# Eight to Late

Sensemaking and Analytics for Organizations

## A gentle introduction to decision trees using R

### Introduction

Most techniques of predictive analytics have their origins in probability or statistical theory (see my post on Naïve Bayes, for example).  In this post I’ll look at one that has more a commonplace origin: the way in which humans make decisions.  When making decisions, we typically identify the options available and then evaluate them based on criteria that are important to us.  The intuitive appeal of such a procedure is in no small measure due to the fact that it can be easily explained through a visual. Consider the following graphic, for example:

(Original image: https://www.flickr.com/photos/dullhunk/7214525854, Credit: Duncan Hull)

The tree structure depicted here provides a neat, easy-to-follow description of the issue under consideration and its resolution. The decision procedure is based on asking a series of questions, each of which serve to further reduce the domain of possibilities. The predictive technique I discuss in this post,classification and regression trees (CART), works in much the same fashion. It was invented by Leo Breiman and his colleagues in the 1970s.

In what follows, I will use the open source software, R. If you are new to R,   you may want to follow this link for more on the basics of setting up and installing it. Note that the R implementation of the CART algorithm is called RPART (Recursive Partitioning And Regression Trees). This is essentially because Breiman and Co. trademarked the term CART. As some others have pointed out, it is somewhat ironical that the algorithm is now commonly referred to as RPART rather than by the term coined by its inventors.

### A bit about the algorithm

The rpart algorithm works by splitting the dataset recursively, which means that the subsets that arise from a split are further split until a predetermined termination criterion is reached.  At each step, the split is made based on the independent variable that results in the largest possible reduction in heterogeneity of the dependent (predicted) variable.

Splitting rules can be constructed in many different ways, all of which are based on the notion of impurity-  a measure of the degree of heterogeneity of the leaf nodes. Put another way, a leaf node that contains a single class is homogeneous and has impurity=0.   There are three popular impurity quantification methods: Entropy (aka information gain), Gini Index and Classification Error.  Check out this article for a simple explanation of the three methods.

The rpart algorithm offers the entropy  and Gini index methods as choices. There is a fair amount of fact and opinion on the Web about which method is better. Here are some of the better articles I’ve come across:

https://www.quora.com/Are-gini-index-entropy-or-classification-error-measures-causing-any-difference-on-Decision-Tree-classification

http://stats.stackexchange.com/questions/130155/when-to-use-gini-impurity-and-when-to-use-information-gain

https://www.garysieling.com/blog/sklearn-gini-vs-entropy-criteria

http://www.salford-systems.com/resources/whitepapers/114-do-splitting-rules-really-matter

The answer as to which method is the best is: it depends.  Given this, it may be prudent to try out a couple of methods and pick the one that works best for your problem.

Regardless of the method chosen, the splitting rules partition the decision space (a fancy word for the entire dataset) into rectangular regions each of which correspond to a split. Consider the following simple example with two predictors x1 and x2. The first split is at x1=1 (which splits the decision space into two regions x11), the second at x2=2, which splits the (x1>1) region into 2 sub-regions, and finally x1=1.5 which splits the (x1>1,x2>2) sub-region further.

It is important to note that the algorithm works by making the best possible choice at each particular stage, without any consideration of whether those choices remain optimal in future stages. That is, the algorithm makes a locally optimal decision at each stage. It is thus quite possible that such a choice at one stage turns out to be sub-optimal in the overall scheme of things.  In other words,  the algorithm does not find a globally optimal tree.

Another important point relates to well-known bias-variance tradeoff in machine learning, which in simple terms is a tradeoff between the degree to which a model fits the training data and its predictive accuracy.  This refers to the general rule that beyond a point, it is counterproductive to improve the fit of a model to the training data as this increases the likelihood of overfitting.  It is easy to see that deep trees are more likely to overfit the data than shallow ones. One obvious way to control such overfitting is to construct shallower trees by stopping the algorithm at an appropriate point based on whether a split significantly improves the fit.  Another is to grow a tree unrestricted and then prune it back using an appropriate criterion. The rpart algorithm takes the latter approach.

Here is how it works in brief:

Essentially one minimises the cost, $C_{\alpha}(T)$, a quantity that is a  linear combination of the error (essentially, the fraction of misclassified instances, or variance in the case of a continuous variable), $R(T)$  and the number of leaf nodes in the tree, $|\tilde{T} |$: $C_{\alpha}(T) = R(T) + \alpha |\tilde{T} |$

First, we note that when $\alpha = 0$, this simply returns the original fully grown tree. As $\alpha$ increases, we incur a penalty that is proportional to the number of leaf nodes.  This tends to cause the minimum cost to occur for a tree that is a subtree of the original one (since a subtree will have a smaller number of leaf nodes). In practice we vary $\alpha$ and pick the value that gives the subtree that results in the smallest cross-validated prediction error.  One does not have to worry about programming this because the rpart algorithm actually computes the errors for different values of $\alpha$ for us. All we need to do is pick the value of the coefficient that gives the lowest cross-validated error. I will illustrate this in detail in the next section.

