CISC 4631 Data Mining

1. CISC 4631Data Mining Lecture 04: Decision Trees Theses slides are based on the slides by Tan, Steinbach and Kumar (textbook authors) EamonnKoegh(UC Riverside) Raymond Mooney (UT Austin)

2. Classification: Definition • Given a collection of records (training set ) • Each record contains a set of attributes, one of the attributes is the class. • Find a model for class attribute as a function of the values of other attributes. • Goal: previously unseen records should be assigned a class as accurately as possible. • A test set is used to determine the accuracy of the model. Usually, the given data set is divided into training and test sets, with training set used to build the model and test set used to validate it.

3. Illustrating Classification Task

4. Classification Techniques • Decision Tree based Methods • Rule-based Methods • Memory based reasoning • Neural Networks • Naïve Bayes and Bayesian Belief Networks • Support Vector Machines

5. categorical categorical continuous class Example of a Decision Tree Splitting Attributes Refund Yes No NO MarSt Married Single, Divorced TaxInc NO < 80K > 80K YES NO Model: Decision Tree Training Data

6. NO Another Example of Decision Tree categorical categorical continuous class Single, Divorced MarSt Married NO Refund No Yes TaxInc < 80K > 80K YES NO There could be more than one tree that fits the same data!

7. Decision Tree Classification Task Decision Tree

8. Refund Yes No NO MarSt Married Single, Divorced TaxInc NO < 80K > 80K YES NO Apply Model to Test Data Test Data Start from the root of tree.

9. Refund Yes No NO MarSt Married Single, Divorced TaxInc NO < 80K > 80K YES NO Apply Model to Test Data Test Data

10. Apply Model to Test Data Test Data Refund Yes No NO MarSt Married Single, Divorced TaxInc NO < 80K > 80K YES NO

11. Apply Model to Test Data Test Data Refund Yes No NO MarSt Married Single, Divorced TaxInc NO < 80K > 80K YES NO

12. Apply Model to Test Data Test Data Refund Yes No NO MarSt Married Single, Divorced TaxInc NO < 80K > 80K YES NO

13. Apply Model to Test Data Test Data Refund Yes No NO MarSt Assign Cheat to “No” Married Single, Divorced TaxInc NO < 80K > 80K YES NO

14. Decision Tree Terminology

15. Decision Tree Classification Task Decision Tree

16. Decision Tree Induction • Many Algorithms: • Hunt’s Algorithm (one of the earliest) • CART • ID3, C4.5 • SLIQ,SPRINT • John Ross Quinlan is a computer science researcher in data mining and decision theory. He has contributed extensively to the development of decision tree algorithms, including inventing the canonical C4.5 and ID3 algorithms. 

17. Decision Tree Classifier 10 9 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 Ross Quinlan Abdomen Length > 7.1? Antenna Length yes no Antenna Length > 6.0? Katydid yes no Katydid Grasshopper Abdomen Length

18. Antennae shorter than body? Yes No 3 Tarsi? Grasshopper Yes No Foretiba has ears? Yes No Cricket Decision trees predate computers Katydids Camel Cricket

19. Definition • Decision tree is a classifier in the form of a tree structure • Decision node: specifies a test on a single attribute • Leaf node: indicates the value of the target attribute • Arc/edge: split of one attribute • Path: a disjunction of test to make the final decision • Decision trees classify instances or examples by starting at the root of the tree and moving through it until a leaf node.

20. Decision Tree Classification • Decision tree generation consists of two phases • Tree construction • At start, all the training examples are at the root • Partition examples recursively based on selected attributes • Tree pruning • Identify and remove branches that reflect noise or outliers • Use of decision tree: Classifying an unknown sample • Test the attribute values of the sample against the decision tree

21. Decision Tree Representation • Each internal node tests an attribute • Each branch corresponds to attribute value • Each leaf node assigns a classification outlook sunny overcast rain humidity yes wind weak normal strong high no yes no yes

22. How do we construct the decision tree? • Basic algorithm (a greedy algorithm) • Tree is constructed in a top-down recursive divide-and-conquer manner • At start, all the training examples are at the root • Attributes are categorical (if continuous-valued, they can be discretized in advance) • Examples are partitioned recursively based on selected attributes. • Test attributes are selected on the basis of a heuristic or statistical measure (e.g., information gain) • Conditions for stopping partitioning • All samples for a given node belong to the same class • There are no remaining attributes for further partitioning – majority voting is employed for classifying the leaf • There are no samples left

23. Top-Down Decision Tree Induction • Main loop: • A  the “best” decision attribute for next node • Assign A as decision attribute for node • For each value of A, create new descendant of node • Sort training examples to leaf nodes • If training examples perfectly classified, Then STOP, Else iterate over new leaf nodes

