An algorithm can be transparent only if its decisions can be read and understood by people clearly. Even though deep learning is superstar of machine learning nowadays, it is an opaque algorithm and we do not know the reason of decision. Herein, Decision tree algorithms still keep their popularity because they can produce transparent decisions. ID3 uses information gain whereas C4.5 uses gain ratio for splitting. Here, CART is an alternative decision tree building algorithm. It can handle both classification and regression tasks. This algorithm uses a new metric named gini index to create decision points for classification tasks. We will mention a step by step CART decision tree example by hand from scratch.
Vlog
Here, you should watch the following video to understand how decision tree algorithms work. No matter which decision tree algorithm you are running: ID3, C4.5, CART, CHAID or Regression Trees. They all look for the feature offering the highest information gain. Then, they add a decision rule for the found feature and build an another decision tree for the sub data set recursively until they reached a decision.
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Besides, regular decision tree algorithms are designed to create branches for categorical features. Still, we are able to build trees with continuous and numerical features. The trick is here that we will convert continuos features into categorical. We will split the numerical feature where it offers the highest information gain.
CART in Python
This blog post mentions the deeply explanation of CART algorithm and we will solve a problem step by step. On the other hand, you might just want to run CART algorithm and its mathematical background might not attract your attention.
Herein, you can find the python implementation of CART algorithm here. You can build CART decision trees with a few lines of code. This package supports the most common decision tree algorithms such as ID3, C4.5, CHAID or Regression Trees, also some bagging methods such as random forest and some boosting methods such as gradient boosting and adaboost.
Objective
Decision rules will be found by GINI index value.
Data set
We will work on same dataset in ID3. There are 14 instances of golf playing decisions based on outlook, temperature, humidity and wind factors.
| Day | Outlook | Temp. | Humidity | Wind | Decision |
|---|---|---|---|---|---|
| 1 | Sunny | Hot | High | Weak | No |
| 2 | Sunny | Hot | High | Strong | No |
| 3 | Overcast | Hot | High | Weak | Yes |
| 4 | Rain | Mild | High | Weak | Yes |
| 5 | Rain | Cool | Normal | Weak | Yes |
| 6 | Rain | Cool | Normal | Strong | No |
| 7 | Overcast | Cool | Normal | Strong | Yes |
| 8 | Sunny | Mild | High | Weak | No |
| 9 | Sunny | Cool | Normal | Weak | Yes |
| 10 | Rain | Mild | Normal | Weak | Yes |
| 11 | Sunny | Mild | Normal | Strong | Yes |
| 12 | Overcast | Mild | High | Strong | Yes |
| 13 | Overcast | Hot | Normal | Weak | Yes |
| 14 | Rain | Mild | High | Strong | No |
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.
Gini = 1 – Σ (Pi)2 for i=1 to number of classes
Outlook
Outlook is a nominal feature. It can be sunny, overcast or rain. I will summarize the final decisions for outlook feature.
| Outlook | Yes | No | Number of instances |
| Sunny | 2 | 3 | 5 |
| Overcast | 4 | 0 | 4 |
| Rain | 3 | 2 | 5 |
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) x 0.48 + (4/14) x 0 + (5/14) x 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.
| Temperature | Yes | No | Number of instances |
| Hot | 2 | 2 | 4 |
| Cool | 3 | 1 | 4 |
| Mild | 4 | 2 | 6 |
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) x 0.5 + (4/14) x 0.375 + (6/14) x 0.445 = 0.142 + 0.107 + 0.190 = 0.439
Humidity
Humidity is a binary class feature. It can be high or normal.
| Humidity | Yes | No | Number of instances |
| High | 3 | 4 | 7 |
| Normal | 6 | 1 | 7 |
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) x 0.489 + (7/14) x 0.244 = 0.367
Wind
Wind is a binary class similar to humidity. It can be weak and strong.
| Wind | Yes | No | Number of instances |
| Weak | 6 | 2 | 8 |
| Strong | 3 | 3 | 6 |
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) x 0.375 + (6/14) x 0.5 = 0.428
Time to decide
We’ve calculated gini index values for each feature. The winner will be outlook feature because its cost is the lowest.
| Feature | Gini index |
| Outlook | 0.342 |
| Temperature | 0.439 |
| Humidity | 0.367 |
| Wind | 0.428 |
We’ll put outlook decision at the top of the tree.
