Lab 4 Solutions
Prof. Chouldechova
library(ggplot2) # graphics library
library(ISLR) # contains code and data from the textbook
library(knitr) # contains kable() function
library(tree) # For the tree-fitting 'tree' function
library(rpart) # For nicer tree fitting
library(partykit) # For nicer tree plotting## Loading required package: gridlibrary(MASS) # For Boston data
options(scipen = 4) # Suppresses scientific notation2. Classification trees
You will need the
Carseatsdata set from theISLRlibrary in order to complete this exercise.
Please run all of the code indicated in §8.3.1 of ISLR, even if I don’t explicitly ask you to do so in this document.
(a) Run the View() command on the Carseats data to see what the data set looks like.
#View(Carseats)The following code construst the High variable for the purpose of classification. Our goal will be to classify whether Carseat sales in a store are high or not.
High <- with(Carseats, ifelse(Sales <= 8, "No", "Yes"))
Carseats <- data.frame(Carseats, High)(b) What proportion of stores have High carseat sales?
prop.table(table(Carseats$High))##
## No Yes
## 0.59 0.41- We see that 59, 41% of locations had high carseat sales.
(c) Use the tree command to fit a decision tree to every other variable in the Carseats data other than Sales. Run summary on your tree object. Run the plot and text commands on your tree object.
# We don't want to use the Sales variable as an input because
# our outcome variable, High, is derived from Sales
tree.carseats <- tree::tree(High ~ . -Sales, Carseats)
summary(tree.carseats)##
## Classification tree:
## tree::tree(formula = High ~ . - Sales, data = Carseats)
## Variables actually used in tree construction:
## [1] "ShelveLoc" "Price" "Income" "CompPrice" "Population"
## [6] "Advertising" "Age" "US"
## Number of terminal nodes: 27
## Residual mean deviance: 0.4575 = 170.7 / 373
## Misclassification error rate: 0.09 = 36 / 400plot(tree.carseats)
text(tree.carseats,pretty=0)
(d) Does the tree make use of all of the available variables, or are there some that the tree does not use? (Look at the tree summary print-out to figure this out.)
summary(tree.carseats)$used## [1] ShelveLoc Price Income CompPrice Population Advertising
## [7] Age US
## 11 Levels: <leaf> CompPrice Income Advertising Population ... USnames(Carseats)[which(!(names(Carseats) %in%summary(tree.carseats)$used))]## [1] "Sales" "Education" "Urban" "High"- The output above shows which variables get used, and which ones do not get used. We see that not all variables are used (Sales is not the only ommitted variable).
(e) What is the misclassification rate of the tree? Is this a good misclassification rate? i.e., Is the rate low relative to the prevalence of High sales in the data?
# This tells us: # misclass, # total
summary(tree.carseats)$misclass## [1] 36 400misclass.rate <- summary(tree.carseats)$misclass[1] / summary(tree.carseats)$misclass[2]
misclass.rate## [1] 0.09- We see that the misclassification rate is just 9%, which is much lower than the prevalence of High sales in the data.
