CART can handle both classification and regression tasks. This algorithm uses a new metric named gini index to create decision points for classification tasks. A CART tree is a binary decision tree that is constructed by splitting a node into two child nodes repeatedly, beginning with the root node that contains the whole learning sample. Decision tree learning uses a decision tree (as a predictive model) to go from observations about an item (represented in the branches) to conclusions about the item's target value (represented in the leaves). It is one of the predictive modeling approaches used in statistics, data mining and machine learning. Tree models where the target variable can take a discrete set of values are called classification trees; in these tree structures, leaves represent class labels and branches represent conjunctions of features that lead to those class labels. Decision trees where the target variable can take continuous values (typically real numbers) are called regression trees.

In decision analysis, a decision tree can be used to visually and explicitly represent decisions and decision making. In data mining, a decision tree describes data (but the resulting classification tree can be an input for decision making).

Gini index

Gini index is a metric for classification tasks in CART. It stores sum of squared probabilities of each class. We can formulate it as illustrated below.

$\text{Gini} = 1 – \sum (Pi)^2   \text{ for i=1 to number of classes}$

Q: Find the decision tree using CART.

• Outlook

Outlook is a nominal feature. It can be sunny, overcast or rain. I will summarize the final decisions for outlook feature.

$Gini(Outlook=Sunny) = 1 – (2/5)^2  – (3/5)^2  = 1 – 0.16 – 0.36 = 0.48$

$Gini(Outlook=Overcast) = 1 – (4/4)^2  – (0/4)^2  = 0$

$Gini(Outlook=Rain) = 1 – (3/5)^2  – (2/5)^2  = 1 – 0.36 – 0.16 = 0.48$

Then, we will calculate weighted sum of gini indexes for outlook feature.

$Gini(Outlook) = (5/14) \times 0.48 + (4/14)\times 0 + (5/14) \times 0.48 = 0.171 + 0 + 0.171 = 0.342$

• Temperature

Similarly, temperature is a nominal feature and it could have 3 different values: Cool, Hot and Mild. Let’s summarize decisions for temperature feature.

$Gini(Temp=Hot) = 1 – (2/4)^2  – (2/4)^2  = 0.5$

$Gini(Temp=Cool) = 1 – (3/4)^2  – (1/4)^2  = 1 – 0.5625 – 0.0625 = 0.375$

$Gini(Temp=Mild) = 1 – (4/6)^2  – (2/6)^2  = 1 – 0.444 – 0.111 = 0.445$

We’ll calculate weighted sum of gini index for temperature feature

$Gini(Temp) = (4/14) \times 0.5 + (4/14) \times 0.375 + (6/14) \times 0.445 = 0.142 + 0.107 + 0.190 = 0.439$

• Humidity

Humidity is a binary class feature. It can be high or normal.

$Gini(Humidity=High) = 1 – (3/7)^2  – (4/7)^2  = 1 – 0.183 – 0.326 = 0.489$

$Gini(Humidity=Normal) = 1 – (6/7)^2  – (1/7)^2  = 1 – 0.734 – 0.02 = 0.244$

Weighted sum for humidity feature will be calculated next

$Gini(Humidity) = (7/14) \times 0.489 + (7/14) \times 0.244 = 0.367$

• Wind

Wind is a binary class similar to humidity. It can be weak and strong.

$Gini(Wind=Weak) = 1 – (6/8)^2  – (2/8)^2  = 1 – 0.5625 – 0.062 = 0.375$

$Gini(Wind=Strong) = 1 – (3/6)^2  – (3/6)^2  = 1 – 0.25 – 0.25 = 0.5$

$Gini(Wind) = (8/14) \times 0.375 + (6/14) \times 0.5 = 0.428$

We’ve calculated gini index values for each feature. The winner will be outlook feature because its cost is the lowest.

We’ll put outlook decision at the top of the tree.

First decision would be outlook feature

You might realize that sub dataset in the overcast leaf has only yes decisions. This means that overcast leaf is over.