You might realize that sub dataset in the overcast leaf has only yes decisions. This means that overcast leaf is over.
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.
| Day | Outlook | Temp. | Humidity | Wind | Decision |
| 1 | Sunny | Hot | High | Weak | No |
| 2 | Sunny | Hot | High | Strong | No |
| 8 | Sunny | Mild | High | Weak | No |
| 9 | Sunny | Cool | Normal | Weak | Yes |
| 11 | Sunny | Mild | Normal | Strong | Yes |
Gini of temperature for sunny outlook
| Temperature | Yes | No | Number of instances |
| Hot | 0 | 2 | 2 |
| Cool | 1 | 0 | 1 |
| Mild | 1 | 1 | 2 |
Gini(Outlook=Sunny and Temp.=Hot) = 1 – (0/2)2 – (2/2)2 = 0
Gini(Outlook=Sunny and Temp.=Cool) = 1 – (1/1)2 – (0/1)2 = 0
Gini(Outlook=Sunny and Temp.=Mild) = 1 – (1/2)2 – (1/2)2 = 1 – 0.25 – 0.25 = 0.5
Gini(Outlook=Sunny and Temp.) = (2/5)x0 + (1/5)x0 + (2/5)x0.5 = 0.2
Gini of humidity for sunny outlook
| Humidity | Yes | No | Number of instances |
| High | 0 | 3 | 3 |
| Normal | 2 | 0 | 2 |
Gini(Outlook=Sunny and Humidity=High) = 1 – (0/3)2 – (3/3)2 = 0
Gini(Outlook=Sunny and Humidity=Normal) = 1 – (2/2)2 – (0/2)2 = 0
Gini(Outlook=Sunny and Humidity) = (3/5)x0 + (2/5)x0 = 0
Gini of wind for sunny outlook
| Wind | Yes | No | Number of instances |
| Weak | 1 | 2 | 3 |
| Strong | 1 | 1 | 2 |
Gini(Outlook=Sunny and Wind=Weak) = 1 – (1/3)2 – (2/3)2 = 0.266
Gini(Outlook=Sunny and Wind=Strong) = 1- (1/2)2 – (1/2)2 = 0.2
Gini(Outlook=Sunny and Wind) = (3/5)x0.266 + (2/5)x0.2 = 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.
| Feature | Gini index |
| Temperature | 0.2 |
| Humidity | 0 |
| Wind | 0.466 |
We’ll put humidity check at the extension of sunny outlook.
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.
Now, we need to focus on rain outlook.
Rain outlook
| Day | Outlook | Temp. | Humidity | Wind | Decision |
| 4 | Rain | Mild | High | Weak | Yes |
| 5 | Rain | Cool | Normal | Weak | Yes |
| 6 | Rain | Cool | Normal | Strong | No |
| 10 | Rain | Mild | Normal | Weak | Yes |
| 14 | Rain | Mild | High | Strong | No |
We’ll calculate gini index scores for temperature, humidity and wind features when outlook is rain.
