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1 Data Mining Classification Techniques: Decision Trees (BUSINESS INTELLIGENCE) Slides prepared by Elizabeth Anglo, DISCS ADMU
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2 Example of a Decision Tree categorical continuous class Refund MarSt TaxInc YES NO YesNo Married Single, Divorced < 80K> 80K Training Data Model: Decision Tree
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3 Structure of a Decision Tree categorical continuous class MarSt Refund TaxInc No Single, Divorced Married YES NO Yes < 80K> 80K There could be more than one tree that fits the same data! Splitting Attributes
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4 Decision Tree Classification Task Decision Tree
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5 Apply Model to Test Data Refund MarSt TaxInc YES NO YesNo Married Single, Divorced < 80K> 80K Test Data Start from the root of tree.
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6 Apply Model to Test Data Refund MarSt TaxInc YES NO YesNo Married Single, Divorced < 80K> 80K Test Data
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7 Apply Model to Test Data Refund MarSt TaxInc YES NO YesNo Married Single, Divorced < 80K> 80K Test Data
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8 Apply Model to Test Data Refund MarSt TaxInc YES NO YesNo Married Single, Divorced < 80K> 80K Test Data
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9 Apply Model to Test Data Refund MarSt TaxInc YES NO YesNo Married Single, Divorced < 80K> 80K Test Data
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10 Apply Model to Test Data Refund MarSt TaxInc YES NO YesNo Married Single, Divorced < 80K> 80K Test Data Assign Cheat to “No”
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11 Decision Tree Classification Task Decision Tree
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12 Decision Tree Induction l Many Algorithms: –Hunt’s Algorithm (one of the earliest) –CART –ID3, C4.5 –SLIQ,SPRINT
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13 General Structure of Hunt’s Algorithm l Let D t be the set of training records that reach a node t l General Procedure: –If D t contains records that belong the same class y t, then t is a leaf node labeled as y t –If D t is an empty set, then t is a leaf node labeled by the default class, y d –If D t contains records that belong to more than one class, use an attribute test to split the data into smaller subsets. Recursively apply the procedure to each subset. DtDt ?
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14 Hunt’s Algorithm Don’t Cheat Refund Don’t Cheat Don’t Cheat YesNo Refund Don’t Cheat YesNo Marital Status Don’t Cheat Single, Divorced Married Taxable Income Don’t Cheat < 80K>= 80K Refund Don’t Cheat YesNo Marital Status Don’t Cheat Single, Divorced Married
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15 Tree Induction l Greedy strategy. –Split the records based on an attribute test that optimizes certain criterion. l 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
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16 Tree Induction l Greedy strategy. –Split the records based on an attribute test that optimizes certain criterion. l 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
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17 How to Specify Test Condition? l Depends on attribute types –Nominal –Ordinal –Continuous l Depends on number of ways to split –2-way split –Multi-way split
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18 Splitting Based on Nominal Attributes l Multi-way split: Use as many partitions as distinct values. l Binary split: Divides values into two subsets. Need to find optimal partitioning. CarType Family Sports Luxury CarType {Family, Luxury} {Sports} CarType {Sports, Luxury} {Family} OR
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19 l Multi-way split: Use as many partitions as distinct values. l Binary split: Divides values into two subsets. Need to find optimal partitioning. l What about this split? Splitting Based on Ordinal Attributes Size Small Medium Large Size {Medium, Large} {Small} Size {Small, Medium} {Large} OR Size {Small, Large} {Medium}
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20 Splitting Based on Continuous Attributes l Different ways of handling –Discretization to form an ordinal categorical attribute Static – discretize once at the beginning Dynamic – ranges can be found by equal interval bucketing, equal frequency bucketing (percentiles), or clustering. –Binary Decision: (A < v) or (A v) consider all possible splits and finds the best cut can be more compute intensive
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21 Splitting Based on Continuous Attributes
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22 Tree Induction l Greedy strategy. –Split the records based on an attribute test that optimizes certain criterion. l 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
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23 How to determine the Best Split Before Splitting: 10 records of class 0, 10 records of class 1
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24 How to determine the Best Split Before Splitting: 10 records of class 0, 10 records of class 1 Own Car? C0: 6 C1: 4 C0: 4 C1: 6 Yes No
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25 How to determine the Best Split Before Splitting: 10 records of class 0, 10 records of class 1 Own Car? C0: 6 C1: 4 C0: 4 C1: 6 C0: 1 C1: 3 C0: 8 C1: 0 C0: 1 C1: 7 Car Type? Yes No Family Sports Luxury
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26 How to determine the Best Split Before Splitting: 10 records of class 0, 10 records of class 1 Which test condition is the best?
