## Decision Tree

### What’s decision tree?

A decision tree is a flowchart-like structure in which each internal node represents a "test" on an attribute (e.g. whether a coin flip comes up heads or tails), each branch represents the outcome of the test and each leaf node represents a class label (decision taken after computing all attributes). The paths from root to leaf represents classification rules.

### Overview

A decision tree is a flowchart-like structure in which each internal node represents a "test" on an attribute (e.g. whether a coin flip comes up heads or tails), each branch represents the outcome of the test and each leaf node represents a class label (decision taken after computing all attributes). The paths from root to leaf represents classification rules.

A decision tree consists of 3 types of nodes:

• Decision nodes - commonly represented by squares
• Chance nodes - represented by circles
• End nodes - represented by triangles

An example of decision tree goes below:

We get a decision tree using training data: Abe, Barb, Colette and Don. And we can get the Home value of Sally using this decision tree.

### Metrics

#### Information Gain

##### What’s entropy ?

The entropy (very common in Information Theory) characterizes the impurity of an arbitrary collection of examples.

We can calculate the entropy as follows:

$entropy(p)=\sum\limits_{i=1}^n{P_i}\log{P_i}$

For example, for the set $R = \{a, a, a, b, b, b, b, b\}$

$entropy(R)=-\frac{3}{8}\log\frac{3}{8}-\frac{5}{8}\log\frac{5}{8}$

##### What’s information entropy?

In general terms, the expected information gain is the change in information entropy H from a prior state to a state that takes some information as given:

$IG(D,A)=\Delta{Entropy}=Entropy(D)-Entropy(D|A)$

Take the image below as an example:

The entropy before spliting: $Entropy(D)=-\frac{14}{30}\log\frac{14}{30}-\frac{16}{30}\log\frac{16}{30}\approx{0.996}$

The entropy after spliting: $Entropy(D|A)=-\frac{17}{30}(\frac{13}{17}\log\frac{13}{17}+\frac{4}{17}\log\frac{4}{17})-\frac{13}{30}(\frac{1}{13}\log\frac{1}{13}+\frac{12}{13}\log\frac{12}{13})\approx{0.615}$

So the information gain should be: $IG(D,A)=Entropy(D)-Entropy(D|A)\approx0.381$

#### Information Gain Ratio

##### Problem of information gain approach
• Biased towards tests with many outcomes (attributes having a large number of values)
• E.g: attribute acting as unique identifier
• Produce a large number of partitions (1 tuple per partition)
• Each resulting partition D is pure entropy(D)=0, the information gain is maximized
##### What’s information gain ratio?

Information gain ratio overcomes the bias of information gain, and it applies a kind of normalization to information gain using a split information value.

The split information value represents the potential information generated by splitting the training data set D into v partitions, corresponding to v outcomes on attribute A:

$SplitInfo_A(D)=-\sum\limits_{j=1}^v\frac{|D_j|}{|D|}\times\log_2(\frac{|D_j|}{|D|})$

The gain ratio is defined as:

$GainRatio(D, A)=\frac{IG(D, A)}{SplitInfo_A(D)}$

• $IG(D, A)\approx0.381$
• $SplitInfo_A(D)=-\frac{17}{30}\log_2\frac{17}{30}-\frac{13}{30}\log_2\frac{13}{30}\approx0.987$
• $GainRatio(D, A)=\frac{IG(D, A)}{SplitInfo_A(D)}\approx0.386$

#### Gini Impurity

The Gini index is used in CART. Using the notation previously described, the Gini index measures the impurity of D, a data partition or set of training tuples, as:

$Gini(D)=1-\sum\limits_{i=1}^m{p_i^2}$

where $p_i$ is the probability that a tuple in D belongs to class $C_i$ and is estimated by $|C_i, D| / |D|$. The sum is computed over m classes.

A Gini index of zero expresses perfect equality, where all values are the same; a Gini index of one (or 100%) expresses maximal inequality among values.

Still take the image above as an example:

• $Gini(D_1)=1-(\frac{13}{17})^2-(\frac{4}{17})^2=\frac{104}{289}\approx0.360$
• $Gini(D_2)=1-(\frac{12}{13})^2-(\frac{1}{13})^2=\frac{24}{169}\approx0.142$
• $Gini(D)=\frac{|D_1|}{D}Gini(D_1)+\frac{|D_2|}{D}Gini(D_2)=\frac{176}{663}\approx0.265$

### Decision Tree Algorithms

#### ID3 Algorithm

ID3 (Iterative Dichotomiser 3) is an algorithm used to generate a decision tree from a dataset.

The ID3 algorithm begins with the original set S as the root node. On each iteration of the algorithm, it iterates through every unused attribute of the set S and calculates the entropy H(S) (or information gain IG(A)) of that attribute. It then selects the attribute which has the smallest entropy (or largest information gain) value. The set S is then split by the selected attribute (e.g. age < 50, 50 <= age < 100, age >= 100) to produce subsets of the data.

The algorithm continues to recur on each subset, considering only attributes never selected before. Recursion on a subset may stop in one of these cases:

• every element in the subset belongs to the same class (+ or -), then the node is turned into a leaf and labelled with the class of the examples;
• there are no more attributes to be selected, but the examples still do not belong to the same class (some are + and some are -), then the node is turned into a leaf and labelled with the most common class of the examples in the subset;

Let's take the following data as an example:

 Day Outlook Temperature Humidity Wind Play ball D1 Sunny Hot High Weak No D2 Sunny Hot High Strong No D3 Overcast Hot High Weak Yes D4 Rain Mild High Weak Yes D5 Rain Cool Normal Weak Yes D6 Rain Cool Normal Strong No D7 Overcast Cool Normal Strong Yes D8 Sunny Mild High Weak No D9 Sunny Cool Normal Weak Yes D10 Rain Mild Normal Weak Yes D11 Sunny Mild Normal Strong Yes D12 Overcast Mild High Strong Yes D13 Overcast Hot Normal Weak Yes D14 Rain Mild High Strong No

The information gain is calculated for all four attributes:

• $IG(S, Outlook)=0.246$
• $IG(S, Temperature)=0.029$
• $IG(S, Humidity)=0.151$
• $IG(S, Wind)=0.048$

So, we'll choose Outlook attribute for the first time.