Gini of temprature for rain outlook
| Temperature | Yes | No | Number of instances |
| Cool | 1 | 1 | 2 |
| Mild | 2 | 1 | 3 |
Gini(Outlook=Rain and Temp.=Cool) = 1 – (1/2)2 – (1/2)2 = 0.5
Gini(Outlook=Rain and Temp.=Mild) = 1 – (2/3)2 – (1/3)2 = 0.444
Gini(Outlook=Rain and Temp.) = (2/5)x0.5 + (3/5)x0.444 = 0.466
Gini of humidity for rain outlook
| Humidity | Yes | No | Number of instances |
| High | 1 | 1 | 2 |
| Normal | 2 | 1 | 3 |
Gini(Outlook=Rain and Humidity=High) = 1 – (1/2)2 – (1/2)2 = 0.5
Gini(Outlook=Rain and Humidity=Normal) = 1 – (2/3)2 – (1/3)2 = 0.444
Gini(Outlook=Rain and Humidity) = (2/5)x0.5 + (3/5)x0.444 = 0.466
Gini of wind for rain outlook
| Wind | Yes | No | Number of instances |
| Weak | 3 | 0 | 3 |
| Strong | 0 | 2 | 2 |
Gini(Outlook=Rain and Wind=Weak) = 1 – (3/3)2 – (0/3)2 = 0
Gini(Outlook=Rain and Wind=Strong) = 1 – (0/2)2 – (2/2)2 = 0
Gini(Outlook=Rain and Wind) = (3/5)x0 + (2/5)x0 = 0
Decision for rain outlook
The winner is wind feature for rain outlook because it has the minimum gini index score in features.
| Feature | Gini index |
| Temperature | 0.466 |
| Humidity | 0.466 |
| Wind | 0 |
Put the wind feature for rain outlook branch and monitor the new sub data sets.
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.
Feature Importance
Decision trees are naturally explainable and interpretable algorithms. Besides, we can find the feature importance values as well to understand how model works.
Gradient Boosting Decision Trees
Nowadays, gradient boosting decision trees are very popular in machine learning community. They are actually not different than the decision tree algorithm mentioned in this blog post. They mainly builds sequantial decision trees based on the errors in the previous loop.
Random Forest vs Gradient Boosting
The both random forest and gradient boosting are an approach instead of a core decision tree algorithm itself. They require to run core decision tree algorithms. They also build many decision trees in the background. So, we will discuss how they are similar and how they are different in the following video.
Adaboost vs Gradient Boosting
The both gradient boosting and adaboost are boosting techniques for decision tree based machine learning models. We will discuss how they are similar and how they are different than each other.
Conclusion
So, decision tree building is over. We have built a decision tree by hand. BTW, you might realize that we’ve created exactly the same tree in ID3 example. This does not mean that ID3 and CART algorithms produce same trees always. We are just lucky. Finally, I believe that CART is easier than ID3 and C4.5, isn’t it?
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But Cart Algorithm isn’t used only with max 2 splits? Because it is a binary tree… In this example we can see three split (Sunny, Overcast and Rainy)… How is that possible? Sorry but I don’t understand.
Thanks
There is no up limit for classes of a feature in CART algorithm. As you mentioned, classes for outlook feature are sunny, overcast and rain. We’ve checked gini score for sunny outlook, overcast outlook and rainy outlook respectively. Then, we will combine these calculations and calculate gini score for outlook feature.
If you want to build a binary tree, you might still construct a tree for multiple classes as illustrated below.
if outlook = ‘sunny’:
#do something
else:
if outlook = ‘overcast’:
#do something
else: #this case refers to rainy outlook
#do something
But Cart Algorithm isn’t used only with max 2 splits? Because it is a binary tree… In this example we can see three split (Sunny, Overcast and Rainy)… How is that possible? Sorry but I don’t understand.
Thanks
Yes, you’re right. I need to fix the tutorial in this way.
For Overcast Outlook, Gini is 0 (Pure subset). Hence no sub-split is required.
Hence the made tree is correct. Therefore in default, Play golf if Outlook is Overcast.
Any suggestions is welcome on this post.
HAVE YOU FIXED THIS TUTOTAIL IN SOME OTHER LINK?
blogs are awesome, but i am confused too, because the representation for
cart model is a binary tree, maybe the calculation need to be slightly adjusted?
just a small question.
