2. 1 Classification with Decision Trees Instructor: Qiang Yang Hong Kong University of Science and Technology Qyang@cs.ust.hk Thanks: Eibe Frank and Jiawei Han

3. 2 Continuous Classes Sometimes, classes are continuous in that they come from a continuous domain, e.g., temperature or stock price. Regression is well suited in this case: Linear and multiple regression Non-Linear regression We shall focus on categorical classes, e.g., colors or Yes/No binary decisions. We will deal with continuous class values later in CART

4. 3 DECISION TREE [Quinlan93] An internal node represents a test on an attribute. A branch represents an outcome of the test, e.g., Color=red. A leaf node represents a class label or class label distribution. At each node, one attribute is chosen to split training examples into distinct classes as much as possible A new case is classified by following a matching path to a leaf node.

5. Training Set

6. Outlook overcast humiditywindy highnormal false true sunny rain NNPP P overcast Example

7. 6 Building Decision Tree [Q93] Top-down tree construction At start, all training examples are at the root. Partition the examples recursively by choosing one attribute each time. Bottom-up tree pruning Remove subtrees or branches, in a bottom-up manner, to improve the estimated accuracy on new cases.

8. 7 Choosing the Splitting Attribute At each node, available attributes are evaluated on the basis of separating the classes of the training examples. A Goodness function is used for this purpose. Typical goodness functions: information gain (ID3/C4.5) information gain ratio gini index

9. 8 Which attribute to select?

10. 9 A criterion for attribute selection Which is the best attribute? The one which will result in the smallest tree Heuristic: choose the attribute that produces the “purest” nodes Popular impurity criterion: information gain Information gain increases with the average purity of the subsets that an attribute produces Strategy: choose attribute that results in greatest information gain

11. 10 Computing information Information is measured in bits Given a probability distribution, the info required to predict an event is the distribution’s entropy Entropy gives the information required in bits (this can involve fractions of bits!) Formula for computing the entropy:

12. 11 Example: attribute “Outlook” “Outlook” = “Sunny”: “Outlook” = “Overcast”: “Outlook” = “Rainy”: Expected information for attribute: Note: this is normally not defined.

13. 12 Computing the information gain Information gain: information before splitting – information after splitting Information gain for attributes from weather data:

14. 13 Continuing to split

15. 14 The final decision tree Note: not all leaves need to be pure; sometimes identical instances have different classes  Splitting stops when data can’t be split any further

16. 15 Highly-branching attributes Problematic: attributes with a large number of values (extreme case: ID code) Subsets are more likely to be pure if there is a large number of values  Information gain is biased towards choosing attributes with a large number of values  This may result in overfitting (selection of an attribute that is non-optimal for prediction) Another problem: fragmentation

17. 16 The gain ratio Gain ratio: a modification of the information gain that reduces its bias on high-branch attributes Gain ratio takes number and size of branches into account when choosing an attribute It corrects the information gain by taking the intrinsic information of a split into account Also called split ratio Intrinsic information: entropy of distribution of instances into branches (i.e. how much info do we need to tell which branch an instance belongs to)

18. Gain Ratio Gain ratio should be Large when data is evenly spread Small when all data belong to one branch Gain ratio (Quinlan’86) normalizes info gain by this reduction:

19. 18 Computing the gain ratio Example: intrinsic information for ID code Importance of attribute decreases as intrinsic information gets larger Example of gain ratio: Example:

20. 19 Gain ratios for weather data OutlookTemperature Info:0.693Info:0.911 Gain: 0.940-0.6930.247Gain: 0.940-0.9110.029 Split info: info([5,4,5])1.577Split info: info([4,6,4])1.362 Gain ratio: 0.247/1.5770.156Gain ratio: 0.029/1.3620.021 HumidityWindy Info:0.788Info:0.892 Gain: 0.940-0.7880.152Gain: 0.940-0.8920.048 Split info: info([7,7])1.000Split info: info([8,6])0.985 Gain ratio: 0.152/10.152Gain ratio: 0.048/0.9850.049

