Assignment #8
Nathaniel Sattler
5/1/2021
Assignment #8
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.
(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(123)
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.
plot(x1[y == 0], x2[y == 0], col = "red", xlab = "X1", ylab = "X2", pch = 5)
points(x1[y == 1], x2[y == 1], col = "blue", pch = 2)
### (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
## -1.227 -1.200 1.133 1.157 1.188
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 0.04792 0.08949 0.535 0.592
## x1 -0.03999 0.31516 -0.127 0.899
## x2 0.11509 0.30829 0.373 0.709
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 692.86 on 499 degrees of freedom
## Residual deviance: 692.71 on 497 degrees of freedom
## AIC: 698.71
##
## 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.
dat = data.frame(x1 = x1, x2 = x2, y = y)
probs = predict(log.fit, dat, type = "response")
#mean(probs)
preds = ifelse(probs >= 0.512, 1, 0)
dat.pos = dat[preds == 1,]
dat.neg = dat[preds == 0,]
plot(dat.pos$x1, dat.pos$x2, col = "blue", xlab = "X1", ylab = "X2", pch = 5, type = 'p')
points(dat.neg$x1, dat.neg$x2, col = "red", pch = 2)
### (e) 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).
poly.fit = glm(y ~ poly(x1, 2) + poly(x2, 2) + I(x1 * x2), dat, family = binomial)## Warning: glm.fit: algorithm did not converge## Warning: glm.fit: fitted probabilities numerically 0 or 1 occurredsummary(poly.fit)##
## Call:
## glm(formula = y ~ poly(x1, 2) + poly(x2, 2) + I(x1 * x2), family = binomial,
## data = dat)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -8.625e-04 -2.000e-08 2.000e-08 2.000e-08 9.604e-04
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -54.24 3335.32 -0.016 0.987
## poly(x1, 2)1 542.33 71411.93 0.008 0.994
## poly(x1, 2)2 20838.39 778877.62 0.027 0.979
## poly(x2, 2)1 2163.06 115506.63 0.019 0.985
## poly(x2, 2)2 -21646.31 811141.33 -0.027 0.979
## I(x1 * x2) 566.19 36927.06 0.015 0.988
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 6.9286e+02 on 499 degrees of freedom
## Residual deviance: 2.3990e-06 on 494 degrees of freedom
## AIC: 12
##
## Number of Fisher Scoring iterations: 25None of the terms are significant, they are all very close to p-values of 1. Plus the model is predicting perfectly 1s or 2s.
(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.
probs = predict(poly.fit, dat, type = "response")
mean(probs)## [1] 0.512preds = ifelse(probs > 0.512, 1, 0)
dat.pos = dat[preds == 1, ]
dat.neg = dat[preds == 0, ]
plot(dat.pos$x1, dat.pos$x2, col = "blue", xlab = "X1", ylab = "X2", pch = 5)
points(dat.neg$x1, dat.neg$x2, col = "red", pch = 2)
Much closer to original shape of x1 and x2 based on y 0s.
(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, dat, kernel = "linear")
svm.preds = predict(svm.fit, dat)
dat.pos = dat[svm.preds == 1, ]
dat.neg = dat[svm.preds == 0, ]
plot(dat.pos$x1, dat.pos$x2, col = "blue", xlab = "X1", ylab = "X2", pch = 5)
points(dat.neg$x1, dat.neg$x2, col = "red", pch = 2)
Linear prediction kernel returned only values equally 1 in prediction.
(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, dat, kernel = "radial", gamma = 1)
svm.preds = predict(svm.fit, dat)
dat.pos = dat[svm.preds == 1, ]
dat.neg = dat[svm.preds == 0, ]
plot(dat.pos$x1, dat.pos$x2, col = "blue", xlab = "X1", ylab = "X2", pch = 5)
points(dat.neg$x1, dat.neg$x2, col = "red", pch = 2)
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.