Hello Sefik,
I am equally confused with this implementation of the CART algorithm on this dataset. I have modeled this dataset by hand by calculating gini scores and I get the exact same tree as you did, where outlook is selected as the root node with three branches – sunny, overcast and rain. However, when I model the same data using sklearn’s implementation of the CART algorithm, I get a different tree structure with Humidity as the root node. Could you please provide me with an example of this model using sklearn?
Thank you in anticipation of your assistance
Bonny
I do not know actually what sklearn does in background. Possibly they add some modification to improve accuracy or performance.
Hi,
I appreciate the efforts you went to creating this informative example, but unfortunately, it’s not really a good example for understanding Breiman’s CART logic. As others posted here, CART is a binary approach.
If you have categorised data here, then the splits that you should be using are Rain vs Not Rain or Mild vs Not Mild. You should calculate the gini index for each example member of the category i.e. for Sunny/Overcast/Rain & Hot/Mild/Cold etc. Each of these binary splits give’s you a value for that member of the category. You then take the minimum of these values as your gini for that attribute. You then compare the lowest gini value for each of the four attributes to decide what should be your first level split. You then have a left and a right quantity.
With these two new quantities you repeat the exercise e.g. if was humidity (as suggested above) then you have one set of humid data and one set of not humid data and you use again all the members of a category i.e. for Sunny/Overcast/Rain & Hot/Mild/Cold etc to find the next split. You stop splitting when you have nothing meaningful to gain i.e. all data will fall 50/50 due to the splits.
There is also the stuff about pruning the tree back up if your decision tree is too deep, but that is no fun for a manually worked example.
I am also unaware of whether you can do a “weighted gini” of the individual ginis as you have done here.
“Gini(Outlook) = (5/14) x 0.48 + (4/14) x 0 + (5/14) x 0.48 = 0.171 + 0 + 0.171 = 0.342”.
This is again, not a binary split of the data. See the section “twoing the data” in Breiman’s book.
Anyhow, I just wanted to correct what you have written. You have made a good attempt to explain it in a friendly way, unfortunately it is not quite accurate.
Regards
Conan
Hello Conan, is there any tutorial that you would recommend solving the (Play Tennis) problem according to your approach?
I’ll check sources while I was writing this blog post. Then, I’ll share that here soon.
Interesting of what you said.. do you have any suggestions for good sources that explain this kind of topics in details? thanks.
I put the together the sources I found and give links in the post. Please follow the links. Besides, tutorials of computer science department professors guided me a lot.
Very neat explanation. Your post doesn’t involve lots of math symbols and confuse the readers like others. Easy to understand.
That’s the concept of this blog 🙂 thank you for your feedback btw.
Nice solving of the example.But i just have this one doubt…What should i do if the values for the attributes are continuous like the humidity and temperature in the c4.5 example?Do i have to convert them to nominal ones using the gini index as threshold?
Yes, you have to convert continuous / numeric features to numerical similar to C4.5 post.
Hi,
Why use gini index instead of
PL, PR, P(j|tL),P(j|tR), Q(s\t), and ϕ(s|t) ?
I see another journal publication use PL, PR, P(j|tL),P(j|tR), Q(s\t), and ϕ(s|t).
The original study uses gini for cart. Studies you’ve mentioned would be extended features.
Thanks for your simple useful explanation..
Thanks for the clear explanation.
one question, there are possibilities as below that you could have in each level of split, then how you will choose which feature goes first?
– In a case that the min cost for 2 features be equal i.e. assume in 2nd level that data gives you same mini gini index for Humidity and Wind and none of the feature gives you straight answer to Yes/No
– In a case on feature is giving the min gini index for different features i.e. assume if Humidity gives you the min gini index for Sunny and Rainy as same time
You can pick randomly if two features or classes have same scores
much appriciated
Why does this have more than 2 splits :((