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27 How to determine the Best Split l Greedy approach: –Nodes with homogeneous class distribution are preferred l Need a measure of node impurity: Non-homogeneous, High degree of impurity Homogeneous, Low degree of impurity
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28 Measures of Node Impurity l Gini Index l Entropy l Misclassification error
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29 How to Find the Best Split B? YesNo Node N3Node N4 A? YesNo Node N1Node N2 Before Splitting: M0 M1 M2M3M4 M12 M34 Gain = M0 – M12 vs M0 – M34
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30 Measure of Impurity: GINI l Gini Index for a given node t : (NOTE: p( j | t) is the relative frequency of class j at node t). –Maximum (1 - 1/n c ) when records are equally distributed among all classes, implying least interesting information –Minimum (0.0) when all records belong to one class, implying most interesting information
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31 Examples for computing GINI P(C1) = 0/6 = 0 P(C2) = 6/6 = 1 Gini = 1 – P(C1) 2 – P(C2) 2 = 1 – 0 – 1 = 0 P(C1) = 1/6 P(C2) = 5/6 Gini = 1 – (1/6) 2 – (5/6) 2 = 0.278 P(C1) = 2/6 P(C2) = 4/6 Gini = 1 – (2/6) 2 – (4/6) 2 = 0.444
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32 Splitting Based on GINI l Used in CART, SLIQ, SPRINT. l When a node p is split into k partitions (children), the quality of split is computed as, where,n i = number of records at child i, n = number of records at node p.
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33 Binary Attributes: Computing GINI Index l Splits into two partitions l Effect of Weighing partitions: –Larger and Purer Partitions are sought for. Yes B? No Node N1Node N2 Gini(N1) = 1 – (5/7) 2 – (2/7) 2 = 0.204 Gini(N2) = 1 – (1/5) 2 – (4/5) 2 = 0.32 Gini(Children) = 7/12 * 0.204 + 5/12 * 0.320 = 0.252
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34 Categorical Attributes: Computing GINI Index l For each distinct value, gather counts for each class in the dataset l Use the count matrix to make decisions Multi-way splitTwo-way split (find best partition of values)
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35 Continuous Attributes: Computing GINI Index l Use Binary Decisions based on one value l Several Choices for the splitting value –Number of possible splitting values = Number of distinct values l Each splitting value has a count matrix associated with it –Class counts in each of the partitions, A < v and A v l Simple method to choose best v –For each v, scan the database to gather count matrix and compute its Gini index –Computationally Inefficient! Repetition of work.
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36 l For efficient computation: for each attribute, –Sort the attribute on values –Linearly scan these values, each time updating the count matrix and computing gini index –Choose the split position that has the least gini index Split Positions Sorted Values Continuous Attributes: Computing GINI Index
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37 Tree Induction l Greedy strategy. –Split the records based on an attribute test that optimizes certain criterion. l 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
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38 Stopping Criteria for Tree Induction l Stop expanding a node when all the records belong to the same class l Stop expanding a node when all the records have similar attribute values l Set a threshold
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39 Decision Tree Based Classification l Advantages: –Inexpensive to construct –Extremely fast at classifying unknown records –Easy to interpret for small-sized trees –In general, does not require domain knowledge; no parameter setting –Useful for all types of data –Can be used for high-dimensional data May be useful with data sets with redundant attributes
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40 Example: C4.5 l Simple depth-first construction. l Uses Information Gain l Sorts Continuous Attributes at each node. l Needs entire data to fit in memory. l Unsuitable for Large Datasets. –Needs out-of-core sorting. l You can download the software from: http://www.cse.unsw.edu.au/~quinlan/c4.5r8.tar.gz http://www.cse.unsw.edu.au/~quinlan/c4.5r8.tar.gz
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