(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)## Warning: package 'ISLR' was built under R version 4.0.3library(e1071)
set.seed(123)
gas.med = median(Auto$mpg)
med.var = ifelse(Auto$mpg > gas.med, 1, 0)
Auto$mpglevel = as.factor(med.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.
lin.svm = tune(svm, mpglevel ~ ., data = Auto, kernel = "linear", ranges = list(cost = c(0.01, 0.1, 1, 5, 10, 100)))
summary(lin.svm)##
## 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 0.07634615 0.03928191
## 2 1e-01 0.04333333 0.03191738
## 3 1e+00 0.01025641 0.01792836
## 4 5e+00 0.01538462 0.01792836
## 5 1e+01 0.01788462 0.01727588
## 6 1e+02 0.03320513 0.02720447The cross validation error rate range from 0.0763 at a cost of 0.01 to 0.033 at a cost of 100. There is a dip, with the lowest error being at a cost of 1, with an error of 0.0102.
(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(123)
poly.svm = tune(svm, mpglevel ~ ., data = Auto, kernel = "polynomial", ranges = list(cost=c(0.1,1,10,100), gamma=c(0.1,0.5,1,2,3,4), degree = c(0.1,1,2,3,4)))
summary(poly.svm)##
## Parameter tuning of 'svm':
##
## - sampling method: 10-fold cross validation
##
## - best parameters:
## cost gamma degree
## 10 0.1 1
##
## - best performance: 0.01025641
##
## - Detailed performance results:
## cost gamma degree error dispersion
## 1 0.1 0.1 0.1 0.58173077 0.04740051
## 2 1.0 0.1 0.1 0.58173077 0.04740051
## 3 10.0 0.1 0.1 0.58173077 0.04740051
## 4 100.0 0.1 0.1 0.58173077 0.04740051
## 5 0.1 0.5 0.1 0.58173077 0.04740051
## 6 1.0 0.5 0.1 0.58173077 0.04740051
## 7 10.0 0.5 0.1 0.58173077 0.04740051
## 8 100.0 0.5 0.1 0.58173077 0.04740051
## 9 0.1 1.0 0.1 0.58173077 0.04740051
## 10 1.0 1.0 0.1 0.58173077 0.04740051
## 11 10.0 1.0 0.1 0.58173077 0.04740051
## 12 100.0 1.0 0.1 0.58173077 0.04740051
## 13 0.1 2.0 0.1 0.58173077 0.04740051
## 14 1.0 2.0 0.1 0.58173077 0.04740051
## 15 10.0 2.0 0.1 0.58173077 0.04740051
## 16 100.0 2.0 0.1 0.58173077 0.04740051
## 17 0.1 3.0 0.1 0.58173077 0.04740051
## 18 1.0 3.0 0.1 0.58173077 0.04740051
## 19 10.0 3.0 0.1 0.58173077 0.04740051
## 20 100.0 3.0 0.1 0.58173077 0.04740051
## 21 0.1 4.0 0.1 0.58173077 0.04740051
## 22 1.0 4.0 0.1 0.58173077 0.04740051
## 23 10.0 4.0 0.1 0.58173077 0.04740051
## 24 100.0 4.0 0.1 0.58173077 0.04740051
## 25 0.1 0.1 1.0 0.07634615 0.03928191
## 26 1.0 0.1 1.0 0.04333333 0.03191738
## 27 10.0 0.1 1.0 0.01025641 0.01792836
## 28 100.0 0.1 1.0 0.01788462 0.01727588
## 29 0.1 0.5 1.0 0.05602564 0.03551922
## 30 1.0 0.5 1.0 0.01788462 0.01727588
## 31 10.0 0.5 1.0 0.01538462 0.01792836
## 32 100.0 0.5 1.0 0.03320513 0.02720447
## 33 0.1 1.0 1.0 0.04333333 0.03191738
## 34 1.0 1.0 1.0 0.01025641 0.01792836
## 35 10.0 1.0 1.0 0.01788462 0.01727588
## 36 100.0 1.0 1.0 0.03320513 0.02720447
## 37 0.1 2.0 1.0 0.02288462 0.01427008
