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 pˆm1. The x-axis should display pˆm1, 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, 0.001)
gini <- 2 * p * (1 - p)
error <- 1 - pmax(p, 1 - p)
entropy <- - (p * log(p) + (1 - p) * log(1 - p))
entropy[is.nan(entropy)] <- 0 # handle log(0)
matplot(p, cbind(gini, error, entropy), type = "l", lty = 1, col = c("green", "blue", "orange"),
ylab = "Value", xlab = expression(hat(p)[m1]))
legend("top", legend = c("Gini", "Error", "Entropy"),
col = c("green", "blue", "orange"), lty = 1)
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(1)
train_idx <- sample(1:nrow(Carseats), nrow(Carseats) / 2)
strain <- Carseats[train_idx, ]
stest <- Carseats[-train_idx, ](b) Fit a regression tree to the training set. Plot the tree, and interpret the results. What test MSE do you obtain?
set.seed(1)
train_idx <- sample(1:nrow(Carseats), nrow(Carseats) / 2)
strain <- Carseats[train_idx, ]
stest <- Carseats[-train_idx, ]
tree.seats <- rpart(Sales ~ ., data = strain)
rpart.plot(tree.seats, type = 2, extra = 1, fallen.leaves = TRUE)
pred.b <- predict(tree.seats, newdata = stest)
mean((pred.b - stest$Sales)^2)[1] 5.165182Test MSE: 5.17 Top splits: ShelveLoc, Price, CompPrice Tree: Deep, uses several predictors, predicts Sales between ~3.9 and ~12
(c) Use cross-validation in order to determine the optimal level of tree complexity. Does pruning the tree improve the test MSE?
set.seed(1)
tree.seats <- tree(Sales ~ ., data = strain)
cv.seats <- cv.tree(tree.seats)
plot(cv.seats$size, cv.seats$dev, type = "b",
xlab = "Tree Size", ylab = "CV Deviance")
prune.seats <- prune.tree(tree.seats, best = 10)
plot(prune.seats)
text(prune.seats, pretty = 0)
pred.c <- predict(prune.seats, newdata = stest)
mean((pred.c - stest$Sales)^2)[1] 4.918134(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(1)
bag.seats <- randomForest(Sales ~ ., data = strain, mtry = ncol(strain) - 1, importance = TRUE)
pred.d <- predict(bag.seats, newdata = stest)
mean((pred.d - stest$Sales)^2)[1] 2.605253importance(bag.seats) %IncMSE IncNodePurity
CompPrice 24.8888481 170.182937
Income 4.7121131 91.264880
Advertising 12.7692401 97.164338
Population -1.8074075 58.244596
Price 56.3326252 502.903407
ShelveLoc 48.8886689 380.032715
Age 17.7275460 157.846774
Education 0.5962186 44.598731
Urban 0.1728373 9.822082
US 4.2172102 18.073863(e) Use random forests to analyze this data. What test MSE do you obtain? Use the importance() function to determine which variables are most important. Describe the effect of m, the number of variables considered at each split, on the error rate obtained.
set.seed(1)
rf.seats <- randomForest(Sales ~ ., data = strain, mtry = 4, importance = TRUE)
pred.e <- predict(rf.seats, newdata = stest)
mean((pred.e - stest$Sales)^2)[1] 2.787584importance(rf.seats) %IncMSE IncNodePurity
CompPrice 15.7891655 160.57944
Income 4.1275509 121.12953
Advertising 9.6425758 111.54581
Population -1.3596645 85.92575
Price 43.4055391 423.06225
ShelveLoc 37.8850232 311.97119
Age 13.8924424 174.18229
Education 0.1960888 62.77782
Urban 0.1393816 12.92952
US 6.3532441 30.42255(f) Now analyze the data using BART, and report your results.
