit = svm(as.factor(y) ~ x1 + x2, data, kernel = "linear", cost = 0.1)
svm.pred = predict(svm.fit, data)
data.pos = data[svm.pred == 1, ]
data.neg = 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)

A linear kernel, even with low cost fails to find non-linear decision boundary and classifies all points to a single class.

Part 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, data, gamma = 1)
svm.pred = predict(svm.fit, data)
data.pos = data[svm.pred == 1, ]
data.neg = 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)

Again, the non-linear decision boundary on predicted labels closely resembles the true decision boundary.

Part I

Comment on your results.

This experiment enforces the idea that SVMs with non-linear kernel are extremely powerful in finding non-linear boundary. Both, logistic regression with non-interactions and SVMs with linear kernels fail to find the decision boundary. Adding interaction terms to logistic regression seems to give them same power as radial-basis kernels. However, there is some manual efforts and tuning involved in picking right interaction terms. This effort can become prohibitive with large number of features. Radial basis kernels, on the other hand, only require tuning of one parameter - gamma - which can be easily done using cross-validation.

Problem 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.

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

Part 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.

library(e1071)
set.seed(3255)
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: 0.01282051 
## 
## - Detailed performance results:
##    cost      error dispersion
## 1 1e-02 0.07384615 0.04219942
## 2 1e-01 0.04083333 0.03008810
## 3 1e+00 0.01282051 0.02179068
## 4 5e+00 0.01538462 0.02477158
## 5 1e+01 0.02044872 0.02354784
## 6 1e+02 0.03070513 0.02357884

We see, using Cross-Validation that to minmize the error of cost we get cost = 1.

Part 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(21)
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: 0.5508974 
## 
## - Detailed performance results:
##    cost degree     error dispersion
## 1   0.1      2 0.5867308 0.07310319
## 2   1.0      2 0.5867308 0.07310319
## 3   5.0      2 0.5867308 0.07310319
## 4  10.0      2 0.5508974 0.11697667
## 5   0.1      3 0.5867308 0.07310319
## 6   1.0      3 0.5867308 0.07310319
## 7   5.0      3 0.5867308 0.07310319
## 8  10.0      3 0.5867308 0.07310319
## 9   0.1      4 0.5867308 0.07310319
## 10  1.0      4 0.5867308 0.07310319
## 11  5.0      4 0.5867308 0.07310319
## 12 10.0      4 0.5867308 0.07310319

The lowest cross-validation error is obtained for cost=10 and degree=2.

set.seed(463)
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: 0.02288462 
## 
## - Detailed performance results:
##    cost gamma      error dispersion
## 1   0.1 1e-02 0.08916667 0.04526330
## 2   1.0 1e-02 0.07397436 0.03896185
## 3   5.0 1e-02 0.05102564 0.03813274
## 4  10.0 1e-02 0.02288462 0.03286718
## 5   0.1 1e-01 0.07903846 0.04724112
## 6   1.0 1e-01 0.05602564 0.03950993
## 7   5.0 1e-01 0.02801282 0.02231663
## 8  10.0 1e-01 0.02551282 0.02093755
## 9   0.1 1e+00 0.55102564 0.03813274
## 10  1.0 1e+00 0.06365385 0.04199145
## 11  5.0 1e+00 0.06108974 0.04358351
## 12 10.0 1e+00 0.06108974 0.04358351
## 13  0.1 5e+00 0.55102564 0.03813274
## 14  1.0 5e+00 0.48717949 0.03963085
## 15  5.0 5e+00 0.49224359 0.04525523
## 16 10.0 5e+00 0.49224359 0.04525523
## 17  0.1 1e+01 0.55102564 0.03813274
## 18  1.0 1e+01 0.50506410 0.04235779
## 19  5.0 1e+01 0.49993590 0.04269277
## 20 10.0 1e+01 0.49993590 0.04269277
## 21  0.1 1e+02 0.55102564 0.03813274
## 22  1.0 1e+02 0.55102564 0.03813274
## 23  5.0 1e+02 0.55102564 0.03813274
## 24 10.0 1e+02 0.55102564 0.03813274

Finally, for radial basis kernel, cost=10 and gamma=0.01.

