STA 6543 : Assignment 7 (ISLR: Chapter 8 - Tree Based Methods)
Mihreteab Teklehaimanot
4/20/2022
Exercise 5.
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.
p = seq(0, 1, 0.001)
gini = p * (1 - p) * 2
entropy = -(p * log(p) + (1 - p) * log(1 - p))
classification.error = 1 - pmax(p, 1 - p)
data = data.frame(p, gini, entropy, classification.error)
matplot(p, cbind(gini, entropy, classification.error), col = c("Blue", "red", "brown"), ylab = "Value", xlab = "p", main = "Plot of metrics as a function of p", pch = 20, lwd = 1, type = "p")
- The Highest values for the three metrics is at 0.5.
Exercise 8.
(a) Split the data set into a training set and a test set.
attach(Carseats)
set.seed(1)
Train = sample(nrow(Carseats), nrow(Carseats)/2)
cseat_train = Carseats[Train, ]
cseat_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?
carseat_tree = tree(Sales ~ ., data = cseat_train)
summary(carseat_tree)##
## Regression tree:
## tree(formula = Sales ~ ., data = cseat_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 4.09900plot(carseat_tree)
text(carseat_tree, pretty = 0, cex=.55)
Predict and compute mse
carseat_pred = predict(carseat_tree, cseat_test)
carseat_mse<-mean((cseat_test$Sales - carseat_pred)^2)
carseat_mse## [1] 4.922039- The test MSE is 4.022039
(c) Use cross-validation in order to determine the optimal level of tree complexity. Does pruning the tree improve the test MSE?
cv_carseat = cv.tree(carseat_tree, FUN = prune.tree)
par(mfrow = c(1, 2))
plot(cv_carseat$size, cv_carseat$dev, type = "b")
plot(cv_carseat$k, cv_carseat$dev, type = "b")
- The best size is 9
Use best = 9 to prune tree
carseat_prune = prune.tree(carseat_tree, best = 9)
par(mfrow = c(1, 1))
plot(carseat_prune)
text(carseat_prune, pretty = 0, cex=.55)
prune_pred = predict(carseat_prune, cseat_test)
prune_mse=mean((cseat_test$Sales - prune_pred)^2)
prune_mse## [1] 4.918134Pruning decreased the MSE error slightly.
(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.
car_bag = randomForest(Sales ~ ., data = cseat_train, mtry = 10, ntree = 100, importance = T)
pred_bag = predict(car_bag, cseat_test)
bag_mse=mean((cseat_test$Sales - pred_bag)^2)
bag_mse## [1] 2.747535importance(car_bag)## %IncMSE IncNodePurity
## CompPrice 12.55183177 170.375825
## Income 0.93971125 90.503890
## Advertising 5.25205647 103.493698
## Population -1.46213439 59.646975
## Price 24.29931798 497.717927
## ShelveLoc 20.34955441 367.886332
## Age 6.75952990 154.875691
## Education -0.01964881 50.794521
## Urban -2.22920227 7.587184
## US 1.64730860 16.342743Price, ShelveLoc, CompPrice and Age are important variables in predicting sales.
(e) Use random forests to analyze this data. What test MSE do you obtain? Use the importance() function to determine which variables are most important. Describe the effect of m, the number of variables considered at each split, on the error rate obtained.
set.seed(1)
carseat_rf = randomForest(Sales ~ ., data = cseat_train, mtry = 9, ntree = 1000,
importance = T)
pred_rf = predict(carseat_rf, cseat_test)
(rf_mse=mean((cseat_test$Sales - pred_rf)^2))## [1] 2.604951importance(carseat_rf)## %IncMSE IncNodePurity
## CompPrice 35.7427013 171.232879
## Income 7.9028240 92.450563
## Advertising 19.3708369 101.291431
## Population -2.0231580 60.542328
## Price 77.4780930 503.635424
## ShelveLoc 69.3276236 367.336695
## Age 21.9503880 158.867412
## Education -0.8694952 43.216765
## Urban -0.9832457 8.928158
## US 7.9185075 17.933675- once again Price, ShelveLoc, CompPrice and Age are important variables in predicting sales. Changing m changes the test RSE as well.