An implication of their tendency to overfit data is that decision trees tend to be sensitive to relatively minor changes in the training datasets. Indeed, small differences can lead to radically different looking trees. Pruning addresses this to an extent, but does not resolve it completely.  A better resolution is offered by the so-called ensemble methods that average over many differently constructed trees. I’ll discuss one such method at length in a future post.

Finally, I should also mention that decision trees can be used for both classification and regression problems (i.e. those in which the predicted variable is discrete and continuous respectively).  I’ll demonstrate both types of problems in the next two sections.

### Classification trees using rpart

To demonstrate classification trees, we’ll use the Ionosphere dataset available in the mlbench package in R. I have chosen this dataset because it nicely illustrates the points I wish to make in this post. In general, you will almost always find that algorithms that work fine on classroom datasets do not work so well in the real world…but of course, you know that already!

#set working directory if needed (modify path as needed)
setwd(“C:/Users/Kailash/Documents/decisiontrees”)
#load required libraries – rpart for classification and regression trees
library(rpart)
#mlbench for Ionosphere dataset
library(mlbench)
data(“Ionosphere”)

Next we separate the data into training and test sets. We’ll use the former to build the model and the latter to test it. To do this, I use a simple scheme wherein I randomly select 80% of the data for the training set and assign the remainder to the test data set. This is easily done in a single R statement that invokes the uniform distribution (runif) and the vectorised function, ifelse. Before invoking runif, I set a seed integer to my favourite integer in order to ensure reproducibility of results.

#set seed to ensure reproducible results
set.seed(42)
#split into training and test sets
Ionosphere[,”train”] <- ifelse(runif(nrow(Ionosphere))<0.8,1,0)
#separate training and test sets
trainset <- Ionosphere[Ionosphere$train==1,] testset <- Ionosphere[Ionosphere$train==0,]
#get column index of train flag
trainColNum <- grep("train",names(trainset))
#remove train flag column from train and test sets
trainset <- trainset[,-trainColNum]
testset <- testset[,-trainColNum]

In the above, I have also removed the training flag from the training and test datasets.

Next we  invoke rpart. I strongly recommend you take some time to go through the documentation and understand the parameters and their defaults values.  Note that we need to remove the predicted variable from the dataset before passing the latter on to the algorithm, which is why we need to find the column index of the  predicted variable (first line below). Also note that we set the method parameter to “class“, which simply tells the algorithm that the predicted variable is discrete.  Finally, rpart uses Gini rule for splitting by default, and we’ll stick with this option.

#get column index of predicted variable in dataset
typeColNum <- grep("Class",names(Ionosphere))
#build model
rpart_model <- rpart(Class~.,data = trainset, method="class")
#plot tree
plot(rpart_model);text(rpart_model)

The resulting plot is shown in Figure 3 below.  It is  quite self-explanatory so I  won’t dwell on it here.

Next we see how good the model is by seeing how it fares against the test data.

#…and the moment of reckoning
rpart_predict <- predict(rpart_model,testset[,-typeColNum],type="class")
mean(rpart_predict==testset$Class)  0.8450704 #confusion matrix table(pred=rpart_predict,true=testset$Class)

Note that we need to verify the above results by doing multiple runs, each using different training and test sets. I will  do this later, after discussing pruning.

Next, we prune the tree using the cost complexity criterion. Basically, the intent is to see if a shallower subtree can give us comparable results. If so, we’d be better of choosing the shallower tree because it reduces the likelihood of overfitting.

As described earlier, we choose the appropriate pruning parameter (aka cost-complexity parameter) $\alpha$ by picking the value that results in the lowest prediction error. Note that all relevant computations have already been carried out by R when we built the original tree (the call to rpart in the code above). All that remains now is to pick the value of $\alpha$:

#cost-complexity pruning
printcp(rpart_model)
 CP nsplit rel error xerror xstd 1 0.57 0 1.00 1.00 0.080178 2 0.20 1 0.43 0.46 0.062002 3 0.02 2 0.23 0.26 0.048565 4 0.01 4 0.19 0.35

It is clear from the above, that the lowest cross-validation error (xerror in the table) occurs for $\alpha =0.02$ (this is CP in the table above).   One can find CP programatically like so:

# get index of CP with lowest xerror
opt <- which.min(rpart_model$cptable[,"xerror"]) #get its value cp <- rpart_model$cptable[opt, "CP"]

Next, we prune the tree based on this value of CP:

#prune tree
pruned_model <- prune(rpart_model,cp)
#plot tree
plot(pruned_model);text(pruned_model)

Note that rpart will use a default CP value of 0.01 if you don’t specify one in prune.