Part 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(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.poly)

plotpairs(svm.radial)

Problem 8

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

Part A

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

library(ISLR)
set.seed(9004)
train = sample(dim(OJ)[1], 800)
OJ.train = OJ[train, ]
OJ.test = OJ[-train, ]

Part 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.

library(e1071)
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 
##       gamma:  0.05555556 
## 
## Number of Support Vectors:  432
## 
##  ( 217 215 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM

Part C

What are the training and test error rates?

train.pred = predict(svm.linear, OJ.train)
table(OJ.train$Purchase, train.pred)
##     train.pred
##       CH  MM
##   CH 439  53
##   MM  82 226
(82 + 53)/(439 + 53 + 82 + 226)
## [1] 0.16875
test.pred = predict(svm.linear, OJ.test)
table(OJ.test$Purchase, test.pred)
##     test.pred
##       CH  MM
##   CH 142  19
##   MM  29  80
(19 + 29)/(142 + 19 + 29 + 80)
## [1] 0.1777778

The training error rate is 16.9% and test error rate is about 17.8%.

Part 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: 0.16875 
## 
## - Detailed performance results:
##           cost   error dispersion
## 1   0.01000000 0.16875 0.03691676
## 2   0.01778279 0.16875 0.03397814
## 3   0.03162278 0.17125 0.03230175
## 4   0.05623413 0.17250 0.03162278
## 5   0.10000000 0.17000 0.03291403
## 6   0.17782794 0.17125 0.03335936
## 7   0.31622777 0.16875 0.03498512
## 8   0.56234133 0.17000 0.03129164
## 9   1.00000000 0.16875 0.03397814
## 10  1.77827941 0.16875 0.03240906
## 11  3.16227766 0.16875 0.03294039
## 12  5.62341325 0.17125 0.03120831
## 13 10.00000000 0.17125 0.03283481

Tuning shows that optimal cost is 0.3162

Part 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$Purchase, train.pred)
##     train.pred
##       CH  MM
##   CH 435  57
##   MM  71 237
(57 + 71)/(435 + 57 + 71 + 237)
## [1] 0.16
test.pred = predict(svm.linear, OJ.test)
table(OJ.test$Purchase, test.pred)
##     test.pred
##       CH  MM
##   CH 141  20
##   MM  29  80
(29 + 20)/(141 + 20 + 29 + 80)
## [1] 0.1814815

The training error decreases to 16% but test error slightly increases to 18.1% by using best cost.

Part F

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

set.seed(410)
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 
##       gamma:  0.05555556 
## 
## Number of Support Vectors:  367
## 
##  ( 184 183 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM
train.pred = predict(svm.radial, OJ.train)
table(OJ.train$Purchase, train.pred)
##     train.pred
##       CH  MM
##   CH 452  40
##   MM  78 230
(40 + 78)/(452 + 40 + 78 + 230)
## [1] 0.1475
test.pred = predict(svm.radial, OJ.test)
table(OJ.test$Purchase, test.pred)
##     test.pred
##       CH  MM
##   CH 146  15
##   MM  27  82
(27 + 15)/(146 + 15 + 27 + 82)
## [1] 0.1555556

The radial basis kernel with default gamma creates 367 support vectors, out of which, 184 belong to level CH and remaining 183 belong to level MM. The classifier has a training error of 14.7% and a test error of 15.6% which is a slight improvement over linear kernel. We now use cross validation to find optimal gamma.

