Support Vector Machines
Sanjana Kumar
2023-04-19
Question 3
Consider the Gini index, classification error, and entropy in a simple classification setting with two classes. Create a single plot that displays each of these quantities as a function of ˆpm1. The x axis should display ˆpm1, ranging from 0 to 1, and the y-axis should display the value of the Gini index, classification error, and entropy. Hint: In a setting with two classes, pˆm1 = 1 − pˆm2. You could make this plot by hand, but it will be much easier to make in R.
p=seq(0,1,0.01)
gini= p*(1-p)*2
class.entropy= -(p*log(p)+(1-p)*log(1-p))
class.error= 1-pmax(p,1-p)
plot(NA,NA,xlim=c(0,1),ylim=c(0,1),xlab='p',ylab='values')
lines(p,gini,col = 'blue')
lines(p,class.error,col='darkgrey')
lines(p,class.entropy,col='lightblue')
legend(x='topright',legend=c('gini','class error','cross entropy'),
col=c('blue','black','green'),lty=1,text.width = 0.22)
Question 8
In the lab, a classification tree was applied to the Carseats data set after converting Sales into a qualitative response variable. Now we will seek to predict Sales using regression trees and related approaches, treating the response as a quantitative variable.
- Split the data set into a training set and a test set.
- Fit a regression tree to the training set. Plot the tree, and interpret the results. What test MSE do you obtain?
##
## Regression tree:
## tree(formula = Sales ~ ., data = carseats.train)
## Variables actually used in tree construction:
## [1] "ShelveLoc" "Price" "Age" "Advertising" "CompPrice"
## [6] "US"
## Number of terminal nodes: 18
## Residual mean deviance: 2.167 = 394.3 / 182
## Distribution of residuals:
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## -3.88200 -0.88200 -0.08712 0.00000 0.89590 4.09900
## [1] 4.922039The MSE obtained is 4.922
- Use cross-validation in order to determine the optimal level of tree complexity. Does pruning the tree improve the test MSE?
carseats.cv <- cv.tree(carseats.tree, FUN = prune.tree)
par(mfrow = c(1, 2))
plot(carseats.cv$size, carseats.cv$dev, type = "b")
plot(carseats.cv$k, carseats.cv$dev, type = "b")
pruned.carseats = prune.tree(carseats.tree, best = 13)
par(mfrow = c(1, 1))
plot(pruned.carseats)
text(pruned.carseats, pretty = 0)
prune.pred <- predict(pruned.carseats, carseats.test)
prune.mse <- mean((carseats.test$Sales - prune.pred)^2)
prune.mse## [1] 4.96547The MSE for the pruned data is 4.965.
- Use the bagging approach in order to analyze this data. What test MSE do you obtain? Use the importance() function to determine which variables are most important.
set.seed(1)
bag.car <- randomForest(Sales ~ ., data = carseats.train, mtry = 10, importance = TRUE)
bag.car##
## Call:
## randomForest(formula = Sales ~ ., data = carseats.train, mtry = 10, importance = TRUE)
## Type of random forest: regression
## Number of trees: 500
## No. of variables tried at each split: 10
##
## Mean of squared residuals: 2.889221
## % Var explained: 63.26bag.pred <- predict(bag.car, carseats.test)
bag.err <- mean((carseats.test$Sales - bag.pred)^2)
bag.err## [1] 2.605253importance(bag.car)## %IncMSE IncNodePurity
## CompPrice 24.8888481 170.182937
## Income 4.7121131 91.264880
## Advertising 12.7692401 97.164338
## Population -1.8074075 58.244596
## Price 56.3326252 502.903407
## ShelveLoc 48.8886689 380.032715
## Age 17.7275460 157.846774
## Education 0.5962186 44.598731
## Urban 0.1728373 9.822082
## US 4.2172102 18.073863The importance output shows that CompPrice, Price and Sheloc are the the most important predictors. Using the bagging approach, it reduces the MSE to 2.605.
- Use random forests to analyze this data. What test MSE do you obtain? Use the importance() function to determine which variables are most important. Describe the effect of m, the number of variables considered at each split, on the error rate obtained.
rf.car <- randomForest(Sales ~ ., data = carseats.train, mtry = 5, importance = TRUE)
rf.pred <- predict(rf.car, carseats.test)
rf.err <- mean((carseats.test$Sales - rf.pred)^2)
rf.err## [1] 2.710806importance(rf.car)## %IncMSE IncNodePurity
## CompPrice 18.3350929 163.77398
## Income 4.7546303 115.94202
## Advertising 10.4995404 98.99790
## Population -0.8865995 78.91872
## Price 45.2942175 451.27503
## ShelveLoc 40.8250417 333.45728
## Age 13.6552105 165.03566
## Education 0.9386481 56.40921
## Urban -0.8796633 11.21732
## US 6.1388918 25.33755Again we see that CompPrice, Price, and Shelveloc are the most important predictors. The MSE here us 2.710. This is slightly worse than in part d, the bagging approach.
Question 9
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.
Fit a tree to the training data, with Purchase as the response and the other variables as predictors. Use the summary() function to produce summary statistics about the tree, and describe the results obtained. What is the training error rate? How many terminal nodes does the tree have?
set.seed(1)
sample.train <- sample(dim(OJ)[1], 800)
oj.train <- OJ[sample.train, ]
oj.test <- OJ[-sample.train, ]
oj.tree <- tree(Purchase~., data= oj.train)
summary(oj.tree)##
## Classification tree:
## tree(formula = Purchase ~ ., data = oj.train)
## Variables actually used in tree construction:
## [1] "LoyalCH" "PriceDiff" "SpecialCH" "ListPriceDiff"
## [5] "PctDiscMM"
## Number of terminal nodes: 9
## Residual mean deviance: 0.7432 = 587.8 / 791
## Misclassification error rate: 0.1588 = 127 / 800The training error rate is 0.1588 for the tree and the tree has 9 terminal nodes.