(f) Study the output from the tree-printout you get when you run the command below. Try to figure out how it connects to the tree diagram you got from running plot. (You do not need to write down an answer)
tree.carseats## node), split, n, deviance, yval, (yprob)
## * denotes terminal node
##
## 1) root 400 541.500 No ( 0.59000 0.41000 )
## 2) ShelveLoc: Bad,Medium 315 390.600 No ( 0.68889 0.31111 )
## 4) Price < 92.5 46 56.530 Yes ( 0.30435 0.69565 )
## 8) Income < 57 10 12.220 No ( 0.70000 0.30000 )
## 16) CompPrice < 110.5 5 0.000 No ( 1.00000 0.00000 ) *
## 17) CompPrice > 110.5 5 6.730 Yes ( 0.40000 0.60000 ) *
## 9) Income > 57 36 35.470 Yes ( 0.19444 0.80556 )
## 18) Population < 207.5 16 21.170 Yes ( 0.37500 0.62500 ) *
## 19) Population > 207.5 20 7.941 Yes ( 0.05000 0.95000 ) *
## 5) Price > 92.5 269 299.800 No ( 0.75465 0.24535 )
## 10) Advertising < 13.5 224 213.200 No ( 0.81696 0.18304 )
## 20) CompPrice < 124.5 96 44.890 No ( 0.93750 0.06250 )
## 40) Price < 106.5 38 33.150 No ( 0.84211 0.15789 )
## 80) Population < 177 12 16.300 No ( 0.58333 0.41667 )
## 160) Income < 60.5 6 0.000 No ( 1.00000 0.00000 ) *
## 161) Income > 60.5 6 5.407 Yes ( 0.16667 0.83333 ) *
## 81) Population > 177 26 8.477 No ( 0.96154 0.03846 ) *
## 41) Price > 106.5 58 0.000 No ( 1.00000 0.00000 ) *
## 21) CompPrice > 124.5 128 150.200 No ( 0.72656 0.27344 )
## 42) Price < 122.5 51 70.680 Yes ( 0.49020 0.50980 )
## 84) ShelveLoc: Bad 11 6.702 No ( 0.90909 0.09091 ) *
## 85) ShelveLoc: Medium 40 52.930 Yes ( 0.37500 0.62500 )
## 170) Price < 109.5 16 7.481 Yes ( 0.06250 0.93750 ) *
## 171) Price > 109.5 24 32.600 No ( 0.58333 0.41667 )
## 342) Age < 49.5 13 16.050 Yes ( 0.30769 0.69231 ) *
## 343) Age > 49.5 11 6.702 No ( 0.90909 0.09091 ) *
## 43) Price > 122.5 77 55.540 No ( 0.88312 0.11688 )
## 86) CompPrice < 147.5 58 17.400 No ( 0.96552 0.03448 ) *
## 87) CompPrice > 147.5 19 25.010 No ( 0.63158 0.36842 )
## 174) Price < 147 12 16.300 Yes ( 0.41667 0.58333 )
## 348) CompPrice < 152.5 7 5.742 Yes ( 0.14286 0.85714 ) *
## 349) CompPrice > 152.5 5 5.004 No ( 0.80000 0.20000 ) *
## 175) Price > 147 7 0.000 No ( 1.00000 0.00000 ) *
## 11) Advertising > 13.5 45 61.830 Yes ( 0.44444 0.55556 )
## 22) Age < 54.5 25 25.020 Yes ( 0.20000 0.80000 )
## 44) CompPrice < 130.5 14 18.250 Yes ( 0.35714 0.64286 )
## 88) Income < 100 9 12.370 No ( 0.55556 0.44444 ) *
## 89) Income > 100 5 0.000 Yes ( 0.00000 1.00000 ) *
## 45) CompPrice > 130.5 11 0.000 Yes ( 0.00000 1.00000 ) *
## 23) Age > 54.5 20 22.490 No ( 0.75000 0.25000 )
## 46) CompPrice < 122.5 10 0.000 No ( 1.00000 0.00000 ) *
## 47) CompPrice > 122.5 10 13.860 No ( 0.50000 0.50000 )
## 94) Price < 125 5 0.000 Yes ( 0.00000 1.00000 ) *
## 95) Price > 125 5 0.000 No ( 1.00000 0.00000 ) *
## 3) ShelveLoc: Good 85 90.330 Yes ( 0.22353 0.77647 )
## 6) Price < 135 68 49.260 Yes ( 0.11765 0.88235 )
## 12) US: No 17 22.070 Yes ( 0.35294 0.64706 )
## 24) Price < 109 8 0.000 Yes ( 0.00000 1.00000 ) *
## 25) Price > 109 9 11.460 No ( 0.66667 0.33333 ) *
## 13) US: Yes 51 16.880 Yes ( 0.03922 0.96078 ) *
## 7) Price > 135 17 22.070 No ( 0.64706 0.35294 )
## 14) Income < 46 6 0.000 No ( 1.00000 0.00000 ) *
## 15) Income > 46 11 15.160 Yes ( 0.45455 0.54545 ) *(g) What does the following line from the output mean? What does the * at the end mean? How many High = yes observations are there at this node?
18) Population < 207.5 16 21.170 Yes ( 0.37500 0.62500 ) *- The
*indicates that this split corresponds to a leaf node. There are 16 observations in this final node.0.62500 * 16 = 10of them haveHigh = yes.