## 38 1.0 2.0 1.0 0.01025641 0.01792836
## 39 10.0 2.0 1.0 0.03320513 0.02720447
## 40 100.0 2.0 1.0 0.03320513 0.02720447
## 41 0.1 3.0 1.0 0.02038462 0.01617396
## 42 1.0 3.0 1.0 0.01025641 0.01792836
## 43 10.0 3.0 1.0 0.03320513 0.02720447
## 44 100.0 3.0 1.0 0.03320513 0.02720447
## 45 0.1 4.0 1.0 0.01788462 0.01727588
## 46 1.0 4.0 1.0 0.01282051 0.01813094
## 47 10.0 4.0 1.0 0.03320513 0.02720447
## 48 100.0 4.0 1.0 0.03320513 0.02720447
## 49 0.1 0.1 2.0 0.30602564 0.10620333
## 50 1.0 0.1 2.0 0.25467949 0.10273137
## 51 10.0 0.1 2.0 0.15782051 0.07365927
## 52 100.0 0.1 2.0 0.16288462 0.08475580
## 53 0.1 0.5 2.0 0.16294872 0.06148264
## 54 1.0 0.5 2.0 0.16551282 0.08747687
## 55 10.0 0.5 2.0 0.17826923 0.07176717
## 56 100.0 0.5 2.0 0.19871795 0.06697346
## 57 0.1 1.0 2.0 0.15782051 0.07365927
## 58 1.0 1.0 2.0 0.16288462 0.08475580
## 59 10.0 1.0 2.0 0.19871795 0.06697346
## 60 100.0 1.0 2.0 0.19871795 0.06697346
## 61 0.1 2.0 2.0 0.16801282 0.08821472
## 62 1.0 2.0 2.0 0.18333333 0.06944214
## 63 10.0 2.0 2.0 0.19871795 0.06697346
## 64 100.0 2.0 2.0 0.19871795 0.06697346
## 65 0.1 3.0 2.0 0.16282051 0.08547489
## 66 1.0 3.0 2.0 0.19871795 0.06697346
## 67 10.0 3.0 2.0 0.19871795 0.06697346
## 68 100.0 3.0 2.0 0.19871795 0.06697346
## 69 0.1 4.0 2.0 0.17564103 0.08212248
## 70 1.0 4.0 2.0 0.19871795 0.06697346
## 71 10.0 4.0 2.0 0.19871795 0.06697346
## 72 100.0 4.0 2.0 0.19871795 0.06697346
## 73 0.1 0.1 3.0 0.23711538 0.09399789
## 74 1.0 0.1 3.0 0.07891026 0.03228186
## 75 10.0 0.1 3.0 0.04083333 0.03008810
## 76 100.0 0.1 3.0 0.03570513 0.03436393
## 77 0.1 0.5 3.0 0.03814103 0.02133598
## 78 1.0 0.5 3.0 0.03570513 0.03436393
## 79 10.0 0.5 3.0 0.03570513 0.02442271
## 80 100.0 0.5 3.0 0.03570513 0.02442271
## 81 0.1 1.0 3.0 0.03570513 0.03436393
## 82 1.0 1.0 3.0 0.03570513 0.02442271
## 83 10.0 1.0 3.0 0.03570513 0.02442271
## 84 100.0 1.0 3.0 0.03570513 0.02442271
## 85 0.1 2.0 3.0 0.03826923 0.02740388
## 86 1.0 2.0 3.0 0.03570513 0.02442271
## 87 10.0 2.0 3.0 0.03570513 0.02442271
## 88 100.0 2.0 3.0 0.03570513 0.02442271
## 89 0.1 3.0 3.0 0.03570513 0.02442271
## 90 1.0 3.0 3.0 0.03570513 0.02442271
## 91 10.0 3.0 3.0 0.03570513 0.02442271
## 92 100.0 3.0 3.0 0.03570513 0.02442271
## 93 0.1 4.0 3.0 0.03570513 0.02442271
## 94 1.0 4.0 3.0 0.03570513 0.02442271
## 95 10.0 4.0 3.0 0.03570513 0.02442271
## 96 100.0 4.0 3.0 0.03570513 0.02442271
## 97 0.1 0.1 4.0 0.31121795 0.09684710
## 98 1.0 0.1 4.0 0.23692308 0.09340281
## 99 10.0 0.1 4.0 0.19089744 0.06434271
## 100 100.0 0.1 4.0 0.18589744 0.07043897
## 101 0.1 0.5 4.0 0.19096154 0.06956790
## 102 1.0 0.5 4.0 0.19365385 0.05673827
## 103 10.0 0.5 4.0 0.19346154 0.07042458
## 104 100.0 0.5 4.0 0.19346154 0.07042458
## 105 0.1 1.0 4.0 0.19615385 0.06565708
## 106 1.0 1.0 4.0 0.19346154 0.07042458
## 107 10.0 1.0 4.0 