Lab 8 – Tree-based Methods

Bagging and Random Forests

We now apply bagging and random forests to improve our prediction. Bagging is the idea that the high variance of a single bushy tree can be reduced by bootstapping samples and averaging over trees that were grown on the samples.

Note: Bagging gets an estimate of the test error for free as it always leaves out some observations when a tree is fit. The reported out-of-bag MSE is thus an estimate of test error. We also estimate test error separately on a test set. This is one particular test set, so the test error may vary.

In our run below the OOB MSE may be a better estimate of test error. It is reported to be lower than our test error estimate.

library(randomForest)

## randomForest 4.6-12

## Type rfNews() to see new features/changes/bug fixes.

bag1 <- randomForest(Income ~ . , mtry = 19, data = bes, subset = train, importance = TRUE)
bag1

## 
## Call:
##  randomForest(formula = Income ~ ., data = bes, mtry = 19, importance = TRUE,      subset = train) 
##                Type of random forest: regression
##                      Number of trees: 500
## No. of variables tried at each split: 19
## 
##           Mean of squared residuals: 3.545845
##                     % Var explained: 38.59

# predict outcome, illustrate, MSE
yhat.bag <- predict(bag1, newdata = bes.test)
plot(yhat.bag, bes.test$Income)
abline(0, 1) # line of 1:1 perfect prediction

mean( (yhat.bag - bes.test$Income)^2 )

## [1] 3.836317

# reduction of error
(3.836317 - 4.200642) / 3.836317

## [1] -0.09496739

We reduce the error rate by \(9\%\) by using bagging.

We examine what happens when we reduce the number of trees we grow. The default is 500.

# dcrease the number of trees (defaults to 500)
bag2 <- randomForest(Income ~ ., mtry = 19, data = bes, subset = train, ntree = 25, importance = TRUE)

# predict outcome
yhat.bag <- predict(bag2, newdata = bes.test)
mean( (yhat.bag - bes.test$Income)^2 )

## [1] 4.001648

The result is that our rate increases substantially again.

We now apply random forest. The trick is to decorrelate the trees by randomly considering only a subset of variables at every split. We thereby reduce variance further. The number of variables argument mtry is a tuning parameter.

## Random Forest: not trying all vars at each split decorrelates the trees
set.seed(123)

# we try to find the optimal tuning parameter for the number of variables to use at each split
oob.error <- NA
val.set.error <- NA
for ( idx in 1:10){
  rf1 <- randomForest(Income ~ ., mtry = idx, data = bes, subset = train, importance = TRUE)
  # record out of bag error
  oob.error[idx] <- rf1$mse[length(rf1$mse)]
  cat(paste("\n", "Use ", idx, " variables at each split", sep=""))
  # record validation set error
  val.set.error[idx] <- mean( (predict(rf1, newdata = bes.test) - bes.test$Income)^2 )
}

## 
## Use 1 variables at each split
## Use 2 variables at each split
## Use 3 variables at each split
## Use 4 variables at each split
## Use 5 variables at each split
## Use 6 variables at each split
## Use 7 variables at each split
## Use 8 variables at each split
## Use 9 variables at each split
## Use 10 variables at each split

# check optimal values for mtry
matplot( 1:idx, cbind(oob.error, val.set.error), pch = 19, col = c("red", "blue"),
         type = "b", ylab = "MSE", frame.plot = FALSE)
legend("topright", legend = c("OOB", "Val. Set"), pch = 19, col = c("red", "blue"),
       bty = "n")

We use 4 as the optimal value for mtry.

rf1 <- randomForest(Income ~ ., mtry = 4, data = bes, subset = train, importance = TRUE)

# predict outcomes
yhat.rf <- predict(rf1, newdata = bes.test)
mean( (yhat.rf - bes.test$Income)^2 )

## [1] 3.738068

# reduction in error
(3.738068 - 3.836317) / 3.836317

## [1] -0.02561024

We reduced the error rate by another \(3\%\) by decorrelating the trees. We can exmine variable importance as well. Variable reduction is obtained by as the average that a predictor reduces error at splits within a tree where it was used and averaged again over all trees. Similarly, node purity is based on the gini index of how heterogenous a group becomes due to a split.

# which varaibles help explain outcome
importance(rf1)

##                     %IncMSE IncNodePurity
## cs_id            2.06598543    1191.62963
## Turnout         10.63267089     187.08688
## Vote2001         4.74392742     134.49639
## Age             61.67167351    2726.49512
## Gender           7.75324398     270.15383
## PartyID         -1.30764676     186.00643
## Influence        8.25220908     622.72510
## Attention       13.48430187     734.16131
## Telephone       41.55335004     420.28959
## LeftrightSelf    3.84799215     701.93854
## CivicDutyIndex  19.09482995     765.92377
## polinfoindex    12.62283324     761.74783
## edu15           22.02752039     856.04335
## edu16           11.78494180     286.72268
## edu17            8.09626481     109.68566
## edu18            8.90440198     102.36531
## edu19plus       53.12112919    2236.08056
## in_school        6.07771141      77.57576
## in_uni          -0.02058015      51.77373
## CivicDutyScores 18.09776217    1145.64391

# importance plot
varImpPlot(rf1)

Boosting

library(gbm)

## Loading required package: survival

## 
## Attaching package: 'survival'

## The following object is masked from 'package:boot':
## 
##     aml

## Loading required package: lattice

## 
## Attaching package: 'lattice'

## The following object is masked from 'package:boot':
## 
##     melanoma

## Loading required package: parallel

## Loaded gbm 2.1.3

set.seed(1234)

We run gradient boosting. The tuning parameters are the tree size. Tree size is directly related to the second tuning parameter: the learning rate. When the learning rate is smaller, we need more trees. The third tuning parameter interaction.depth determines how bushy the tree is. Common choices are 1, 2, 4, 8. When interaction depth is 1, each tree is a stump. If we increase to two we can get bivariate interactions with 2 splits and so. A final parameter that is related to the complexity of the tree could be minimum number of observations in the terminal node which defaults to 10.