The pruned tree is shown in Figure 4 below.

Let’s see how this tree stacks up against the fully grown one shown in Fig 3.

#find proportion of correct predictions using test set
rpart_pruned_predict <- predict(pruned_model,testset[,-typeColNum],type="class")
mean(rpart_pruned_predict==testset$Class)  0.8873239 This seems like an improvement over the unpruned tree, but one swallow does not a summer make. We need to check that this holds up for different training and test sets. This is easily done by creating multiple random partitions of the dataset and checking the efficacy of pruning for each. To do this efficiently, I’ll create a function that takes the training fraction, number of runs (partitions) and the name of the dataset as inputs and outputs the proportion of correct predictions for each run. It also optionally prunes the tree. Here’s the code: #function to do multiple runs multiple_runs_classification <- function(train_fraction,n,dataset,prune_tree=FALSE){ fraction_correct <- rep(NA,n) set.seed(42) for (i in 1:n){ dataset[,”train”] <- ifelse(runif(nrow(dataset))<0.8,1,0) trainColNum <- grep("train",names(dataset)) typeColNum <- grep("Class",names(dataset)) trainset <- dataset[dataset$train==1,-trainColNum]
testset <- dataset[dataset$train==0,-trainColNum] rpart_model <- rpart(Class~.,data = trainset, method="class") if(prune_tree==FALSE) { rpart_test_predict <- predict(rpart_model,testset[,-typeColNum],type="class") fraction_correct[i] <- mean(rpart_test_predict==testset$Class)
}else{
opt <- which.min(rpart_model$cptable[,"xerror"]) cp <- rpart_model$cptable[opt, "CP"]
pruned_model <- prune(rpart_model,cp)
rpart_pruned_predict <- predict(pruned_model,testset[,-typeColNum],type="class")
fraction_correct[i] <- mean(rpart_pruned_predict==testset$Class) } } return(fraction_correct) } Note that in the above, I have set the default value of the prune_tree to FALSE, so the function will execute the first branch of the if statement unless the default is overridden. OK, so let’s do 50 runs with and without pruning, and check the mean and variance of the results for both sets of runs. #50 runs, no pruning unpruned_set <- multiple_runs_classification(0.8,50,Ionosphere) mean(unpruned_set)  0.8772763 sd(unpruned_set)  0.03168975 #50 runs, with pruning pruned_set <- multiple_runs_classification(0.8,50,Ionosphere,prune_tree=TRUE) mean(pruned_set)  0.9042914 sd(pruned_set)  0.02970861 So we see that there is an improvement of about 3% with pruning. Also, if you were to plot the trees as we did earlier, you would see that this improvement is achieved with shallower trees. Again, I point out that this is not always the case. In fact, it often happens that pruning results in worse predictions, albeit with better reliability – a classic illustration of the bias-variance tradeoff. ### Regression trees using rpart In the previous section we saw how one can build decision trees for situations in which the predicted variable is discrete. Let’s now look at the case in which the predicted variable is continuous. We’ll use the Boston Housing dataset from the mlbench package. Much of the discussion of the earlier section applies here, so I’ll just display the code, explaining only the differences. #load Boston Housing dataset data(“BostonHousing”) #set seed to ensure reproducible results set.seed(42) #split into training and test sets BostonHousing[,”train”] <- ifelse(runif(nrow(BostonHousing))<0.8,1,0) #separate training and test sets trainset <- BostonHousing[BostonHousing$train==1,]
testset <- BostonHousing[BostonHousing$train==0,] #get column index of train flag trainColNum <- grep("train",names(trainset)) #remove train flag column from train and test sets trainset <- trainset[,-trainColNum] testset <- testset[,-trainColNum] Next we invoke rpart, noting that the predicted variable is medv (median value of owner-occupied homes in$1000 units) and that we need to set the method parameter to “anova“. The latter tells rpart that the predicted variable is continuous (i.e that this is a regression problem).

#build model
rpart_model <- rpart(medv~.,data = trainset, method="anova")
#plot tree
plot(rpart_model);text(rpart_model)

The plot of the tree is shown in Figure 5 below.

Next, we need to see how good the predictions are. Since the dependent variable is continuous, we cannot compare the predictions directly against the test set. Instead, we calculate the root mean square (RMS) error. To do this, we request rpart to output the predictions as a vector – one prediction per record in the test dataset. The RMS error can then easily be calculated by comparing this vector with the medv column in the test dataset.