0.19346154 0.07042458
## 108 100.0 1.0 4.0 0.19346154 0.07042458
## 109 0.1 2.0 4.0 0.19346154 0.07042458
## 110 1.0 2.0 4.0 0.19346154 0.07042458
## 111 10.0 2.0 4.0 0.19346154 0.07042458
## 112 100.0 2.0 4.0 0.19346154 0.07042458
## 113 0.1 3.0 4.0 0.19346154 0.07042458
## 114 1.0 3.0 4.0 0.19346154 0.07042458
## 115 10.0 3.0 4.0 0.19346154 0.07042458
## 116 100.0 3.0 4.0 0.19346154 0.07042458
## 117 0.1 4.0 4.0 0.19346154 0.07042458
## 118 1.0 4.0 4.0 0.19346154 0.07042458
## 119 10.0 4.0 4.0 0.19346154 0.07042458
## 120 100.0 4.0 4.0 0.19346154 0.07042458poly.svm$best.parameters## cost gamma degree
## 27 10 0.1 1The best parameters come from using a cost of 10, a gamma of 0.1, and a degree of 1.
set.seed(123)
rad.svm = tune(svm, mpglevel ~ ., data = Auto, kernel = "radial", ranges = list(cost=c(0.1,1,10,100), gamma=c(0.1,0.5,1,2,3,4), degree = c(0.1,1,2,3,4)))
summary(rad.svm)##
## Parameter tuning of 'svm':
##
## - sampling method: 10-fold cross validation
##
## - best parameters:
## cost gamma degree
## 10 0.1 0.1
##
## - best performance: 0.03314103
##
## - Detailed performance results:
## cost gamma degree error dispersion
## 1 0.1 0.1 0.1 0.07634615 0.03928191
## 2 1.0 0.1 0.1 0.05852564 0.03960325
## 3 10.0 0.1 0.1 0.03314103 0.02942215
## 4 100.0 0.1 0.1 0.03326923 0.02434857
## 5 0.1 0.5 0.1 0.08147436 0.03707182
## 6 1.0 0.5 0.1 0.04576923 0.03903092
## 7 10.0 0.5 0.1 0.05339744 0.03440111
## 8 100.0 0.5 0.1 0.05339744 0.03440111
## 9 0.1 1.0 0.1 0.58173077 0.04740051
## 10 1.0 1.0 0.1 0.05865385 0.04942437
## 11 10.0 1.0 0.1 0.05608974 0.04595880
## 12 100.0 1.0 0.1 0.05608974 0.04595880
## 13 0.1 2.0 0.1 0.58173077 0.04740051
## 14 1.0 2.0 0.1 0.11474359 0.06630201
## 15 10.0 2.0 0.1 0.11474359 0.06630201
## 16 100.0 2.0 0.1 0.11474359 0.06630201
## 17 0.1 3.0 0.1 0.58173077 0.04740051
## 18 1.0 3.0 0.1 0.42878205 0.17823496
## 19 10.0 3.0 0.1 0.40839744 0.18573046
## 20 100.0 3.0 0.1 0.40839744 0.18573046
## 21 0.1 4.0 0.1 0.58173077 0.04740051
## 22 1.0 4.0 0.1 0.51538462 0.06959451
## 23 10.0 4.0 0.1 0.50012821 0.07022396
## 24 100.0 4.0 0.1 0.50012821 0.07022396
## 25 0.1 0.1 1.0 0.07634615 0.03928191
## 26 1.0 0.1 1.0 0.05852564 0.03960325
## 27 10.0 0.1 1.0 0.03314103 0.02942215
## 28 100.0 0.1 1.0 0.03326923 0.02434857
## 29 0.1 0.5 1.0 0.08147436 0.03707182
## 30 1.0 0.5 1.0 0.04576923 0.03903092
## 31 10.0 0.5 1.0 0.05339744 0.03440111
## 32 100.0 0.5 1.0 0.05339744 0.03440111
## 33 0.1 1.0 1.0 0.58173077 0.04740051
## 34 1.0 1.0 1.0 0.05865385 0.04942437
## 35 10.0 1.0 1.0 0.05608974 0.04595880
## 36 100.0 1.0 1.0 0.05608974 0.04595880
## 37 0.1 2.0 1.0 0.58173077 0.04740051
## 38 1.0 2.0 1.0 0.11474359 0.06630201
## 39 10.0 2.0 1.0 0.11474359 0.06630201
## 40 100.0 2.0 1.0 0.11474359 0.06630201
## 41 0.1 3.0 1.0 0.58173077 0.04740051
## 42 1.0 3.0 1.0 0.42878205 0.17823496