Here is some code for splitting the data into training and testing, and for fitting a new tree.carseats model to just the training data.
set.seed(2)
train <- sample(1:nrow(Carseats), 200)
Carseats.test <- Carseats[-train,]
High.test <- High[-train]
tree.carseats <- tree(High~.-Sales,Carseats,subset=train)
tree.pred <- predict(tree.carseats,Carseats.test,type="class")
table(tree.pred,High.test)## High.test
## tree.pred No Yes
## No 86 27
## Yes 30 57(h) Use the cv.tree command to carry out 10-Fold Cross-Validation pruning on tree.subsets. You’ll need to supply FUN = prune.misclass as an argument to ensure that the error metric is taken to be the number of misclassifications instead of the deviance.
How many leaf nodes does the lowest CV error tree have? How many misclassifications does it make? What is its misclassification rate?
cv.carseats <- cv.tree(tree.carseats, FUN=prune.misclass)
names(cv.carseats)## [1] "size" "dev" "k" "method"cv.carseats## $size
## [1] 19 17 14 13 9 7 3 2 1
##
## $dev
## [1] 53 53 50 50 48 51 67 66 80
##
## $k
## [1] -Inf 0.0000000 0.6666667 1.0000000 1.7500000 2.0000000
## [7] 4.2500000 5.0000000 23.0000000
##
## $method
## [1] "misclass"
##
## attr(,"class")
## [1] "prune" "tree.sequence"# Index of tree with minimum error
min.idx <- which.min(cv.carseats$dev)
min.idx## [1] 5# Number of leaves in that tree
cv.carseats$size[min.idx]## [1] 9# Number of misclassifications (this is a count)
cv.carseats$dev[min.idx]## [1] 48# Misclassification rate
cv.carseats$dev[min.idx] / length(train)## [1] 0.24The following code produces CV error plots indexed by tree size and k (essentially what we called alpha in class).
par(mfrow = c(1,2))
plot(cv.carseats$size, cv.carseats$dev, type="b")
plot(cv.carseats$k, cv.carseats$dev, type="b")
(i) Use the prune.misclass command to prune tree.carseats down to the subtree that has the lowest CV error. Plot this tree, and overlay text. Compare the variables that get used in this tree compared to those that get used in the unpruned tree. Does the pruned tree wind up using fewer variables?
par(mfrow = c(1,1))
prune.carseats <- prune.misclass(tree.carseats, best = 9)
plot(prune.carseats)
text(prune.carseats, pretty=0)
# Variables used
summary(prune.carseats)$used## [1] ShelveLoc Price Advertising Age CompPrice
## 11 Levels: <leaf> CompPrice Income Advertising Population ... US# Variables that are used in one model but not the other
c(setdiff(summary(prune.carseats)$used, summary(tree.carseats)$used),
setdiff(summary(tree.carseats)$used, summary(prune.carseats)$used))## [1] "Income" "Population"- As we can see, there are two variables that appear in the larger tree but not the pruned tree.
(j) The code below produces a Test set confusion matrix for the pruned tree. What is the Test misclassification rate of this tree? How does this compare to the error you got in part (e)? Is there a big difference? Explain.
tree.pred <- predict(prune.carseats, Carseats.test, type="class")
confusion.pred <- table(tree.pred, High.test)
confusion.pred## High.test
## tree.pred No Yes
## No 94 24
## Yes 22 60# Misclassification rate
1 - sum(diag(confusion.pred)) / sum(confusion.pred)## [1] 0.23- The mislcassification rate on the test set is over 2x larger than the full data misclassification rate we saw from the full tree. There are two factors at play: First, the model that achieved 0.09 error rate was trained on the full data set, which has twice as many obseravations as
Carseats.train. Second, unpruned trees can greatly overfit data.
(k) The following code produces a pruned decision tree using the rpart command instead of tree. This results in a fit that can be converted into a party object, and plotted in a more aesthetically pleasing way. The code below illustrates how to perform this conversion and how to get a tree out of it. There are some accompanying questions below the code.