Tree is over for overcast outlook leaf

We will apply same principles to those sub datasets in the following steps.

Focus on the sub dataset for sunny outlook. We need to find the gini index scores for temperature, humidity and wind features respectively.

• Gini of temperature for sunny outlook

$\text{Gini(Outlook=Sunny and Temp.=Hot)} = 1 – (0/2)^2  – (2/2)^2  = 0$

$\text{Gini(Outlook=Sunny and Temp.=Cool)} = 1 – (1/1)^2  – (0/1)^2  = 0$

$\text{Gini(Outlook=Sunny and Temp.=Mild)} = 1 – (1/2)^2  – (1/2)^2  = 1 – 0.25 – 0.25 = 0.5$

$\text{Gini(Outlook=Sunny and Temp.)} = (2/5)\times 0 + (1/5)\times 0 + (2/5)\times 0.5 = 0.2$

• Gini of humidity for sunny outlook

$\text{Gini(Outlook=Sunny and Humidity=High)} = 1 – (0/3)^2  – (3/3)^2  = 0$

$\text{Gini(Outlook=Sunny and Humidity=Normal)} = 1 – (2/2)^2  – (0/2)^2  = 0$

$\text{Gini(Outlook=Sunny and Humidity)} = (3/5)\times 0 + (2/5)\times 0 = 0$

• Gini of wind for sunny outlook

$\text{Gini(Outlook=Sunny and Wind=Weak)} = 1 – (1/3)^2  – (2/3)^2  = 0.266$

$\text{Gini(Outlook=Sunny and Wind=Strong)} = 1- (1/2)^2  – (1/2)^2  = 0.2$

$\text{Gini(Outlook=Sunny and Wind)} = (3/5)\times 0.266 + (2/5)\times 0.2 = 0.466$

• Decision for sunny outlook

We’ve calculated gini index scores for feature when outlook is sunny. The winner is humidity because it has the lowest value.

We’ll put humidity check at the extension of sunny outlook.

Sub datasets for high and normal humidity

As seen, decision is always no for high humidity and sunny outlook. On the other hand, decision will always be yes for normal humidity and sunny outlook. This branch is over.

Decisions for high and normal humidity

Now, we need to focus on rain outlook.

Rain outlook

We’ll calculate gini index scores for temperature, humidity and wind features when outlook is rain.

• Gini of temperature for rain outlook

$\text{Gini(Outlook=Rain and Temp.=Cool)} = 1 – (1/2)^2  – (1/2)^2  = 0.5$

$\text{Gini(Outlook=Rain and Temp.=Mild)} = 1 – (2/3)^2  – (1/3)^2  = 0.444$

$\text{Gini(Outlook=Rain and Temp.)} = (2/5)\times 0.5 + (3/5)\times 0.444 = 0.466$

• Gini of humidity for rain outlook

$\text{Gini(Outlook=Rain and Humidity=High)} = 1 – (1/2)^2  – (1/2)^2  = 0.5$

$\text{Gini(Outlook=Rain and Humidity=Normal)} = 1 – (2/3)^2  – (1/3)^2  = 0.444$

$\text{Gini(Outlook=Rain and Humidity)} = (2/5)\times 0.5 + (3/5)\times 0.444 = 0.466$

• Gini of wind for rain outlook

$\text{Gini(Outlook=Rain and Wind=Weak)} = 1 – (3/3)^2  – (0/3)^2  = 0$

$\text{Gini(Outlook=Rain and Wind=Strong)} = 1 – (0/2)^2  – (2/2)^2  = 0$

$\text{Gini(Outlook=Rain and Wind)} = (3/5)\times0 + (2/5)\times 0 = 0$

• Decision for rain outlook

The winner is wind feature for rain outlook because it has the minimum gini index score in features.

Put the wind feature for rain outlook branch and monitor the new sub data sets.

Sub data sets for weak and strong wind and rain outlook

As seen, decision is always yes when wind is weak. On the other hand, decision is always no if wind is strong. This means that this branch is over.

Final form of the decision tree built by CART algorithm