21. 20 More on the gain ratio “Outlook” still comes out top However: “ID code” has greater gain ratio Standard fix: ad hoc test to prevent splitting on that type of attribute Problem with gain ratio: it may overcompensate May choose an attribute just because its intrinsic information is very low Standard fix: First, only consider attributes with greater than average information gain Then, compare them on gain ratio

22. Gini Index If a data set T contains examples from n classes, gini index, gini(T) is defined as where pj is the relative frequency of class j in T. gini(T) is minimized if the classes in T are skewed. After splitting T into two subsets T1 and T2 with sizes N1 and N2, the gini index of the split data is defined as The attribute providing smallest ginisplit(T) is chosen to split the node.

23. 22 Discussion Consider the following variations of decision trees

24. 23 1. Apply KNN to each leaf node Instead of choosing a class label as the majority class label, use KNN to choose a class label

25. 24 2. Apply Naïve Bayesian at each leaf node For each leave node, use all the available information we know about the test case to make decisions Instead of using the majority rule, use probability/likelihood to make decisions

26. 25 3. Use error rates instead of entropy If a node has N1 positive class labels P, and N2 negative class labels N, If N1> N2, then choose P The error rate = N2/(N1+N2) at this node The expected error at a parent node can be calculated as weighted sum of the error rates at each child node The weights are the proportion of training data in each child

27. 26 4. When there is missing value, allow tests to be done Attribute selection criterion: minimal total cost (C total = C mc + C test ) instead of minimal entropy in C4.5 If growing a tree has a smaller total cost, then choose an attribute with minimal total cost. Otherwise, stop and form a leaf. Label leaf also according to minimal total cost: Suppose the leaf have P positive examples and N negative examples FP denotes the cost of a false positive example and FN false negative If (P×FN  N×FP) THEN label = positive ELSE label = negative More in the next lecture slides…

28. 27 Missing Values Missing values in test data Humidity={High, Normal}, but which one? Allow splitting of the values down to each branch of the decision tree Methods 1. equal proportion ½ to each side, 2. unequal proportion: use proportion = training data % Weighted result:

29. 28 Dealing with Continuous Class Values 1.Use the mean of a set as a predicted value 2.Use a linear regression formula to compute the predicted value In linear algebra

30. 29 Using Entropy Reduction to Discretize Continuous Variables Given the following data sorted by increasing Temperature values, and associated Play attribute values: Task: Partition the continuous ranged temperature into discrete values: Cold and Warm Hint: decision of boundary by entropy reduction! 1014152022252627293032363940 FFFFTTTTTTTTTF

31. 30 Entropy-Based Discretization Given a set of samples S, if S is partitioned into two intervals S1 and S2 using boundary T, the entropy after partitioning is The boundary that minimizes the entropy function over all possible boundaries is selected as a binary discretization. The process is recursively applied to partitions obtained until some stopping criterion is met, e.g., Experiments show that it may reduce data size and improve classification accuracy

32. 31 How to Calculate ent(S)? Given two classes Yes and No, in a set S, Let p1 be the proportion of Yes Let p2 be the proportion of No, p1 + p2 = 100% Entropy is: ent(S) = -p1*log(p1) – p2*log(p2) When p1=1, p2=0, ent(S)=0, When p1=50%, p2=50%, ent(S)=maximum! See TA ’ s tutorial notes for an Example.