Here is the relevant code:

#…the moment of reckoning
rpart_test_predict <- predict(rpart_model,testset[,-typeColNum],type = "vector" )
#calculate RMS error
rmsqe <- sqrt(mean((rpart_test_predict-testset$medv)^2))) rmsqe  4.586388 Again, we need to do multiple runs to check on the reliability of the predictions. However, you already know how to do that so I will leave it to you. Moving on, we prune the tree using the cost complexity criterion as before. The code is exactly the same as in the classification problem. # get index of CP with lowest xerror opt <- which.min(rpart_model$cptable[,"xerror"])
#get its value
cp <- rpart_model\$cptable[opt, "CP"]
#prune tree
pruned_model <- prune(rpart_model,cp)
#plot tree
plot(pruned_model);text(pruned_model)

The tree is unchanged so I won’t show it here. This means, as far as the cost complexity pruning is concerned, the optimal subtree is the same as the original tree. To confirm this, we’d need to do multiple runs as before – something that I’ve already left as as an exercise for you :).  Basically, you’ll need to write a function analogous to the one above, that computes the root mean square error instead of the proportion of correct predictions.

### Wrapping up

This brings us to the end of my introduction to classification and regression trees using R.  Unlike some articles on the topic I have attempted to describe each of the steps in detail and provide at least some kind of a rationale for them. I hope you’ve found the description and code snippets useful.

I’ll end by reiterating a couple points I made early in this piece. The nice thing about decision trees is that they are easy to explain to the users of our predictions. This is primarily because they  reflect the way we think about how decisions are made in real life – via a set of binary choices based on appropriate criteria. That  said, in many practical situations decision trees turn out to be unstable:  small changes in the dataset can lead to wildly different trees. It turns out that this limitation can be addressed by building a variety of trees using different starting points and then averaging over  them.  This is the domain of the so-called random forest algorithm.We’ll make the journey from decision trees to random forests in a future post.

Postscript, 20th September 2016: I finally got around to finishing my article on random forests.

Written by K

February 16, 2016 at 6:33 pm

### 11 Responses

1. Add probabilistic branches and you’ve got all you need to make decisions in presence of uncertainty

Like DoubleDuce

February 16, 2016 at 11:19 pm

2. Great tutorial. I have a question though, I am trying to understand the code where you are calling “rpart” below:

#get column index of predicted variable in dataset
typeColNum <- grep(“Class”,names(Ionosphere))
#build model
rpart_model <- rpart(Class~.,data = trainset, method=”class”)

In my code, I am calling rpart as:
fit <- rpart(frmla, method="class", data=redwine, control=rpart.control(minsplit=300, cp=0.001)).
How then do I call rpart similar to what you have in your code?

Thanks

Like George C

May 13, 2016 at 4:54 am

Like

4. […] a previous post, I described how decision tree algorithms work and demonstrated their use via the rpart library in […]

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5. […] in last line of previous section) and then move on to clustering, topic modelling, naive Bayes, decision trees, random forests and support vector machines. I’m slowly adding to the list as I find the time, so […]

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6. […] using grid search with cross-validation, a process akin to the one described in my discussion on cost-complexity parameter  estimation in decision trees. Most canned algorithms provide methods to do this; the one we’ll use in the next section is […]

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7. Nice ! Would be great if each value in the tree is explained in detail and also what decisions can be made wrt to each value at the nodes Thanks

Like KS Abhi

September 26, 2017 at 10:08 pm

8. Thanks for this – very helpful.

I’m trying to replicate the code for multiple_runs_classification and it’s giving a few errors, one being that prune_tree is not found – I’m just a little confused on how to fix it – is anyone else having this problem?

And just to clarify, this multiple runs code will work for numerical values too? I’ve found a lot of tutorials for categorical variables, but much fewer for numerical, and finding it difficult to figure out how to determine the accuracy of the model. Any help would be much appreciated

Like Laura Brannelly

January 25, 2019 at 12:15 pm

• Hi Laura,

I think the problem could be that the double quote (“) is rendered incorrectly in HTML. To fix this you will need to manually replace incorrectly rendered double quotes with the correct ones, after cutting and pasting the code from the article.

When working with continuous numerical variables, you will need to use the appropriate error metric. Two common possibilities are root mean square error or mean absolute error. You will then also need to change the code in multiple_runs appropriately.

Hope this helps.

Regards,

Kailash.

Like K

January 29, 2019 at 12:13 pm

9. In the regression tree example, what is resultColNum? I followed your code and it gets stuck there. Many Thanks

Like Erika

October 31, 2019 at 4:13 am

• Thanks for catching that Erika. It should be typeColNum (the column index of the target variable) not resultColNum.

Like K

October 31, 2019 at 4:20 am

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