# gradient boosting
gb1 <- gbm(Income ~ ., data = bes[train, ], 
           distribution = "gaussian", 
           n.trees = 5000, 
           interaction.depth = 4,
           shrinkage = 0.001)

summary(gb1)

##                             var    rel.inf
## edu19plus             edu19plus 41.4726490
## Age                         Age 28.6747586
## Telephone             Telephone  7.0882505
## polinfoindex       polinfoindex  5.3053648
## Gender                   Gender  2.5211165
## edu15                     edu15  2.4673348
## cs_id                     cs_id  2.1982736
## LeftrightSelf     LeftrightSelf  1.7370627
## Turnout                 Turnout  1.4784065
## in_school             in_school  1.2178698
## CivicDutyScores CivicDutyScores  1.1787687
## Attention             Attention  1.1600579
## edu16                     edu16  0.9406745
## Influence             Influence  0.9211903
## CivicDutyIndex   CivicDutyIndex  0.6809589
## Vote2001               Vote2001  0.3880845
## edu17                     edu17  0.2379545
## edu18                     edu18  0.1226292
## PartyID                 PartyID  0.1054473
## in_uni                   in_uni  0.1031473

# partial dependence plots
plot(gb1, i = "edu19plus")

plot(gb1, i = "Age")

# predict outcome
yhat.gb <- predict(gb1, newdata = bes.test, n.trees = 5000)
mean( (yhat.gb - bes.test$Income)^2 )

## [1] 3.720392

# reduction in error
(3.716085 - 3.738321) / 3.738321

## [1] -0.005948125

We reduce the error rate by roughly another half percent.

Q1

We applied random forests for varying mtry. In addition vary the number of trees and create a plot that displays both test error and OOB error.

Q2

Looking at the Carseats data from the ISLR package we will seek to predict Sales using regression trees and related approaches, treating the response as a quantitative variable.

1. Split the data set into a training set and a test set.
2. Fit a regression tree to the training set. Plot the tree, and interpret the results. What test error rate do you obtain?
3. Use cross-validation in order to determine the optimal level of tree complexity. Does pruning the tree improve the test error rate?
4. Use the bagging approach in order to analyze this data. What test error rate do you obtain? Use the importance() function to determine which variables are most important.
5. Use random forests to analyze this data. What test error rate do you obtain? Use the importance() function to determine which variables are most important. Describe the effect of mm, the number of variables considered at each split, on the error rate obtained.

Q3

This problem involves the OJ dataset which is part of the ISLR package.

1. Create a training set containing a random sample of 800 observations, and a test set containing the remaining observations.
2. Fit a tree to the training data, with Purchase as the response and the other variables except for Buy 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?
3. 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.
4. Create a plot of the tree, and interpret the results.
5. 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?
6. Produce a plot with tree size on the \(x\)-axis and cross-validated classification error rate on the \(y\)-axis.
7. Which tree size corresponds to the lowest cross-validated classification error rate?
8. 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.
9. Compare the training error rates between the pruned and unpruned trees. Which is higher?
10. Compare the test error rates between the pruned and unpruned trees. Which is higher?

Q4

We now use boosting to predict Salary in the Hitters dataset, which is part of the ISLR package.

1. Remove the observations for whom the salary information is unknown, and then log-transform the salaries.
2. Create a training set consisting of the first 200 observations, and a test set consisting of the remaining observations.
3. Perform boosting on the training set with \(1,000\) trees for a range of values of the shrinkage parameter \(\lambda\). Produce a plot with different shrinkage values on the \(x\)-axis and the corresponding training set MSE on the \(y\)-axis.
4. Produce a plot with different shrinkage values on the \(x\)-axis and the corresponding test set MSE on the \(y\)-axis.
5. Compare the test MSE of boosting to the test MSE that results from applying two of the regression approaches seen in our discussions of regression models.
6. Which variables appear to be the most important predictors in the boosted model?
7. Now apply bagging to the training set. What is the test set MSE for this approach?

Q5

This question uses the Caravan dataset, part of the ISRL package.

1. Create a training set consisting of the first 1,000 observations, and a test set consisting of the remaining observations.
2. Fit a boosting model to the training set with Purchase as the response and the other variables as predictors. Use 1,000 trees, and a shrinkage value of 0.01. Which predictors appear to be the most important?
3. Use the boosting model to predict the response on the test data. Predict that a person will make a purchase if the estimated probability of purchase is greater than 20%. Form a confusion matrix. What fraction of the people predicted to make a purchase do in fact make one? How does this compare with the results obtained from applying KNN or logistic regression to this data set?