33. 32 Numeric attributes Standard method: binary splits (i.e. temp < 45) Difference to nominal attributes: every attribute offers many possible split points Solution is straightforward extension: Evaluate info gain (or other measure) for every possible split point of attribute Choose “best” split point Info gain for best split point is info gain for attribute Computationally more demanding

34. 33 An example Split on temperature attribute from weather data: Eg. 4 yes es and 2 no s for temperature < 71.5 and 5 yes es and 3 no s for temperature  71.5 Info([4,2],[5,3]) = (6/14)info([4,2]) + (8/14)info([5,3]) = 0.939 bits Split points are placed halfway between values All split points can be evaluated in one pass! 64 65 68 69 70 71 72 72 75 75 80 81 83 85 Yes No Yes Yes Yes No No Yes Yes Yes No Yes Yes No

35. 34 Missing values C4.5 splits instances with missing values into pieces (with weights summing to 1) A piece going down a particular branch receives a weight proportional to the popularity of the branch Info gain etc. can be used with fractional instances using sums of weights instead of counts During classification, the same procedure is used to split instances into pieces Probability distributions are merged using weights

36. 35 Stopping Criteria When all cases have the same class. The leaf node is labeled by this class. When there is no available attribute. The leaf node is labeled by the majority class. When the number of cases is less than a specified threshold. The leaf node is labeled by the majority class.

37. 36 Pruning Pruning simplifies a decision tree to prevent overfitting to noise in the data Two main pruning strategies: 1.Postpruning: takes a fully-grown decision tree and discards unreliable parts 2.Prepruning: stops growing a branch when information becomes unreliable  Postpruning preferred in practice because of early stopping in prepruning

38. 37 Prepruning Usually based on statistical significance test Stops growing the tree when there is no statistically significant association between any attribute and the class at a particular node Most popular test: chi-squared test ID3 used chi-squared test in addition to information gain Only statistically significant attributes where allowed to be selected by information gain procedure

39. 38 The Weather example: Observed Count Play  Outlook YesNoOutlook Subtotal Sunny202 Cloudy011 Play Subtotal: 21Total count in table =3

40. 39 The Weather example: Expected Count Play  Outlook YesNoSubtotal Sunny 2*2/6=4/3= 1.3 2*1/6=2/3= 0.6 2 Cloudy 2*1/3=0.61*1/3=0.3 1 Subtotal:21Total count in table =3 If attributes were independent, then the subtotals would be Like this

41. 40 Question: How different between observed and expected? If Chi-squared value is very large, then A1 and A2 are not independent  that is, they are dependent! Degrees of freedom: if table has n*m items, then freedom = (n-1)*(m-1) If all attributes in a node are independent with the class attribute, then stop splitting further.

42. 41 Postpruning Builds full tree first and prunes it afterwards Attribute interactions are visible in fully-grown tree Problem: identification of subtrees and nodes that are due to chance effects Two main pruning operations: 1.Subtree replacement 2.Subtree raising Possible strategies: error estimation, significance testing, MDL principle

43. 42 Subtree replacement Bottom-up: tree is considered for replacement once all its subtrees have been considered

44. 43 Subtree raising Deletes node and redistributes instances Slower than subtree replacement (Worthwhile?)

45. 44 Estimating error rates Pruning operation is performed if this does not increase the estimated error Of course, error on the training data is not a useful estimator (would result in almost no pruning) One possibility: using hold-out set for pruning (reduced-error pruning) C4.5’s method: using upper limit of 25% confidence interval derived from the training data Standard Bernoulli-process-based method

46. Training Set

47. 46 Post-pruning in C4.5 Bottom-up pruning: at each non-leaf node v, if merging the subtree at v into a leaf node improves accuracy, perform the merging. Method 1: compute accuracy using examples not seen by the algorithm. Method 2: estimate accuracy using the training examples: Consider classifying E examples incorrectly out of N examples as observing E events in N trials in the binomial distribution. For a given confidence level CF, the upper limit on the error rate over the whole population is with CF% confidence.