## 43 10.0 3.0 1.0 0.40839744 0.18573046
## 44 100.0 3.0 1.0 0.40839744 0.18573046
## 45 0.1 4.0 1.0 0.58173077 0.04740051
## 46 1.0 4.0 1.0 0.51538462 0.06959451
## 47 10.0 4.0 1.0 0.50012821 0.07022396
## 48 100.0 4.0 1.0 0.50012821 0.07022396
## 49 0.1 0.1 2.0 0.07634615 0.03928191
## 50 1.0 0.1 2.0 0.05852564 0.03960325
## 51 10.0 0.1 2.0 0.03314103 0.02942215
## 52 100.0 0.1 2.0 0.03326923 0.02434857
## 53 0.1 0.5 2.0 0.08147436 0.03707182
## 54 1.0 0.5 2.0 0.04576923 0.03903092
## 55 10.0 0.5 2.0 0.05339744 0.03440111
## 56 100.0 0.5 2.0 0.05339744 0.03440111
## 57 0.1 1.0 2.0 0.58173077 0.04740051
## 58 1.0 1.0 2.0 0.05865385 0.04942437
## 59 10.0 1.0 2.0 0.05608974 0.04595880
## 60 100.0 1.0 2.0 0.05608974 0.04595880
## 61 0.1 2.0 2.0 0.58173077 0.04740051
## 62 1.0 2.0 2.0 0.11474359 0.06630201
## 63 10.0 2.0 2.0 0.11474359 0.06630201
## 64 100.0 2.0 2.0 0.11474359 0.06630201
## 65 0.1 3.0 2.0 0.58173077 0.04740051
## 66 1.0 3.0 2.0 0.42878205 0.17823496
## 67 10.0 3.0 2.0 0.40839744 0.18573046
## 68 100.0 3.0 2.0 0.40839744 0.18573046
## 69 0.1 4.0 2.0 0.58173077 0.04740051
## 70 1.0 4.0 2.0 0.51538462 0.06959451
## 71 10.0 4.0 2.0 0.50012821 0.07022396
## 72 100.0 4.0 2.0 0.50012821 0.07022396
## 73 0.1 0.1 3.0 0.07634615 0.03928191
## 74 1.0 0.1 3.0 0.05852564 0.03960325
## 75 10.0 0.1 3.0 0.03314103 0.02942215
## 76 100.0 0.1 3.0 0.03326923 0.02434857
## 77 0.1 0.5 3.0 0.08147436 0.03707182
## 78 1.0 0.5 3.0 0.04576923 0.03903092
## 79 10.0 0.5 3.0 0.05339744 0.03440111
## 80 100.0 0.5 3.0 0.05339744 0.03440111
## 81 0.1 1.0 3.0 0.58173077 0.04740051
## 82 1.0 1.0 3.0 0.05865385 0.04942437
## 83 10.0 1.0 3.0 0.05608974 0.04595880
## 84 100.0 1.0 3.0 0.05608974 0.04595880
## 85 0.1 2.0 3.0 0.58173077 0.04740051
## 86 1.0 2.0 3.0 0.11474359 0.06630201
## 87 10.0 2.0 3.0 0.11474359 0.06630201
## 88 100.0 2.0 3.0 0.11474359 0.06630201
## 89 0.1 3.0 3.0 0.58173077 0.04740051
## 90 1.0 3.0 3.0 0.42878205 0.17823496
## 91 10.0 3.0 3.0 0.40839744 0.18573046
## 92 100.0 3.0 3.0 0.40839744 0.18573046
## 93 0.1 4.0 3.0 0.58173077 0.04740051
## 94 1.0 4.0 3.0 0.51538462 0.06959451
## 95 10.0 4.0 3.0 0.50012821 0.07022396
## 96 100.0 4.0 3.0 0.50012821 0.07022396
## 97 0.1 0.1 4.0 0.07634615 0.03928191
## 98 1.0 0.1 4.0 0.05852564 0.03960325
## 99 10.0 0.1 4.0 0.03314103 0.02942215
## 100 100.0 0.1 4.0 0.03326923 0.02434857
## 101 0.1 0.5 4.0 0.08147436 0.03707182
## 102 1.0 0.5 4.0 0.04576923 0.03903092
## 103 10.0 0.5 4.0 0.05339744 0.03440111
## 104 100.0 0.5 4.0 0.05339744 0.03440111
## 105 0.1 1.0 4.0 0.58173077 0.04740051
## 106 1.0 1.0 4.0 0.05865385 0.04942437
## 107 10.0 1.0 4.0 0.05608974 0.04595880
## 108 100.0 1.0 4.0 0.05608974 0.04595880
## 109 0.1 2.0 4.0 0.58173077 0.04740051
## 110 1.0 2.0 4.0 0.11474359 0.06630201
## 111 10.0 2.0 4.0 0.11474359 0.06630201
## 112 100.0 