For the node where Outlook = Overcast, we'll find that all the attribute(Play ball) of end nodes are the same.
For another two nodes, we should split them once more.
Let's take the node where Outlook = Sunny as an example:

• $IG(S_{sunny}, Temperature)=0.570$
• $IG(S_{sunny}, Humidity)=0.970$
• $IG(S_{sunny}, Wind)=0.019$

So, we'll choose Humidity attribute for this node.
And repeat these steps, you'll get an decision tree as below:

#### C4.5 Algorithm

C4.5 builds decision trees from a set of training data in the same way as ID3, using an extension to information gain known as gain ratio.

##### Improvements from ID3 algorithm
• Handling both continuous and discrete attributes;
• Handling training data with missing attribute values - C4.5 allows attribute values to be marked as ? for missing;
• Pruning trees after creation

#### CART Algorithm

The CART(Classification & Regression Trees) algorithm is a binary decision tree algorithm. It recursively partitions data into 2 subsets so that cases within each subset are more homogeneous. Allows consideration of misclassification costs, prior distributions, cost-complexity pruning.

Let's take following data as an example:

 TID Buy House Marital Status Taxable Income Cheat 1 Yes Single 125K No 2 No Married 100K No 3 No Single 70K No 4 Yes Married 120K No 5 No Divorced 95K Yes 6 No Married 60K No 7 Yes Divorced 220K No 8 No Single 85K Yes 9 No Married 75K No 10 No Single 90K Yes

Firstly, we calculate the Gini Index for Buy House.

 House=Yes House=No Cheat=Yes 0 4 Cheat=No 3 3

So the Gini Index for this attribute should be:

• $Gini(House=Yes)=1-(\frac{3}{3})^2-(\frac{0}{3})^2=0$
• $Gini(House=Yes)=1-(\frac{4}{7})^2-(\frac{3}{7})^2=\frac{24}{49}$
• $Gini(House)=\frac{3}{10}\times{0}+\frac{7}{10}\times\frac{24}{49}=\frac{12}{35}$
##### How to deal with attributes with more than two values?

We calculate the Gini Index for following splits:

• $Gini(Martial Status=Single/Divorced)$ and $Gini(Martial Status=Married)$
• $Gini(Martial Status=Single/Married)$ and $Gini(Martial Status=Divorced)$
• $Gini(Martial Status=Divorced/Married)$ and $Gini(Martial Status=Single)$

Then choose the split with minimum Gini index.

##### How to deal with continuous attributes?

For efficient computation: for each attribute,

• Sort the attribute on values
• Linearly scan these values, each time updating the count matrix and computing Gini index
• Choose the split position that has the least Gini index
 Cheat No No No Yes Yes Yes No No No No Tax Income Sorted 60 70 75 85 90 95 100 120 125 220 Split 55 65 72 80 87 92 97 110 122 172 230 <= > <= > <= > <= > <= > <= > <= > <= > <= > <= > <= > Yes 0 3 0 3 0 3 0 3 1 2 2 1 3 0 3 0 3 0 3 0 3 0 No 0 7 1 6 2 5 3 4 3 4 3 4 3 4 4 3 5 2 6 1 7 0 Gini 0.420 0.400 0.375 0.343 0.417 0.400 0.300 0.343 0.375 0.400 0.420

So the value(Tax Income = 95) with minimum Gini index will be chose as split node.

The values will be split into two nodes: Tax Income > 97 and Tax Income <= 97.

### Notes on Overfitting

Overfitting results in decision trees that are more complex than necessary. Training error no longer provides a good estimate of how well the tree will perform on previously unseen records.

#### Estimating Generalization Errors

Need new ways for estimating errors.

• Re‐substitution errors: error on training (Σe(t))
• Generalization errors: error on testing (Σe’(t))
##### Optimistic approach

Generalization errors: error on testing: e’(t) = (e(t)

##### Pessimistic approach

For each leaf node: e’(t) = (e(t)+0.5)

Total errors: e’(T) = e(T) + N × 0.5 (N: number of leaf nodes)

For a tree with 30 leaf nodes and 10 errors on training (out of 1000 instances):

• Training error = 10 / 1000 = 1%

Generalization error = (10 + 30 × 0.5) / 1000 = 2.5%

##### Pre‐Pruning (Early Stopping Rule)
• Stop the algorithm before it becomes a fully‐grown tree
• Typical stopping conditions for a node
• Stop if all instances belong to the same class
• Stop if all the attribute values are the same
• More restrictive conditions
• Stop if number of instances is less than some user‐specified threshold
• Stop if class distribution of instances are independent of the available features (e.g., using χ2 test)
• Stop if expanding the current node does not improve impurity measures (e.g., Gini or information gain)
##### Post‐pruning
• Grow decision tree to its entirety
• Trim the nodes of the decision tree in a bottom‐up fashion
• If generalization error improves after trimming, replace sub‐tree by a leaf node
• Class label of leaf node is determined from majority class of instances in the sub‐tree