1. Constructing Decision Trees

2. A Decision Tree Example The weather data example. ID codeOutlookTemperatureHumidityWindyPlay abcdefghijklmnabcdefghijklmn Sunny Overcast Rainy Overcast Sunny Rainy Sunny Overcast Rainy Hot Mild Cool Mild Cool Mild Hot Mild High Normal High Normal High Normal High False True False True False True False True No Yes No Yes No Yes No

3. ~continues Outlook humiditywindy yes noyes no sunnyovercastrainy highnormalfalsetrue Decision tree for the weather data.

4. The Process of Constructing a Decision Tree Select an attribute to place at the root of the decision tree and make one branch for every possible value. Repeat the process recursively for each branch.

5. Which Attribute Should Be Placed at a Certain Node One common approach is based on the information gained by placing a certain attribute at this node.

6. Information Gained by Knowing the Result of a Decision In the weather data example, there are 9 instances of which the decision to play is “yes” and there are 5 instances of which the decision to play is “no’. Then, the information gained by knowing the result of the decision is

7. The General Form for Calculating the Information Gain Entropy of a decision = P 1, P 2, …, P n are the probabilities of the n possible outcomes.

8. Information Further Required If “Outlook” Is Placed at the Root Outlook yes no yes no sunnyovercastrainy

9. Information Gained by Placing Each of the 4 Attributes Gain(outlook) = 0.940 bits – 0.693 bits = 0.247 bits. Gain(temperature) = 0.029 bits. Gain(humidity) = 0.152 bits. Gain(windy) = 0.048 bits.

10. The Strategy for Selecting an Attribute to Place at a Node Select the attribute that gives us the largest information gain. In this example, it is the attribute “Outlook”. Outlook 2 “yes” 3 “no” 4 “yes” 3 “yes” 2 “no” sunnyovercastrainy

11. The Recursive Procedure for Constructing a Decision Tree The operation discussed above is applied to each branch recursively to construct the decision tree. For example, for the branch “Outlook = Sunny”, we evaluate the information gained by applying each of the remaining 3 attributes. Gain(Outlook=sunny;Temperature) = 0.971 – 0.4 = 0.571 Gain(Outlook=sunny;Humidity) = 0.971 – 0 = 0.971 Gain(Outlook=sunny;Windy) = 0.971 – 0.951 = 0.02

12. Similarly, we also evaluate the information gained by applying each of the remaining 3 attributes for the branch “Outlook = rainy”. Gain(Outlook=rainy;Temperature) = 0.971 – 0.951 = 0.02 Gain(Outlook=rainy;Humidity) = 0.971 – 0.951 = 0.02 Gain(Outlook=rainy;Windy) =0.971 – 0 = 0.971

13. The Over-fitting Issue Over-fitting is caused by creating decision rules that work accurately on the training set based on insufficient quantity of samples. As a result, these decision rules may not work well in more general cases.

14. Example of the Over-fitting Problem in Decision Tree Construction 11 “Yes” and 9 “No” samples; prediction = “Yes” 8 “Yes” and 9 “No” samples; prediction = “No” 3 “Yes” and 0 “No” samples; prediction = “Yes” A i =0A i =1

15. Hence, with the binary split, we gain more information. However, if we look at the pessimistic error rate, i.e. the upper bound of the confidence interval of the error rate, we may get different conclusion. The formula for the pessimistic error rate is Note that the pessimistic error rate is a function of the confidence level used.

16. The pessimistic error rates under 95% confidence are

17. Therefore, the average pessimistic error rate at the children is Since the pessimistic error rate increases with the split, we do not want to keep the children. This practice is called “tree pruning”.

18. Tree Pruning based on  2 Test of Independence We construct the corresponding contingency table Ai=0Ai=0 Ai=1Ai=1 Yes3811 No099 31720 11 “Yes” and 9 “No” samples; 8 “Yes” and 9 “No” samples; 3 “Yes” and 0 “No samples; A i =0A i =1

19. Therefore, we should not split the subroot node, if we require that the  2 statistic must be larger than  2 k,0.05, where k is the degree of freedom of the corresponding contingency table.

20. Constructing Decision Trees based on  2 test of Independence Using the following example, we can construct a contingency table accordingly. 75 “Yes”s out of 100 samples; Prediction = “Yes” 45 “Yes”s out of 50 samples; 20 “Yes”s out of 25 samples; 10 “Yes”s out of 25 samples; A i =0A i =1 A i =2

21. Therefore, we may say that the split is statistically robust.

22. Assume that we have another attribute A j to consider A j =0A j =1 Yes255075 No025 75100 75 “Yes” out of 100 samples; 50 “Yes” out of 75 samples; 25 “Yes” out of 25 samples; A j =0A j =1

23. Now, both A i and A j pass our criterion. How should we make our selection? We can make our selection based on the significance levels of the two contingency tables.

24. Therefore, A i is preferred over A j.

25. If a subtree is as follows  2 = 4.543 < 5.991 In this case, we do not want to carry out the split. 15 “Yes”s out of 20 samples; 9 “Yes”s out of 10 samples; 4 “Yes”s out of 5 samples; 2 “Yes”s out of 5 samples; Termination of Split due to Low Significance level

26. A More Realistic Example and Some Remarks In the following example, a bank wants to derive a credit evaluation tree for future use based on the records of existing customers. As the data set shows, it is highly likely that the training data set contains inconsistencies. Furthermore, some values may be missing. Therefore, for most cases, it is impossible to derive perfect decision trees, i.e. decision trees with 100% accuracy.

27. ~continues AttributesClass EducationAnnual IncomeAgeOwn HouseSexCredit ranking CollegeHighOldYesMaleGood High school-----MiddleYesMaleGood High schoolMiddleYoungNoFemaleGood CollegeHighOldYesMalePoor CollegeHighOldYesMaleGood CollegeMiddleYoungNoFemaleGood High schoolHighOldYesMalePoor CollegeMiddle -----FemaleGood High schoolMiddleYoungNoMalePoor

28. ~continues A quality measure of decision trees can be based on the accuracy. There are alternative measures depending on the nature of applications. Overfitting is a problem caused by making the derived decision tree work accurately for the training set. As a result, the decision tree may work less accurately in the real world.

29. ~continues There are two situations in which overfitting may occur: insufficient number of samples at the subroot. some attributes are highly branched. A conventional practice for handling missing values is to treat them as possible attribute values. That is, each attribute has one additional attribute value corresponding to the missing value.

30. Alternative Measures of Quality of Decision Trees The recall rate and precision are two widely used measures. where C is the set of samples in the class and C’ is the set of samples which the decision tree puts into the class.

31. ~continues A situation in which the recall rate is the main concern: “A bank wants to find all the potential credit card customers”. A situation in which precision is the main concern: “A bank wants to find a decision tree for credit approval.”