set.seed(755)
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: 0.165 
## 
## - Detailed performance results:
##           cost   error dispersion
## 1   0.01000000 0.38500 0.06258328
## 2   0.01778279 0.38500 0.06258328
## 3   0.03162278 0.37625 0.06908379
## 4   0.05623413 0.21000 0.03855011
## 5   0.10000000 0.18625 0.03143004
## 6   0.17782794 0.18375 0.03230175
## 7   0.31622777 0.17125 0.03438447
## 8   0.56234133 0.16500 0.03763863
## 9   1.00000000 0.17500 0.03584302
## 10  1.77827941 0.17375 0.04059026
## 11  3.16227766 0.17625 0.03747684
## 12  5.62341325 0.17625 0.03839216
## 13 10.00000000 0.17375 0.03458584
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$Purchase, train.pred)
##     train.pred
##       CH  MM
##   CH 452  40
##   MM  77 231
(77 + 40)/(452 + 40 + 77 + 231)
## [1] 0.14625
test.pred = predict(svm.radial, OJ.test)
table(OJ.test$Purchase, test.pred)
##     test.pred
##       CH  MM
##   CH 146  15
##   MM  28  81
(28 + 15)/(146 + 15 + 28 + 81)
## [1] 0.1592593

Tuning slightly decreases training error to 14.6% and slightly increases test error to 16% which is still better than linear kernel.

Part G

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

set.seed(8112)
svm.poly = svm(Purchase ~ ., data = OJ.train, kernel = "poly", degree = 2)
summary(svm.poly)
## 
## Call:
## svm(formula = Purchase ~ ., data = OJ.train, kernel = "poly", 
##     degree = 2)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  polynomial 
##        cost:  1 
##      degree:  2 
##       gamma:  0.05555556 
##      coef.0:  0 
## 
## Number of Support Vectors:  452
## 
##  ( 232 220 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM
train.pred = predict(svm.poly, OJ.train)
table(OJ.train$Purchase, train.pred)
##     train.pred
##       CH  MM
##   CH 460  32
##   MM 105 203
(32 + 105)/(460 + 32 + 105 + 203)
## [1] 0.17125
test.pred = predict(svm.poly, OJ.test)
table(OJ.test$Purchase, test.pred)
##     test.pred
##       CH  MM
##   CH 149  12
##   MM  37  72
(12 + 37)/(149 + 12 + 37 + 72)
## [1] 0.1814815

Summary shows that polynomial kernel produces 452 support vectors, out of which, 232 belong to level CH and remaining 220 belong to level MM. This kernel produces a train error of 17.1% and a test error of 18.1% which are slightly higher than the errors produces by radial kernel but lower than the errors produced by linear kernel.

set.seed(322)
tune.out = tune(svm, Purchase ~ ., data = OJ.train, kernel = "poly", degree = 2, 
    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
##  5.623413
## 
## - best performance: 0.18375 
## 
## - Detailed performance results:
##           cost   error dispersion
## 1   0.01000000 0.38500 0.05426274
## 2   0.01778279 0.36750 0.05075814
## 3   0.03162278 0.35750 0.05177408
## 4   0.05623413 0.34250 0.04937104
## 5   0.10000000 0.31500 0.05230785
## 6   0.17782794 0.24875 0.03928617
## 7   0.31622777 0.20875 0.05684103
## 8   0.56234133 0.20875 0.05653477
## 9   1.00000000 0.20000 0.06095308
## 10  1.77827941 0.19375 0.04497299
## 11  3.16227766 0.18625 0.04185375
## 12  5.62341325 0.18375 0.03335936
## 13 10.00000000 0.18375 0.04041881
svm.poly = svm(Purchase ~ ., data = OJ.train, kernel = "poly", degree = 2, cost = tune.out$best.parameters$cost)
train.pred = predict(svm.poly, OJ.train)
table(OJ.train$Purchase, train.pred)
##     train.pred
##       CH  MM
##   CH 455  37
##   MM  84 224
(37 + 84)/(455 + 37 + 84 + 224)
## [1] 0.15125
test.pred = predict(svm.poly, OJ.test)
table(OJ.test$Purchase, test.pred)
##     test.pred
##       CH  MM
##   CH 148  13
##   MM  34  75
(13 + 34)/(148 + 13 + 34 + 75)
## [1] 0.1740741

Tuning reduces the training error to 15.12% and test error to 17.4% which is worse than radial kernel but slightly better than linear kernel.

Part H

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

Overall, radial basis kernel seems to be producing minimum misclassification error on both train and test data.