Data Algorithms II
Adrianne Kristianto
5/4/2021
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 xaxis should display ˆpm1, ranging from 0 to 1, and the y-axis should display the value of the Gini index, classification error, and entropy
p <- seq(0, 1, 0.01)
gini.index <- 2 * p * (1 - p)
class.error <- 1 - pmax(p, 1 - p)
cross.entropy <- - (p * log(p) + (1 - p) * log(1 - p))
par(bg = "papayawhip")
matplot(p, cbind(gini.index, class.error, cross.entropy), pch=c(15,17,19) ,ylab = "gini.index, class.error, cross.entropy",col = c("darkolivegreen4" , "wheat", "tomato2"), type = 'b')
legend('bottom', inset=.01, legend = c('gini.index', 'class.error', 'cross.entropy'), col = c("darkolivegreen4" , "wheat", "tomato2"), pch=c(15,17,19))
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.
a: Split the data set into a training set and a test set.
set.seed(9)
train <- sample(1:nrow(Carseats), nrow(Carseats)/2)
car.train <- Carseats[train, ]
car.test <- Carseats[-train, ]b: Fit a regression tree to the training set. Plot the tree, and interpret the results. What test MSE do you obtain?
car.tree <- tree(Sales ~ ., data = car.train)
plot(car.tree)
text(car.tree, pretty=0)
summary(car.tree)##
## Regression tree:
## tree(formula = Sales ~ ., data = car.train)
## Variables actually used in tree construction:
## [1] "ShelveLoc" "Price" "CompPrice" "Age" "Advertising"
## Number of terminal nodes: 16
## Residual mean deviance: 2.331 = 428.8 / 184
## Distribution of residuals:
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## -4.04200 -0.90650 0.05187 0.00000 0.99000 4.44700car.pred <- predict(car.tree, newdata = car.test)
mean((car.pred - car.test$Sales)^2)## [1] 4.908031c: Use cross-validation in order to determine the optimal level of tree complexity. Does pruning the tree improve the test MSE?
set.seed(9)
car.cv <- cv.tree(car.tree)
par(mfrow = c(1, 2))
plot(car.cv$size, car.cv$dev, type = "b")
plot(car.cv$k, car.cv$dev, type = "b")
# It looks like 7 is the best size.
prune.car <- prune.tree(car.tree, best = 7)
plot(prune.car)
text(prune.car, pretty = 0)
predict.prune <- predict(prune.car, newdata = car.test)
mean((predict.prune - car.test$Sales)^2)## [1] 5.09613
Comment: In this case, pruning actually makes the MSE better. It goes from 4.92 to 4.86.
d: 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(9)
bag.car <- randomForest(Sales ~ ., data = Carseats, subset = train, mtry = 10,
importance = TRUE)
bag.car##
## Call:
## randomForest(formula = Sales ~ ., data = Carseats, mtry = 10, importance = TRUE, subset = train)
## Type of random forest: regression
## Number of trees: 500
## No. of variables tried at each split: 10
##
## Mean of squared residuals: 2.742449
## % Var explained: 64.99predict.bag <- predict(bag.car, newdata = car.test)
mean((predict.bag - car.test$Sales)^2)## [1] 2.786846importance(bag.car)## %IncMSE IncNodePurity
## CompPrice 26.3543703 195.923764
## Income 5.1080067 73.140474
## Advertising 18.1866875 178.127342
## Population -2.4973779 57.821401
## Price 54.4216001 466.508363
## ShelveLoc 52.2979673 372.757219
## Age 13.9185606 123.663561
## Education 1.4945399 40.441604
## Urban -0.5639776 5.540011
## US 1.0388702 7.700311Comment: In this case, the MSE greatly reduced to 2.61. This is because the bagging method results in low bias and reduced variance. And it looks like ComPrice is the most important variable. This is because it has the highest Mean Decrease Gini, or IncNodePurity compared to all the other variables.
e: Use random forests to analyze this data. What test MSE do you obtain? Use the importance() function to determine which variables are most important. Describe the effect of m, the number of variables considered at each split, on the error rate obtained.
set.seed(9)
rf.car1 <- randomForest(Sales ~ ., data = Carseats, subset = train, mtry = 1,
importance = TRUE)
set.seed(9)
pred.rf1 <- predict(rf.car1, newdata = car.test)
mean((pred.rf1 - car.test$Sales)^2)## [1] 4.868426Comment: When m is equal to 1, MSE is 4.82
rf.carseats <- randomForest(Sales ~ ., data = car.train, mtry = 3, ntree = 500, importance = TRUE)
yhat.rf <- predict(rf.carseats, newdata = car.test)
mean((yhat.rf - car.test$Sales)^2)## [1] 3.201258Comment: In this case, with \[m = \sqrt{p}\], we have MSE = 2.97
importance(rf.carseats)## %IncMSE IncNodePurity
## CompPrice 14.1241022 164.60718
## Income 4.2380268 121.58845
## Advertising 16.5389439 164.30667
## Population -1.3439611 107.93677
## Price 39.2919648 367.66652
## ShelveLoc 39.1313149 305.66983
## Age 9.8914491 153.89346
## Education 0.9465834 65.88861
## Urban -0.4255006 14.41895
## US 2.6321970 16.74755Comment: In this case, the most two important variables are Price and ShelveLoc
Question 9
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.