- Type in the name of the tree object in order to get a detailed text output. Pick one of the terminal nodes, and interpret the information displayed.
oj.tree## node), split, n, deviance, yval, (yprob)
## * denotes terminal node
##
## 1) root 800 1073.00 CH ( 0.60625 0.39375 )
## 2) LoyalCH < 0.5036 365 441.60 MM ( 0.29315 0.70685 )
## 4) LoyalCH < 0.280875 177 140.50 MM ( 0.13559 0.86441 )
## 8) LoyalCH < 0.0356415 59 10.14 MM ( 0.01695 0.98305 ) *
## 9) LoyalCH > 0.0356415 118 116.40 MM ( 0.19492 0.80508 ) *
## 5) LoyalCH > 0.280875 188 258.00 MM ( 0.44149 0.55851 )
## 10) PriceDiff < 0.05 79 84.79 MM ( 0.22785 0.77215 )
## 20) SpecialCH < 0.5 64 51.98 MM ( 0.14062 0.85938 ) *
## 21) SpecialCH > 0.5 15 20.19 CH ( 0.60000 0.40000 ) *
## 11) PriceDiff > 0.05 109 147.00 CH ( 0.59633 0.40367 ) *
## 3) LoyalCH > 0.5036 435 337.90 CH ( 0.86897 0.13103 )
## 6) LoyalCH < 0.764572 174 201.00 CH ( 0.73563 0.26437 )
## 12) ListPriceDiff < 0.235 72 99.81 MM ( 0.50000 0.50000 )
## 24) PctDiscMM < 0.196196 55 73.14 CH ( 0.61818 0.38182 ) *
## 25) PctDiscMM > 0.196196 17 12.32 MM ( 0.11765 0.88235 ) *
## 13) ListPriceDiff > 0.235 102 65.43 CH ( 0.90196 0.09804 ) *
## 7) LoyalCH > 0.764572 261 91.20 CH ( 0.95785 0.04215 ) *Looking at Node 2, where LoyalCH is a terminal node with a split of 0.5036. There are 365 observations, the final prediction for this is MM. There is a probability range of ( 0.29315 0.70685 ), meaning about 29% of the observations in the branch have a CH value and the other 71% have a MM value.
- Create a plot of the tree, and interpret the results.
plot(oj.tree)
text(oj.tree, pretty = 0)
The tree shows that LoyalCH is the most important predictor in the tree
and in the top three nodes.
- Predict the response on the test data, and produce a confusion matrix comparing the test labels to the predicted test labels. What is the test error rate?
oj.pred <- predict(oj.tree, oj.test, type = "class")
table(oj.test$Purchase, oj.pred)## oj.pred
## CH MM
## CH 160 8
## MM 38 64(38+8)/270## [1] 0.1703704The test error rate is 0.170
Apply the cv.tree() function to the training set in order to determine the optimal tree size.
Produce a plot with tree size on the x-axis and cross-validated classification error rate on the y-axis.
cv.oj <- cv.tree(oj.tree, FUN = prune.tree)
cv.oj## $size
## [1] 9 8 7 6 5 4 3 2 1
##
## $dev
## [1] 685.6493 698.8799 702.8083 702.8083 714.1093 725.4734 780.2099
## [8] 790.0301 1074.2062
##
## $k
## [1] -Inf 12.62207 13.94616 14.35384 26.21539 35.74964 43.07317
## [8] 45.67120 293.15784
##
## $method
## [1] "deviance"
##
## attr(,"class")
## [1] "prune" "tree.sequence"plot(cv.oj$size, cv.oj$dev, type = "b", xlab = "Tree Size", ylab = "Cross-Validation Classification Error")
(h) Which tree size corresponds to the lowest cross-validated
classification error rate?
The plot above shows a tree size of 6 appears to have the lowest cross-validated classfication error.
- Produce a pruned tree corresponding to the optimal tree size obtained using cross-validation. If cross-validation does not lead to selection of a pruned tree, then create a pruned tree with five terminal nodes.
oj.prune <- prune.misclass(oj.tree, best = 5)
plot(oj.prune)
text(oj.prune, pretty = 0)
(j) Compare the training error rates between the pruned and unpruned
trees. Which is higher?
summary(oj.tree) ##
## Classification tree:
## tree(formula = Purchase ~ ., data = oj.train)
## Variables actually used in tree construction:
## [1] "LoyalCH" "PriceDiff" "SpecialCH" "ListPriceDiff"
## [5] "PctDiscMM"
## Number of terminal nodes: 9
## Residual mean deviance: 0.7432 = 587.8 / 791
## Misclassification error rate: 0.1588 = 127 / 800summary(oj.prune) ##
## Classification tree:
## snip.tree(tree = oj.tree, nodes = c(4L, 10L))
## Variables actually used in tree construction:
## [1] "LoyalCH" "PriceDiff" "ListPriceDiff" "PctDiscMM"
## Number of terminal nodes: 7
## Residual mean deviance: 0.7748 = 614.4 / 793
## Misclassification error rate: 0.1625 = 130 / 800The prunes trees has a higher training error rate, 0.7748, when compared to the unpruned tree’s test error rate, 0.7432.
- Compare the test error rates between the pruned and unpruned trees. Which is higher?
prune.pred <- predict(oj.prune, oj.test, type = "class")
table(prune.pred, oj.test$Purchase)##
## prune.pred CH MM
## CH 160 36
## MM 8 661-(160+66)/270## [1] 0.162963The pruned trees, 0.1629, increase the test error as opposed to the unpruned model.