2.0 4.0 0.11474359 0.06630201
## 113 0.1 3.0 4.0 0.58173077 0.04740051
## 114 1.0 3.0 4.0 0.42878205 0.17823496
## 115 10.0 3.0 4.0 0.40839744 0.18573046
## 116 100.0 3.0 4.0 0.40839744 0.18573046
## 117 0.1 4.0 4.0 0.58173077 0.04740051
## 118 1.0 4.0 4.0 0.51538462 0.06959451
## 119 10.0 4.0 4.0 0.50012821 0.07022396
## 120 100.0 4.0 4.0 0.50012821 0.07022396rad.svm$best.parameters## cost gamma degree
## 3 10 0.1 0.1The best parameters for the radial model have a cost of 10, a gamma of 0.1, and a degree of 0.1.
(d) Make some plots to back up your assertions in (b) and (c).
svm.lin = svm(mpglevel ~ ., data = Auto, kernel = "linear", cost = 1)
svm.poly = svm(mpglevel ~ ., data = Auto, kernel = "polynomial", cost = 10, gamma = 0.1, degree = 1)
svm.rad = svm(mpglevel ~ ., data = Auto, kernel = "radial", cost = 10, gamma = 0.1, degree = 0.1)
#plot(svm.lin, Auto, mpg~acceleration)
plotpairs = function(fit) {
for (name in names(Auto)[!(names(Auto) %in% c("mpg", "mpglevel", "name"))]) {
plot(fit, Auto, as.formula(paste("mpg~", name, sep = "")))
}
}
plotpairs(svm.lin)
plotpairs(svm.poly)
plotpairs(svm.rad)
The above plots divide each variable between 1 and 0 from our obtained, and then plots the data according to input variable and actual mpg.
8. This problem involves the OJ data set which is part of the ISLR package
(a) Create a training set containing a random sample of 800 observations, and a test set containing the remaining observations.
library(caret)## Warning: package 'caret' was built under R version 4.0.3## Loading required package: lattice## Loading required package: ggplot2set.seed(123)
data("OJ")
trainrows = createDataPartition(OJ$Purchase, p = (799/1070), list = FALSE)
traind = OJ[trainrows,]
testd = OJ[-trainrows,](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.lin = svm(Purchase ~ ., kernel = "linear", data = traind, cost = 0.01)
summary(svm.lin)##
## Call:
## svm(formula = Purchase ~ ., data = traind, kernel = "linear", cost = 0.01)
##
##
## Parameters:
## SVM-Type: C-classification
## SVM-Kernel: linear
## cost: 0.01
##
## Number of Support Vectors: 431
##
## ( 216 215 )
##
##
## Number of Classes: 2
##
## Levels:
## CH MMUsing a linear SVM to predict the purchase of CH or MM at a cost of 1 returns a model that used 431 Support Vectors. 216 of these vectors fall under the CH purchase while the other 215 are CH.
(c) What are the training and test error rates?
train.preds = predict(svm.lin, traind)
table(traind$Purchase, train.preds)## train.preds
## CH MM
## CH 433 55
## MM 75 237(75 + 55) / (433 + 237 + 75 + 55)## [1] 0.1625test.preds = predict(svm.lin, testd)
table(testd$Purchase, test.preds)## test.preds
## CH MM
## CH 136 29
## MM 25 80(25 + 29) / (136 + 80 + 25 + 29)## [1] 0.2The training misclassification rate is 0.1625 while the testing misclassification rate is 0.2.