24. Tree Induction • Greedy strategy. • Split the records based on an attribute test that optimizes certain criterion. • Issues • Determine how to split the records • How to specify the attribute test condition? • How to determine the best split? • Determine when to stop splitting

25. How To Split Records • Random Split • The tree can grow huge • These trees are hard to understand. • Larger trees are typically less accurate than smaller trees. • Principled Criterion • Selection of an attribute to test at each node - choosing the most useful attribute for classifying examples. • How? • Information gain • measures how well a given attribute separates the training examples according to their target classification • This measure is used to select among the candidate attributes at each step while growing the tree

26. Tree Induction • Greedy strategy: • Split the records based on an attribute test that optimizes certain criterion: • Hunt’s algorithm: recursively partition training records into successively purer subsets. How to measure purity/impurity • Entropy and information gain (covered in the lectures slides) • Gini (covered in the textbook) • Classification error

27. How to determine the Best Split Before Splitting: 10 records of class 0, 10 records of class 1 Gender Which test condition is the best? Why is student id a bad feature to use?

28. How to determine the Best Split • Greedy approach: • Nodes with homogeneous class distribution are preferred • Need a measure of node impurity: Non-homogeneous, High degree of impurity Homogeneous, Low degree of impurity

29. Picking a Good Split Feature • Goal is to have the resulting tree be as small as possible, per Occam’s razor. • Finding a minimal decision tree (nodes, leaves, or depth) is an NP-hard optimization problem. • Top-down divide-and-conquer method does a greedy search for a simple tree but does not guarantee to find the smallest. • General lesson in Machine Learning and Data Mining: “Greed is good.” • Want to pick a feature that creates subsets of examples that are relatively “pure” in a single class so they are “closer” to being leaf nodes. • There are a variety of heuristics for picking a good test, a popular one is based on information gain that originated with the ID3 system of Quinlan (1979). R. Mooney, UT Austin

30. Information Theory • Think of playing "20 questions": I am thinking of an integer between 1 and 1,000 -- what is it? What is the first question you would ask? • What question will you ask? • Why? • Entropy measures how much more information you need before you can identify the integer. • Initially, there are 1000 possible values, which we assume are equally likely. • What is the maximum number of question you need to ask?

31. Entropy • Entropy (disorder, impurity) of a set of examples, S, relative to a binary classification is: where p1 is the fraction of positive examples in S and p0 is the fraction of negatives. • If all examples are in one category, entropy is zero (we define 0log(0)=0) • If examples are equally mixed (p1=p0=0.5), entropy is a maximum of 1. • Entropy can be viewed as the number of bits required on average to encode the class of an example in S where data compression (e.g. Huffman coding) is used to give shorter codes to more likely cases. • For multi-class problems with c categories, entropy generalizes to: R. Mooney, UT Austin

32. Entropy Plot for Binary Classification • The entropy is 0 if the outcome is certain. • The entropy is maximum if we have no knowledge of the system (or any outcome is equally possible). Entropy of a 2-class problem with regard to the portion of one of the two groups

33. Information Gain • Is the expected reduction in entropy caused by partitioning the examples according to this attribute. • is the number of bits saved when encoding the target value of an arbitrary member of S, by knowing the value of attribute A.

34. Information Gain in Decision Tree Induction • Assume that using attribute A, a current set will be partitioned into some number of child sets • The encoding information that would be gained by branching on A Note: entropy is at its minimum if the collection of objects is completely uniform

35. Examples for Computing Entropy • NOTE: p( j | t) is computed as the relative frequency of class j at node t P(C1) = 0/6 = 0 P(C2) = 6/6 = 1 Entropy = – 0 log20– 1 log21 = – 0 – 0 = 0 P(C1) = 1/6 P(C2) = 5/6 Entropy = – (1/6) log2 (1/6)– (5/6) log2 (5/6) = 0.65 P(C1) = 2/6 P(C2) = 4/6 Entropy = – (2/6) log2 (2/6)– (4/6) log2 (4/6) = 0.92 P(C1) = 3/6=1/2 P(C2) = 3/6 = 1/2 Entropy = – (1/2) log2(1/2)– (1/2) log2(1/2) = -(1/2)(-1) – (1/2)(-1) = ½ + ½ = 1

36. How to Calculate log2x • Many calculators only have a button for log10x and logex (note log typically means log10) • You can calculate the log for any base b as follows: • logb(x) = logk(x) / logk(b) • Thus log2(x) = log10(x) / log10(2) • Since log10(2) = .301, just calculate the log base 10 and divide by .301 to get log base 2. • You can use this for HW if needed

37. Splitting Based on INFO... • Information Gain: Parent Node, p is split into k partitions; ni is number of records in partition i • Measures Reduction in Entropy achieved because of the split. Choose the split that achieves most reduction (maximizes GAIN) • Used in ID3 and C4.5 • Disadvantage: Tends to prefer splits that result in large number of partitions, each being small but pure.