library(BART)
x.train <- data.matrix(strain[, -which(names(strain) == "Sales")])
y.train <- strain$Sales
x.test <- data.matrix(stest[, -which(names(stest) == "Sales")])
y.test <- stest$Sales
set.seed(1)
bart.mod <- wbart(x.train, y.train, x.test)*****Into main of wbart
*****Data:
data:n,p,np: 200, 10, 200
y1,yn: 2.781850, 1.091850
x1,x[n*p]: 107.000000, 2.000000
xp1,xp[np*p]: 111.000000, 2.000000
*****Number of Trees: 200
*****Number of Cut Points: 63 ... 1
*****burn and ndpost: 100, 1000
*****Prior:beta,alpha,tau,nu,lambda: 2.000000,0.950000,0.273474,3.000000,0.692269
*****sigma: 1.885179
*****w (weights): 1.000000 ... 1.000000
*****Dirichlet:sparse,theta,omega,a,b,rho,augment: 0,0,1,0.5,1,10,0
*****nkeeptrain,nkeeptest,nkeeptestme,nkeeptreedraws: 1000,1000,1000,1000
*****printevery: 100
*****skiptr,skipte,skipteme,skiptreedraws: 1,1,1,1
MCMC
done 0 (out of 1100)
done 100 (out of 1100)
done 200 (out of 1100)
done 300 (out of 1100)
done 400 (out of 1100)
done 500 (out of 1100)
done 600 (out of 1100)
done 700 (out of 1100)
done 800 (out of 1100)
done 900 (out of 1100)
done 1000 (out of 1100)
time: 3s
check counts
trcnt,tecnt,temecnt,treedrawscnt: 1000,1000,1000,1000mean((bart.mod$yhat.test.mean - y.test)^2)[1] 1.4652549. This problem involves the OJ data set which is part of the ISLR2 package
(a) Create a training set containing a random sample of 800 observations, and a test set containing the remaining observations.
set.seed(1)
train_idx <- sample(1:nrow(OJ), 800)
OJtrain <- OJ[train_idx, ]
OJtest <- OJ[-train_idx, ](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 = OJtrain)
summary(tree.OJ)
Classification tree:
tree(formula = Purchase ~ ., data = OJtrain)
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 / 800 plot(tree.OJ)
text(tree.OJ, pretty = 0)
(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.OJnode), 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 ) *Node 9 has 118 observations where LoyalCH > 0.0356. The predicted class is MM, with 80.5% MM and 19.5% CH.
(d) Create a plot of the tree, and interpret the results.
plot(tree.OJ)
text(tree.OJ, pretty = 0)
(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, newdata = OJtest, type = "class")
table(tree.pred, OJtest$Purchase)
tree.pred CH MM
CH 160 38
MM 8 64mean(tree.pred != OJtest$Purchase)[1] 0.1703704(f) Apply the cv.tree() function to the training set in order to determine the optimal tree size.
cv.OJ <- cv.tree(tree.OJ, FUN = prune.misclass)
cv.OJ$size
[1] 9 8 7 4 2 1
$dev
[1] 150 150 149 158 172 315
$k
[1] -Inf 0.000000 3.000000 4.333333 10.500000 151.000000
$method
[1] "misclass"
attr(,"class")
[1] "prune" "tree.sequence"(g) Produce a plot with tree size on the x-axis and cross-validated classification error rate on the y-axis.
plot(cv.OJ$size, cv.OJ$dev, type = "b",
xlab = "Tree Size", ylab = "CV Classification Error")
(h) Which tree size corresponds to the lowest cross-validated classification error rate?
Size = 7
(i) Produce a pruned tree corresponding to the optimal tree size obtained using cross-validation. If cross-validation does not lead to selection of a pruned tree, then create a pruned tree with five terminal nodes.
prune.OJ <- prune.tree(tree.OJ, best = 7)
plot(prune.OJ)
text(prune.OJ, pretty = 0)
(j) Compare the training error rates between the pruned and pruned trees. Which is higher?
summary(tree.OJ)$misclass[1] 127 800summary(prune.OJ)$misclass[1] 130 800Unpruned training error: 127 / 800 = 15.88% Pruned training error: 130 / 800 = 16.25%
(k) Compare the test error rates between the pruned and unpruned trees. Which is higher?
pred.unpruned <- predict(tree.OJ, newdata = OJtest, type = "class")
mean(pred.unpruned != OJtest$Purchase)[1] 0.1703704pred.pruned <- predict(prune.OJ, newdata = OJtest, type = "class")
mean(pred.pruned != OJtest$Purchase)[1] 0.162963Unpruned test error: 17.04% Pruned test error: 16.30%