# Fit a decision tree using rpart
# Note: when you fit a tree using rpart, the fitting routine automatically
# performs 10-fold CV and stores the errors for later use
# (such as for pruning the tree)
carseats.rpart <- rpart(High ~ . -Sales , Carseats, method="class", subset=train)
# Plot the CV error curve for the tree
plotcp(carseats.rpart)
# Identify the value of the complexity parameter that produces
# the lowest CV error
cp.min <- carseats.rpart$cptable[which.min(carseats.rpart$cptable[,"xerror"]),"CP"]
# Prune using the CV error minimizing choice of the complexity parameter cp
carseats.rpart.pruned <- prune(carseats.rpart, cp = cp.min)
# Convert pruned tree to a party object
carseats.party <- as.party(carseats.rpart.pruned)
# Plot
plot(carseats.party)
- How many terminal nodes are in the tree?
- 5
- Which node corresponds to stores with the greatest likelihood of High sales of the product?
- Node 9. The height of the dark (high sales = “Yes”) bar is highest here. It looks like roughly 85% of stores in the training set falling in this node have high sales of the product.
- How does the tree characterize the stores in this node?
- These are the stores where the product is placed on a “Good” shelf location and the price of the product is set at under $142.50
3. Regression trees
These questions prompt you to follow along with §8.3.2 in ISL. We’ll once again be working with the
Bostondata set.
The code below splits the data into training and testing, and then fits a regression tree to the training data.
set.seed(1)
train <- sample(1:nrow(Boston), nrow(Boston)/2)
tree.boston <- tree(medv ~ . ,Boston,subset=train)
summary(tree.boston)##
## Regression tree:
## tree(formula = medv ~ ., data = Boston, subset = train)
## Variables actually used in tree construction:
## [1] "lstat" "rm" "dis"
## Number of terminal nodes: 8
## Residual mean deviance: 12.65 = 3099 / 245
## Distribution of residuals:
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## -14.10000 -2.04200 -0.05357 0.00000 1.96000 12.60000plot(tree.boston)
text(tree.boston, pretty=0)
(a) What is the first variable that is split on? Where is the split for this variable? Next, ook at the terminal nodes (leaves) in the tree. What do the numbers appearing in the leaves of the tree mean?
lstat (percentage of the population that is lower socioeconomic status) is the first variable that is split on. The split occurs at
lstat = 9.715.The numbers refer to the average
medv(median home value) in that leaf node.
(b) The following code repeats the process of randomly splitting the data and fitting a tree for 10 iterations. It produces a list of all of the variables that get used in the tree fit, in the order in which they appear. Does this list change a lot from one random split to the next? Do all of the trees wind up using the same variables? Is the first split always on the same variable?
set.seed(2)
for(i in 1:10) {
train.b <- sample(1:nrow(Boston), nrow(Boston)/2)
tree.boston.b <- tree(medv ~ . , Boston, subset=train.b)
print(as.character(summary(tree.boston.b)$used))
}## [1] "lstat" "rm" "dis" "nox" "crim"
## [1] "lstat" "rm" "dis" "nox" "crim"
## [1] "rm" "lstat" "crim" "ptratio"
## [1] "rm" "lstat" "age" "crim"
## [1] "lstat" "rm" "dis" "ptratio" "nox" "crim"
## [1] "lstat" "rm" "nox"
## [1] "lstat" "rm" "ptratio" "tax" "nox"
## [1] "lstat" "rm" "dis" "indus" "nox" "crim"
## [1] "lstat" "rm" "dis" "age" "crim"
## [1] "rm" "lstat" "dis"- There’s a lot of variability in both which variables get used, and how many variables get used. The first split is always either on
lstatorrm.
The code below performs cross-validation to prune the tree. Make sure that you understand what each line of code is doing. A written reply is not required for this section. Note: The y-axis on the CV error plot is MSE. dev stands for “deviance”, which is the same as MSE for regression (prediction) problems.
# Run CV to find best level at which to prune
cv.boston <- cv.tree(tree.boston)
# Construct a plot (dev = MSE on y-axis)
plot(cv.boston$size,cv.boston$dev,type='b')
# Prune the tree, display pruned tree
prune.boston <- prune.tree(tree.boston,best=5)
plot(prune.boston)
text(prune.boston,pretty=0)
# Get predictions from pruned tree
yhat.tree <- predict(tree.boston, newdata=Boston[-train,])
boston.test <- Boston[-train,"medv"]
# Construct plot of observed values (x-axis) vs predicted values (y-axis)
plot(yhat.tree, boston.test)
# Add a diagonal line
abline(0, 1)
# Calculate test set MSE
mean((yhat.tree-boston.test)^2)## [1] 25.04559