# Decision Tree Classifier – Part 1

To continue my blogging on machine learning (ML) classifiers, I am turning to decision trees. The post on decision trees will be in two parts. Part 1 will provide an introduction to how decision trees work and how they are build. Part 2 will contain the C# implementation of an example decision trees classifier. As in all my posts, I prefer clear and informal explanation over the terse mathematical one. So, any pedants out there – look away now!

Decision trees are a great choice for inductive inference and have been widely used for a long time in the field of Artificial Intelligence (AI). In this post we will cover the decision tree algorithm known as ID3.
There are several reasons why decision trees are great classifiers:

• decision trees are easy to understand;
• decision trees work well with messy data or missing values;
• decision trees give insight into complex data;
• decision trees can be visualized to allow for inference inspection or correction;
• decision tree building algorithms can be as simple or sophisticated as required (e.g. they can incorporate pruning, weights, etc.);

Decision trees work best with discrete classes. That is, the output class for each instance is either a string, boolean or an integer. If you are working with continuous values, you may consider rounding and mapping to a discrete output class, or may need to look for another classifier. Decision trees are used for classification problems. For example, the taxonomy of organisms, plants, minerals, etc. lends itself naturally to decision tree classifiers. This is because in a field of taxonomy we are dealing with a set of records containing values for some attributes. The values come from a finite set of known features. Each record may or may not have a classification. Medicine is another field that makes use of decision trees. Almost all illnesses can be categorized by symptoms, thus decision trees aid doctors in illness diagnosis.

It is time for an example, which I am borrowing from . The tennis playing example in ML is like the ‘Hello World’ in programming languages. We are given some data about the weather conditions that are appropriate for playing tennis. Our task is to construct a decision tree based on this data, and use the tree to classify unknown weather conditions. The learning data is:

Table 1
Outlook Temperature Humidity Wind Play Tennis
sunny hot high strong no
sunny hot high weak no
overcast hot high weak yes
rain mild high weak yes
rain cool normal weak yes
rain cool normal strong no
overcast cool normal strong yes
sunny mild high weak no
sunny cool normal weak yes
rain mild normal weak yes
sunny mild normal strong yes
overcast mild high strong yes
overcast hot normal weak yes
rain mild high strong no

Table 1 tells us which weather conditions are good for playing tennis outdoors. For example, if it is sunny, and the wind is weak, and the temperature is mild but it is very humid, then playing tennis outdoors is not advisable. On the other hand, on an overcast day, regardless of the wind, playing tennis outdoors should be OK.

The most fundamental concept behind the decision tree classifiers is that of order. The amount of order in the data can be measured by assessing its consistency. For example, imagine that every row in Table 1 had a ‘yes’ associated with the Play Tennis column. This would tell us that regardless of the wind, temperature, humidity, we can always play tennis outside. The data would have perfect order, and we would be able to build a perfect classifier for the data – the one that always says ‘yes’ and is always correct. The other extreme would be where the outcome class differs for every observation. For an inductive learner like a decision tree, this would mean that it is impossible to classify new instance unless it perfectly matches some instance in the training set. This is why decision tree classifier won’t work for continuous class problems.

In information theory the concept of order (actually, lack of it), is often represented by entropy. It is a scary word for something that is rather simple. Entropy definition was proposed by the founder of information theory – Claude Shannon, and it is a probabilistic measure of categorical data defined as: $Entropy(S) = -\sum_{i=1}^{n}p_i \log_2 p_i$

where S is the training data set with some target classification that takes on n different values. The discrete target values are assigned their probabilities p, and $\log_2$ is logarithm base 2. The base two is due to the entropy being a measure of the expected encoding in binary format (i.e. 0s and 1s). Note that the smaller are individual probabilities, the greater is the absolute value of entropy. Also, the closer is entropy to zero, the more order there is in the data.