38. Continuous Attribute?(more on it later) • Each non-leaf node is a test, its edge partitioning the attribute into subsets (easy for discrete attribute). • For continuous attribute • Partition the continuous value of attribute A into a discrete set of intervals • Create a new boolean attribute Ac , looking for a threshold c, How to choose c ?

40. Entropy(4F,5M) = -(4/9)log2(4/9) - (5/9)log2(5/9) = 0.9911 no yes Hair Length <= 5? Let us try splitting on Hair length Entropy(3F,2M) = -(3/5)log2(3/5) - (2/5)log2(2/5) = 0.9710 Entropy(1F,3M) = -(1/4)log2(1/4) - (3/4)log2(3/4) = 0.8113 Gain(Hair Length <= 5) = 0.9911 – (4/9 * 0.8113 + 5/9 * 0.9710 ) = 0.0911

41. Entropy(4F,5M) = -(4/9)log2(4/9) - (5/9)log2(5/9) = 0.9911 no yes Weight <= 160? Let us try splitting on Weight Entropy(0F,4M) = -(0/4)log2(0/4) - (4/4)log2(4/4) = 0 Entropy(4F,1M) = -(4/5)log2(4/5) - (1/5)log2(1/5) = 0.7219 Gain(Weight <= 160) = 0.9911 – (5/9 * 0.7219 + 4/9 * 0 ) = 0.5900

42. Entropy(4F,5M) = -(4/9)log2(4/9) - (5/9)log2(5/9) = 0.9911 no yes age <= 40? Let us try splitting on Age Entropy(1F,2M) = -(1/3)log2(1/3) - (2/3)log2(2/3) = 0.9183 Entropy(3F,3M) = -(3/6)log2(3/6) - (3/6)log2(3/6) = 1 Gain(Age <= 40) = 0.9911 – (6/9 * 1 + 3/9 * 0.9183 ) = 0.0183

43. Of the 3 features we had, Weight was best. But while people who weigh over 160 are perfectly classified (as males), the under 160 people are not perfectly classified… So we simply recurse! no yes Weight <= 160? This time we find that we can split on Hair length, and we are done! no yes Hair Length <= 2?

44. We don’t need to keep the data around, just the test conditions. Weight <= 160? yes no How would these people be classified? Hair Length <= 2? Male yes no Male Female

45. It is trivial to convert Decision Trees to rules… Weight <= 160? yes no Hair Length <= 2? Male no yes Male Female Rules to Classify Males/Females IfWeightgreater than 160, classify as Male Elseif Hair Lengthless than or equal to 2, classify as Male Else classify as Female

46. Once we have learned the decision tree, we don’t even need a computer! This decision tree is attached to a medical machine, and is designed to help nurses make decisions about what type of doctor to call. Decision tree for a typical shared-care setting applying the system for the diagnosis of prostatic obstructions.

47. The worked examples we have seen were performed on small datasets. However with small datasets there is a great danger of overfitting the data… When you have few datapoints, there are many possible splitting rules that perfectly classify the data, but will not generalize to future datasets. Yes No Wears green? Male Female For example, the rule “Wears green?” perfectly classifies the data, so does “Mothers name is Jacqueline?”, so does “Has blue shoes”…

48. M0 M2 M3 M4 M1 M12 M34 How to Find the Best Split: GINI Before Splitting: A? B? Yes No Yes No Node N1 Node N2 Node N3 Node N4 Gain = M0 – M12 vs M0 – M34

49. Measure of Impurity: GINI (at node t) • Gini Index for a given node t with classes j NOTE: p( j | t) is computed as the relative frequency of class j at node t • Example: Two classes C1 & C2 and node t has 5 C1 and 5 C2 examples. Compute Gini(t) • 1 – [p(C1|t) + p(C2|t)] = 1 – [(5/10)2 + [(5/10)2 ] • 1 – [¼ + ¼] = ½. • Do you think this Gini value indicates a good split or bad split? Is it an extreme value?

50. More on Gini • Worst Gini corresponds to probabilities of 1/nc, where nc is the number of classes. • For 2-class problems the worst Gini will be ½ • How do we get the best Gini? Come up with an example for node t with 10 examples for classes C1 and C2 • 10 C1 and 0 C2 • Now what is the Gini? • 1 – [(10/10)2 + (0/10)2 = 1 – [1 + 0] = 0 • So 0 is the best Gini • So for 2-class problems: • Gini varies from 0 (best) to ½ (worst).