(d) Use the tune() function to select an optimal “cost”. Consider values in the range 0.01 to 10.
set.seed(123)
svm.lin.tuned = tune(svm, Purchase ~ ., data = traind, kernel = "linear", ranges = list(cost = c(0.01, 0.05, 0.1, 0.5, (seq(1, 10, length.out = 10)))))
summary(svm.lin.tuned)##
## Parameter tuning of 'svm':
##
## - sampling method: 10-fold cross validation
##
## - best parameters:
## cost
## 0.05
##
## - best performance: 0.16625
##
## - Detailed performance results:
## cost error dispersion
## 1 0.01 0.16875 0.05376453
## 2 0.05 0.16625 0.05622685
## 3 0.10 0.16625 0.05337563
## 4 0.50 0.17125 0.04966904
## 5 1.00 0.17250 0.05361903
## 6 2.00 0.17250 0.05614960
## 7 3.00 0.17000 0.05533986
## 8 4.00 0.17000 0.05533986
## 9 5.00 0.16875 0.05408648
## 10 6.00 0.16750 0.05277047
## 11 7.00 0.16750 0.05277047
## 12 8.00 0.16750 0.05277047
## 13 9.00 0.16750 0.05277047
## 14 10.00 0.16750 0.05277047The best cost was 0.05 with an error rate of .16625.
(e) Compute the training and test error rates using this new value for cost.
svm.lin.twoned = svm(Purchase~., kernel = 'linear', data = traind, cost = svm.lin.tuned$best.parameters$cost)
train.preds = predict(svm.lin.twoned, traind)
table(traind$Purchase, train.preds)## train.preds
## CH MM
## CH 433 55
## MM 75 237(75 + 55) / (433 + 237 + 75 + 55)## [1] 0.1625test.preds = predict(svm.lin.twoned, testd)
table(testd$Purchase, test.preds)## test.preds
## CH MM
## CH 136 29
## MM 21 84(21 + 29) / (136 + 84 + 21 + 29)## [1] 0.1851852The training error rate stated the same but there was a slight decrease from 0.2 to 0.185 in the testing error rate.
(f) Repeat parts (b) through (e) using a support vector machine with a radial kernel. Use the default value for gamma.
set.seed(123)
svm.rad = svm(Purchase ~ ., kernel = "radial", data = traind, cost = 0.01)
summary(svm.rad)##
## Call:
## svm(formula = Purchase ~ ., data = traind, kernel = "radial", cost = 0.01)
##
##
## Parameters:
## SVM-Type: C-classification
## SVM-Kernel: radial
## cost: 0.01
##
## Number of Support Vectors: 627
##
## ( 315 312 )
##
##
## Number of Classes: 2
##
## Levels:
## CH MMtrain.preds = predict(svm.rad, traind)
table(traind$Purchase, train.preds)## train.preds
## CH MM
## CH 488 0
## MM 312 0(312) / (488 + 312)## [1] 0.39test.preds = predict(svm.rad, testd)
table(testd$Purchase, test.preds)## test.preds
## CH MM
## CH 165 0
## MM 105 0(105) / (165 + 105)## [1] 0.3888889The radial model did not predict any purchases of MM, despite the support vectors for the two being equally distributed. The misclassification rate for training was 0.39 and testing was 0.389.