Problem 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(1)
ttrain = sample(nrow(OJ), 800)
OJ.train = OJ[ttrain,]
OJ.test = OJ[-ttrain,](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?
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 / 800- Variables actually used in tree construction: LoyalCH, PriceDiff, SpecialCH, ListPriceDiff, and PctDiscMM.
- Training error rate, which is the misclassification error in summary, for the tree is 0.1588.
- It has 7 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.
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 < 0.196197 55 73.14 CH ( 0.61818 0.38182 ) *
## 25) PctDiscMM > 0.196197 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 ) *Terminal node “20”. The splitting variable at this node is SpecialCH. The splitting value of this node is 0.5. There are 64 points in the subtree below this node.The deviance for all points contained in region below this node is 51.98. It is a terminal node denoted by the astrix sign.The prediction at this node is Sales = MM. 40.4% points in this node have CH as value of Sales. Remaining 59.6% points have MM as value of Sales.
(d) Create a plot of the tree, and interpret the results.
plot(OJ.tree)
text(OJ.tree, pretty = 0,cex=0.55)
- LoyalCH is the most important variable of the tree. If LoyalCH< 0.28, the tree predicts MM. If LoyalCH>0.76, the tree predicts CH. For intermediate values of LoyalCH, the decision also depends on the value of the PriceDiff, SpecialCH, ListPriceDiff and PctDiscMM variables.
(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?
pred.oj= predict(OJ.tree, OJ.test, type = "class")
caret:: confusionMatrix(OJ.test$Purchase, pred.oj)## Confusion Matrix and Statistics
##
## Reference
## Prediction CH MM
## CH 160 8
## MM 38 64
##
## Accuracy : 0.8296
## 95% CI : (0.7794, 0.8725)
## No Information Rate : 0.7333
## P-Value [Acc > NIR] : 0.0001259
##
## Kappa : 0.6154
##
## Mcnemar's Test P-Value : 1.904e-05
##
## Sensitivity : 0.8081
## Specificity : 0.8889
## Pos Pred Value : 0.9524
## Neg Pred Value : 0.6275
## Prevalence : 0.7333
## Detection Rate : 0.5926
## Detection Prevalence : 0.6222
## Balanced Accuracy : 0.8485
##
## 'Positive' Class : CH
## (f) Apply the cv.tree() function to the training set in order to determine the optimal tree size.
cv.oj = cv.tree(OJ.tree, FUN = prune.tree)(g) Produce a plot with tree size on the x-axis and cross-validated classification error rate on the y-axis.
plot(cv.oj$size, cv.oj$dev, type = "b", xlab = "Tree Size", ylab = "Deviance")
(h) Which tree size corresponds to the lowest cross-validated classification error rate?
- Size of 9 gives lowest cross-validation error.
(i) Produce a pruned tree corresponding to the optimal tree size obtained using cross-validation. If cross-validation does not lead to selection of a pruned tree, then create a pruned tree with five terminal nodes.
OJ.prune = prune.tree(OJ.tree, best = 9)####(j) Compare the training error rates between the pruned and unpruned trees. Which is higher?
summary(OJ.prune)##
## Classification tree:
## tree(formula = Purchase ~ ., data = OJ.train)
## Variables actually used in tree construction:
## [1] "LoyalCH" "PriceDiff" "SpecialCH" "ListPriceDiff"
## [5] "PctDiscMM"
## Number of terminal nodes: 9
## Residual mean deviance: 0.7432 = 587.8 / 791
## Misclassification error rate: 0.1588 = 127 / 800The results are exactly the same the original tree.
(k) Compare the test error rates between the pruned and unpruned trees. Which is higher?
unpruned_pred = predict(OJ.tree, OJ.test, type = "class")
unpruned_error = sum(OJ.test$Purchase != unpruned_pred)
unpruned_error/length(unpruned_pred)## [1] 0.1703704pruned_pred = predict(OJ.prune, OJ.test, type = "class")
pruned_error = sum(OJ.test$Purchase != pruned_pred)
pruned_error/length(pruned_pred)## [1] 0.1703704The same result here as well.