We now can calculate the entropy of our data, one column at a time. Let’s take the Play Tennis column. We have 9 ‘yes’ and 5 ‘no’, this gives us the entropy of $Entropy(S)=-\frac{9}{14} \log_2(\frac{9}{14}) + -\frac{5}{14}\log_2(\frac{5}{14})=0.940$, accurate to 3 d.p. There is a reason why we started with the class column. It will be used to measure the information gain we can achieve by splitting the data one attribute/column at a time, which is the core idea behind building decision trees. At each step we will be picking the attribute that achieves the greatest information gain to split the tree recursively. The information gain is, essentially, a positive difference in entropy; and the root node of any decision tree is the attribute with the greatest information gain. Information gain is formally defined as: $InfoGain(S,A) = Entropy(S) -\sum_{i=1}^{A}\frac{|S_i|}{|S|}Entropy(S_i)$

where A is the set of all values some attribute takes. For example, the Wind column takes values in {strong, weak}, and the Humidity column takes on values in {high, normal}. $|S_i|$ and $|S|$ are the number of records taking on the value i and the total number of records respectively. Finally, $Entropy(S_i)$ is the corresponding entropy of all records where the attribute A takes on value i with respect to the target class. The last sentence is important. It means that, given all the records where attribute A takes on the same value i we are interested to know the dispersion of the target class. For example, take a look at all the records where Wind is strong. Here we have 3 records where the target class is ‘no’ and 3 records where it is ‘yes’. Thus, the corresponding entropy is $Entropy(S_{wind=strong})=-\frac{3}{6} \log_2(\frac{3}{6}) -\frac{3}{6} \log_2(\frac{3}{6})=1.0$

So, which attribute/column should be the tree’s root node? It should be the one that achieves the greatest information gain: $InfoGain(S,Wind) = 0.940 - \frac{8}{14}\times 0.811 - \frac{6}{14}\times 1.0=0.048$ $InfoGain(S,Humidity) = 0.940 - \frac{7}{14}\times 0.985 - \frac{7}{14}\times 0.592=0.151$ $InfoGain(S,Temperature) = 0.940 - \frac{4}{14}\times 1.0 - \frac{4}{14}\times 0.811 - \frac{6}{14}\times 0.918=0.029$ $InfoGain(S,Outlook) = 0.940 - \frac{5}{14}\times 0.971 - \frac{4}{14}\times 0.0 - \frac{5}{14}\times 0.971=0.246$

It is clear that Outlook achieves the greatest information gain and should be the root of the tree. The worst information gain would be achieved with Temperature attribute. This is what our tree now looks:

At this stage we have the training data split into three clusters. Where outlook is overcast, we have a target class of ‘yes’. We now proceed recursively building the sub-tree in the other two clusters. Where outlook is sunny, we have a total of 5 records. These are reproduced in Table 2 below. Out of three possible attributes for the node we again pick the one that achieves the greatest information gain. If you inspect the data in Table 2, it is easy to see that humidity is that attribute since all five records can be split as following: if humidity is normal, then ‘yes’, else ‘no’. Its corresponding information gain is $InfoGain(S_{sunny},Humidity) = 0.970 - \frac{3}{5}\times 0 - \frac{2}{5}\times 0=0.970$

where 0.970 is entropy of Outlook is sunny, again, with respect to the Play Tennis class: $Entropy(S_{sunny}) = - \frac{3}{5}\log_2(\frac{3}{5})- \frac{2}{5}\log_2(\frac{2}{5})=0.970$.

Table 2
Outlook Temperature Humidity Wind Play Tennis
sunny hot high strong no
sunny hot high weak no
sunny mild high weak no
sunny cool normal weak yes
sunny mild normal strong yes

At this stage we have classified 9 records and our tree has 9 terminal descendant nodes (leaves):

We can now deal with the records in the last outlook cluster, i.e. where outlook is rain. Here we have five records, which are reproduced below in Table 3:

Table 3
Outlook Temperature Humidity Wind Play Tennis
rain mild high weak yes
rain cool normal weak yes
rain cool normal strong no
rain mild normal weak yes
rain mild high strong no

We apply the same algorithm as before by selecting the attribute for the next node that gives the greatest information gain. It is easy to spot that the Wind attribute achieves just that. Its total information gain is $InfoGain(S_{rain},Wind) = 0.970 - \frac{3}{5}\times 0 - \frac{2}{5}\times 0=0.970$

The complete tree looks like this: