- 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. 400
- 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)
- 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[y == 0], x2[y == 0], col = "red", xlab = "X1", ylab = "X2")
points(x1[y == 1], x2[y == 1], )
- Fit a logistic regression model to the data, using X1 and X2 as
predictors.
lm.fit = glm(y ~ x1 + x2, family = binomial)
summary(lm.fit)
## glm(formula = y ~ x1 + x2, family = binomial)
## (Intercept) -0.06401 0.08985 -0.712 0.476
## x1 -0.03169 0.32131 -0.099 0.921
## x2 -0.37902 0.30517 -1.242 0.214
## Number of Fisher Scoring iterations: 3
- 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 > 0.50, 1, 0)
plot(data.pos$x1, data.pos$x2, xlab = "X1", ylab = "X2")
points(data.neg$x1, data.neg$x2, col = "red")
- 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).
lm.fit = glm(y ~ poly(x1, 2) + poly(x2, 2) + I(x1 * x2), data = data, family = binomial)
- 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)
plot(data.pos$x1, data.pos$x2, xlab = "X1", ylab = "X2")
points(data.neg$x1, data.neg$x2, col = "red")
- Comment on your results cross validation would have been a better
option however this method still worked.
- 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.
- 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)
- 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.
library(e1071)
tune.out = tune(svm, mpglevel ~ ., data = Auto, kernel = "linear"
summary(tune.out)
## - sampling method: 10-fold cross validation
## - Detailed performance results:
## cost error dispersion
## 1 1e-02 0.07397436 0.06863413
## 2 1e+02 0.03294872 0.02898463
- 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)
## - sampling method: 10-fold cross validation
## - Detailed performance results:
## cost degree error dispersion
## 1 0.1 2 0.5587821 0.04538579
## 2 1.0 2 0.5587821 0.04538579
## 3 5.0 2 0.5587821 0.04538579
## 4 10.0 2 0.5435897 0.05611162
## 5 0.1 3 0.5587821 0.04538579
## 6 1.0 3 0.5587821 0.04538579
## 7 5.0 3 0.5587821 0.04538579
## 8 10.0 3 0.5587821 0.04538579
## 9 0.1 4 0.5587821 0.04538579
## 10 1.0 4 0.5587821 0.04538579
## 11 5.0 4 0.5587821 0.04538579
## 12 10.0 4 0.5587821 0.04538579
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)
## - Detailed performance results:
## cost gamma error dispersion
## 1 0.1 1e-02 0.09429487 0.04814900
## 2 1.0 1e-02 0.07897436 0.03875105
## 3 5.0 1e-02 0.05352564 0.02532795
## 4 10.0 1e-02 0.02551282 0.02417610
## 5 0.1 1e-01 0.07891026 0.03847631
## 6 1.0 1e-01 0.05602564 0.02881876
## 7 5.0 1e-01 0.03826923 0.03252085
## 8 10.0 1e-01 0.03320513 0.02964746
## 9 0.1 1e+00 0.57660256 0.05479863
## 10 1.0 1e+00 0.06628205 0.02996211
## 11 5.0 1e+00 0.06115385 0.02733573
## 12 10.0 1e+00 0.06115385 0.02733573
## 13 0.1 5e+00 0.57660256 0.05479863
## 14 1.0 5e+00 0.51538462 0.06642516
## 15 5.0 5e+00 0.50775641 0.07152757
## 16 10.0 5e+00 0.50775641 0.07152757
## 17 0.1 1e+01 0.57660256 0.05479863
## 18 1.0 1e+01 0.53833333 0.05640443
## 19 5.0 1e+01 0.53070513 0.05708644
## 20 10.0 1e+01 0.53070513 0.05708644
## 21 0.1 1e+02 0.57660256 0.05479863
## 22 1.0 1e+02 0.57660256 0.05479863
## 23 5.0 1e+02 0.57660256 0.05479863
## 24 10.0 1e+02 0.57660256 0.05479863
- 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.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)
- This problem involves the OJ data set which is part of the ISLR2
package.
- 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, ]
- 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)
## svm(formula = Purchase ~ ., data = OJ.train, kernel = "linear", cost = 0.01)
## Number of Support Vectors: 442
## Number of Classes: 2
## CH MM
- 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 432 51
## MM 80 237
test.pred = predict(svm.linear, OJ.test)
table(OJ.test$Purchase, test.pred)
## test.pred
## CH MM
## CH 146 24
## MM 22 78
- Use the tune() function to select an optimal cost. Consider values
in the range 0.01 to 10.
set.seed(1554)
tune.out = tune(svm, Purchase ~ ., data = OJ.train, kernel = "linear", ranges = list(cost = 10^seq(-2,
1, by = 0.25)))
summary(tune.out)
## Parameter tuning of 'svm':
## - best parameters:
## cost
## 3.162278
## - Detailed performance results:
## cost error dispersion
## 1 0.01000000 0.16750 0.03395258
## 2 0.01778279 0.16875 0.02960973
## 3 0.03162278 0.16625 0.02638523
## 4 0.05623413 0.16875 0.03076005
## 5 0.10000000 0.16875 0.02901748
## 6 0.17782794 0.16750 0.02838231
## 7 0.31622777 0.17000 0.02898755
## 8 0.56234133 0.16875 0.02841288
## 9 1.00000000 0.16500 0.03106892
## 10 1.77827941 0.16500 0.03106892
## 11 3.16227766 0.16250 0.03118048
## 12 5.62341325 0.16375 0.02664713
## 13 10.00000000 0.16750 0.02581989
- Compute the training and test error rates using this new value for
cost.
set.seed(410)
svm.radial = svm(Purchase ~ ., data = OJ.train, kernel = "radial")
summary(svm.radial)
test.pred = predict(svm.linear, OJ.test)
table(OJ.test$Purchase, test.pred)
- 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)
train.pred = predict(svm.radial, OJ.train)
table(OJ.train$Purchase, train.pred)
test.pred = predict(svm.radial, OJ.test)
table(OJ.test$Purchase, test.pred)
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)
## test.pred
## CH MM
## CH 139 15
## MM 25 91
## - best parameters:
## cost
## 0.3162278
##
## - best performance: 0.1675
##
## - Detailed performance results:
## cost error dispersion
## 1 0.01000000 0.39625 0.06615691
## 2 0.01778279 0.39625 0.06615691
## 3 0.03162278 0.35375 0.09754807
## 4 0.05623413 0.20000 0.04249183
## 5 0.10000000 0.17750 0.04073969
## 6 0.17782794 0.17125 0.03120831
## 7 0.31622777 0.16750 0.04216370
## 8 0.56234133 0.16750 0.03782269
## 9 1.00000000 0.17250 0.03670453
## 10 1.77827941 0.17750 0.03374743
## 11 3.16227766 0.18000 0.04005205
## 12 5.62341325 0.18000 0.03446012
## 13 10.00000000 0.18625 0.04427267
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)
- Overall, which approach seems to give the best results on this data?
the linear basis gave me the best results