48. 47 Usage in Statistics: Sampling error estimation Example: population: 1,000,000 people, could be regarded as infinite population mean: percentage of the left handed people sample: 100 people sample mean: 6 left-handed How to estimate the REAL population mean? Pessimistic Estimate U 0.25 (100,6)L 0.25 (100,6) 6 Possibility(%) 210 75% confidence interval 15%

49. 48 Usage in Decision Tree (DT): error estimation for some node in the DT example: unknown testing data: could be regarded as infinite universe population mean: percentage of error made by this node sample: 100 examples from training data set sample mean: 6 errors for the training data set How to estimate the REAL average error rate? Pessimistic Estimate U 0.25 (100,6)L 0.25 (100,6) 6 Possibility(%) 210 75% confidence interval Heuristic! But works well...

50. 49 C4.5’s method Error estimate for subtree is weighted sum of error estimates for all its leaves Error estimate for a node: If c = 25% then z = 0.69 (from normal distribution) f is the error on the training data N is the number of instances covered by the leaf

51. 50 Example for Estimating Error Consider a subtree rooted at Outlook with 3 leaf nodes: Sunny: Play = yes : (0 error, 6 instances) Overcast: Play= yes: (0 error, 9 instances) Cloudy: Play = no (0 error, 1 instance) The estimated error for this subtree is 6*0.074+9*0.050+1*0.323=1.217 If the subtree is replaced with the leaf “yes”, the estimated error is So no pruning is performed

52. 51 Example continued Outlook yes no sunny overcast cloudy yes ?

53. 52 Another Example f=0.33 e=0.47 f=0.5 e=0.72 f=0.33 e=0.47 f=5/14 e=0.46 Combined using ratios 6:2:6 this gives 0.51

54. 53 Continuous Case: The CART Algorithm

55. 54 Numeric prediction Counterparts exist for all schemes that we previously discussed Decision trees, rule learners, SVMs, etc. All classification schemes can be applied to regression problems using discretization Prediction: weighted average of intervals’ midpoints (weighted according to class probabilities) Regression more difficult than classification (i.e. percent correct vs. mean squared error)

56. 55 Regression trees Differences to decision trees: Splitting criterion: minimizing intra-subset variation Pruning criterion: based on numeric error measure Leaf node predicts average class values of training instances reaching that node Can approximate piecewise constant functions Easy to interpret More sophisticated version: model trees

57. 56 Model trees Regression trees with linear regression functions at each node Linear regression applied to instances that reach a node after full regression tree has been built Only a subset of the attributes is used for LR Attributes occurring in subtree (+maybe attributes occurring in path to the root) Fast: overhead for LR not large because usually only a small subset of attributes is used in tree

58. 57 Smoothing Naïve method for prediction outputs value of LR for corresponding leaf node Performance can be improved by smoothing predictions using internal LR models Predicted value is weighted average of LR models along path from root to leaf Smoothing formula: Same effect can be achieved by incorporating the internal models into the leaf nodes

59. 58 Building the tree Splitting criterion: standard deviation reduction Termination criteria (important when building trees for numeric prediction): Standard deviation becomes smaller than certain fraction of sd for full training set (e.g. 5%) Too few instances remain (e.g. less than four)

60. 59 Model tree for servo data

61. 60 Variations of CART Applying Logistic Regression predict probability of “True” or “False” instead of making a numerical valued prediction predict a probability value (p) rather than the outcome itself Probability= odds ratio

62. 61 Other Trees Classification Trees Current node Children nodes (L, R): Decision Trees Current node Children nodes (L, R): GINI index used in CART (STD= ) Current node Children nodes (L, R):

63. Scalability: Previous works Incremental tree construction [Quinlan 1993] using partial data to build a tree. testing other examples and mis-classified ones are used to rebuild the tree interactively. still a main-memory algorithm. Best known algorithms: ID3 C4.5 C5

64. 63 Efforts on Scalability Most algorithms assume data can fit in memory. Recent efforts focus on disk-resident implementation for decision trees. Random sampling Partitioning Examples SLIQ (EDBT ’ 96 -- [MAR96]) SPRINT (VLDB96 -- [SAM96]) PUBLIC (VLDB98 -- [RS98]) RainForest (VLDB98 -- [GRG98])