set.seed(123)
svm.rad.tuned = tune(svm, Purchase ~ ., data = traind, kernel = "rad", ranges = list(cost = c(0.01, 0.05, 0.1, 0.5, (seq(1, 10, length.out = 10)))))
summary(svm.rad.tuned)##
## Parameter tuning of 'svm':
##
## - sampling method: 10-fold cross validation
##
## - best parameters:
## cost
## 1
##
## - best performance: 0.17
##
## - Detailed performance results:
## cost error dispersion
## 1 0.01 0.39000 0.07518496
## 2 0.05 0.22125 0.08760906
## 3 0.10 0.17375 0.05726704
## 4 0.50 0.17000 0.05075814
## 5 1.00 0.17000 0.05342440
## 6 2.00 0.18000 0.05041494
## 7 3.00 0.18375 0.05272110
## 8 4.00 0.18750 0.04787136
## 9 5.00 0.19250 0.04609772
## 10 6.00 0.19375 0.04759858
## 11 7.00 0.19625 0.05138701
## 12 8.00 0.19750 0.05096295
## 13 9.00 0.19625 0.05138701
## 14 10.00 0.19625 0.05239076svm.rad.twoned = svm(Purchase~., kernel = 'radial', data = traind, cost = svm.rad.tuned$best.parameters$cost)
train.preds = predict(svm.rad.twoned, traind)
table(traind$Purchase, train.preds)## train.preds
## CH MM
## CH 446 42
## MM 76 236(76 + 42) / (466 + 236 + 76 + 42)## [1] 0.1439024test.preds = predict(svm.rad.twoned, testd)
table(testd$Purchase, test.preds)## test.preds
## CH MM
## CH 141 24
## MM 25 80(25 + 24) / (141 + 80 + 25 + 24)## [1] 0.1814815The tuned model using the best cost parameter returns a training misclassification rate of 0.144 and a testing misclassification rate of .181. This model performed much better than the untuned radial model and is about on par with the tuned linear model.
###(g) Repeat parts (b) through (e) using a support vector machine with a polynomial kernel. Set degree=2.
set.seed(123)
svm.poly = svm(Purchase ~ ., kernel = "polynomial", data = traind, cost = 0.01, degree = 2)
summary(svm.poly)##
## Call:
## svm(formula = Purchase ~ ., data = traind, 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: 630
##
## ( 318 312 )
##
##
## Number of Classes: 2
##
## Levels:
## CH MMtrain.preds = predict(svm.poly, traind)
table(traind$Purchase, train.preds)## train.preds
## CH MM
## CH 488 0
## MM 312 0(312) / (488 + 312)## [1] 0.39test.preds = predict(svm.poly, testd)
table(testd$Purchase, test.preds)## test.preds
## CH MM
## CH 165 0
## MM 105 0(105) / (165 + 105)## [1] 0.3888889Using an SVM with the polynomial kernel, the results are identical to the untuned radial method.
set.seed(123)
svm.poly.tuned = tune(svm, Purchase ~ ., data = traind, kernel = "polynomial", ranges = list(cost = c(0.01, 0.05, 0.1, 0.5, (seq(1, 10, length.out = 10)))), degree = 2)
summary(svm.poly.tuned)##
## Parameter tuning of 'svm':
##
## - sampling method: 10-fold cross validation
##
## - best parameters:
## cost
## 5
##
## - best performance: 0.17375
##
## - Detailed performance results:
## cost error dispersion
## 1 0.01 0.39000 0.07518496
## 2 0.05 0.33750 0.06744339
## 3 0.10 0.33125 0.07482619
## 4 0.50 0.20375 0.07096801
## 5 1.00 0.18375 0.05239076
## 6 2.00 0.18125 0.05810969
## 7 3.00 0.17750 0.06061032
## 8 4.00 0.17875 0.05622685
## 9 5.00 0.17375 0.04466309
## 10 6.00 0.17875 0.04715886
## 11 7.00 0.18125 0.05212498
## 12 8.00 0.18000 0.05688683
## 13 9.00 0.18375 0.05804991
## 14 10.00 0.18500 0.05583955svm.poly.twoned = svm(Purchase~., kernel = 'polynomial', data = traind, cost = svm.poly.tuned$best.parameters$cost)
train.preds = predict(svm.poly.twoned, traind)
table(traind$Purchase, train.preds)## train.preds
## CH MM
## CH 460 28
## MM 79 233(79 + 28) / (460 + 233 + 79 + 28)## [1] 0.13375test.preds = predict(svm.poly.twoned, testd)
table(testd$Purchase, test.preds)## test.preds
## CH MM
## CH 142 23
## MM 28 77(23 + 28) / (142 + 77 + 23 + 28)## [1] 0.1888889(h) Overall, which approach seems to give the best results on this data?
The Tuned SVM radial returned the best testing misclassification rate of 0.1814815, while the Tuned SVM for Linear and Polynomial kernels returned results fairly close at 0.1851852 and 0.1888889 respectively.
Comment on your results.
The Radial SVM worked much better than the Linear Kernel in predicting nonlinear results. Even on the same data, the linear model failed to predict properly. The model with interactions and squared inputs also performed fairly well.