set.seed(9)
train <- sample(1:nrow(OJ), 800)
OJ.train <- OJ[train,]
OJ.test <- OJ[-train,]b: 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?
tree.oj <- tree(Purchase ~ ., data = OJ.train)
summary(tree.oj)##
## Classification tree:
## tree(formula = Purchase ~ ., data = OJ.train)
## Variables actually used in tree construction:
## [1] "LoyalCH" "PriceDiff" "ListPriceDiff"
## Number of terminal nodes: 8
## Residual mean deviance: 0.7319 = 579.7 / 792
## Misclassification error rate: 0.1512 = 121 / 800Comment: The tree has a misclassification error rate of 15.12%, which isn’t bad at all, and it has 8 terminal nodes.
c: 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.
tree.oj## node), split, n, deviance, yval, (yprob)
## * denotes terminal node
##
## 1) root 800 1053.000 CH ( 0.63125 0.36875 )
## 2) LoyalCH < 0.48285 288 321.700 MM ( 0.24653 0.75347 )
## 4) LoyalCH < 0.280875 159 120.300 MM ( 0.12579 0.87421 )
## 8) LoyalCH < 0.035047 53 9.922 MM ( 0.01887 0.98113 ) *
## 9) LoyalCH > 0.035047 106 99.690 MM ( 0.17925 0.82075 ) *
## 5) LoyalCH > 0.280875 129 173.100 MM ( 0.39535 0.60465 )
## 10) PriceDiff < 0.25 84 98.620 MM ( 0.27381 0.72619 ) *
## 11) PriceDiff > 0.25 45 59.670 CH ( 0.62222 0.37778 ) *
## 3) LoyalCH > 0.48285 512 437.000 CH ( 0.84766 0.15234 )
## 6) LoyalCH < 0.764572 252 297.800 CH ( 0.72222 0.27778 )
## 12) PriceDiff < -0.165 39 46.400 MM ( 0.28205 0.71795 ) *
## 13) PriceDiff > -0.165 213 211.500 CH ( 0.80282 0.19718 )
## 26) ListPriceDiff < 0.135 36 49.800 CH ( 0.52778 0.47222 ) *
## 27) ListPriceDiff > 0.135 177 144.200 CH ( 0.85876 0.14124 ) *
## 7) LoyalCH > 0.764572 260 71.450 CH ( 0.96923 0.03077 ) *Comment: The 8th node is picked. It has a split criterino of < 0.035, 57 observations in the branch with a deviance of 9.922. only 1.7% of the observations in that branch take the value of CH, and the remaining 98% take the value of MM.
d: Create a plot of the tree, and interpret the results
plot(tree.oj)
text(tree.oj, pretty = 0)
Comment: From the plot above, we can see that the most important predictor for Purchase is LoyalCH because it’s set as the first branch which determines the intensity of customer brand loyalty.
e: 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?
tree.pred <- predict(tree.oj, OJ.test, type = "class")
table(tree.pred, OJ.test$Purchase)##
## tree.pred CH MM
## CH 132 37
## MM 16 85tree.test.error = round(mean(tree.pred != OJ.test$Purchase)*100,2)
tree.test.error## [1] 19.63f: Apply the cv.tree() function to the training set in order to determine the optimal tree size.
cv.oj <- cv.tree(tree.oj, FUN = prune.misclass)
cv.oj## $size
## [1] 8 6 4 2 1
##
## $dev
## [1] 143 143 149 154 295
##
## $k
## [1] -Inf 0.0 5.5 8.5 146.0
##
## $method
## [1] "misclass"
##
## attr(,"class")
## [1] "prune" "tree.sequence"g: Produce a plot with tree size on the x-axis and cross-validated classification error rate on the y-axis.
plot(cv.oj$size, cv.oj$dev, type = "b", xlab = "Tree size", ylab = "Deviance")
h: Which tree size corresponds to the lowest cross-validated classification error rate?
Answer: We can see that tree size 8 corresponds to the lowest cross-validated classification error rate.
j: Compare the training error rates between the pruned and unpruned trees. Which is higher?
prune.oj = prune.misclass(tree.oj, best=2)
summary(prune.oj)##
## Classification tree:
## snip.tree(tree = tree.oj, nodes = 2:3)
## Variables actually used in tree construction:
## [1] "LoyalCH"
## Number of terminal nodes: 2
## Residual mean deviance: 0.9507 = 758.7 / 798
## Misclassification error rate: 0.1862 = 149 / 800summary(tree.oj)##
## Classification tree:
## tree(formula = Purchase ~ ., data = OJ.train)
## Variables actually used in tree construction:
## [1] "LoyalCH" "PriceDiff" "ListPriceDiff"
## Number of terminal nodes: 8
## Residual mean deviance: 0.7319 = 579.7 / 792
## Misclassification error rate: 0.1512 = 121 / 800Comment: The misclassification error rate for tree.oj (unpruned) is slightly lower (15%) than prune.oj (18%)
k: Compare the test error rates between the pruned and unpruned trees. Which is higher?
prune.pred <- predict(prune.oj, OJ.test, type = "class")
table(prune.pred, OJ.test$Purchase)##
## prune.pred CH MM
## CH 125 32
## MM 23 90pruned.test.error = round(mean(prune.pred != OJ.test$Purchase)*100,2)
pruned.test.error## [1] 20.37Comment: The test error rates is slightly higher for pruned tree than unpruned (20.37%)