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Data Mining Classification: Basic Concepts, Decision Trees, and Model Evaluation

Data Mining Classification: Basic Concepts, Decision Trees, and Model Evaluation
Data Mining Classification: Basic Concepts, Decision Trees, and Model Evaluation © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 1Classification: 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. © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Illustrating Classification Task Tid Attrib1 Attrib2 Attrib3 Class Learning No 1 Yes Large 125K algorithm 2 No Medium 100K No 3 No Small 70K No 4 Yes Medium 120K No Induction 5 No Large 95K Yes No 6 No Medium 60K Learn 7 Yes Large 220K No 8 No Small 85K Yes Model 9 No Medium 75K No 10 No Small 90K Yes 10 Model Training Set Apply Model Attrib1 Attrib2 Attrib3 Class Tid 11 No Small 55K 12 Yes Medium 80K Deduction 13 Yes Large 110K 14 No Small 95K 15 No Large 67K 10 Test Set © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Examples of Classification Task  Predicting tumor cells as benign or malignant  Classifying credit card transactions as legitimate or fraudulent  Classifying secondary structures of protein as alphahelix, betasheet, or random coil  Categorizing news stories as finance, weather, entertainment, sports, etc © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Classification Techniques  Decision Tree based Methods  Rulebased Methods  Memory based reasoning  Neural Networks  Naïve Bayes and Bayesian Belief Networks  Support Vector Machines © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Example of a Decision Tree Splitting Attributes Tid Refund Marital Taxable Cheat Status Income 1 Yes Single 125K No Refund 2 No Married 100K No Yes No 3 No Single 70K No 4 Yes Married 120K No NO MarSt 5 No Divorced 95K Yes Married Single, Divorced 6 No Married 60K No TaxInc NO No 7 Yes Divorced 220K 80K 80K Yes 8 No Single 85K 9 No Married 75K No NO YES 10 No Single 90K Yes 10 Model: Decision Tree Training Data © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Another Example of Decision Tree Single, MarSt Married Divorced Tid Refund Marital Taxable Cheat Status Income NO Refund 1 Yes Single 125K No No Yes No 2 No Married 100K 3 No Single 70K No NO TaxInc 4 Yes Married 120K No 80K 80K 5 No Divorced 95K Yes YES NO 6 No Married 60K No 7 Yes Divorced 220K No 8 No Single 85K Yes No There could be more than one tree that 9 No Married 75K fits the same data 10 No Single 90K Yes 10 © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Decision Tree Classification Task Tree Tid Attrib1 Attrib2 Attrib3 Class Induction No 1 Yes Large 125K algorithm 2 No Medium 100K No 3 No Small 70K No 4 Yes Medium 120K No Induction 5 No Large 95K Yes No 6 No Medium 60K Learn 7 Yes Large 220K No 8 No Small 85K Yes Model 9 No Medium 75K No 10 No Small 90K Yes 10 Model Training Set Apply Decision Model Tree Tid Attrib1 Attrib2 Attrib3 Class 11 No Small 55K 12 Yes Medium 80K Deduction 13 Yes Large 110K 14 No Small 95K 15 No Large 67K 10 Test Set © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Apply Model to Test Data Test Data Start from the root of tree. Refund Marital Taxable Cheat Status Income No Married 80K 10 Refund Yes No NO MarSt Married Single, Divorced TaxInc NO 80K 80K NO YES © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Apply Model to Test Data Test Data Refund Marital Taxable Cheat Status Income No Married 80K 10 Refund Yes No NO MarSt Married Single, Divorced TaxInc NO 80K 80K NO YES © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Apply Model to Test Data Test Data Refund Marital Taxable Cheat Status Income No Married 80K 10 Refund Yes No NO MarSt Married Single, Divorced TaxInc NO 80K 80K NO YES © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Apply Model to Test Data Test Data Refund Marital Taxable Cheat Status Income No Married 80K 10 Refund Yes No NO MarSt Married Single, Divorced TaxInc NO 80K 80K NO YES © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Apply Model to Test Data Test Data Refund Marital Taxable Cheat Status Income No Married 80K 10 Refund Yes No NO MarSt Married Single, Divorced TaxInc NO 80K 80K NO YES © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Apply Model to Test Data Test Data Refund Marital Taxable Cheat Status Income No Married 80K 10 Refund Yes No NO MarSt Assign Cheat to “No” Married Single, Divorced TaxInc NO 80K 80K NO YES © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Decision Tree Classification Task Tree Tid Attrib1 Attrib2 Attrib3 Class Induction 1 Yes Large 125K No algorithm 2 No Medium 100K No 3 No Small 70K No 4 Yes Medium 120K No Induction 5 No Large 95K Yes 6 No Medium 60K No Learn 7 Yes Large 220K No Yes 8 No Small 85K Model 9 No Medium 75K No 10 No Small 90K Yes 10 Model Training Set Apply Decision Model Tree Tid Attrib1 Attrib2 Attrib3 Class 11 No Small 55K 12 Yes Medium 80K Deduction 13 Yes Large 110K 14 No Small 95K 15 No Large 67K 10 Test Set © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Decision Tree Induction  Many Algorithms: – Hunt’s Algorithm (one of the earliest) – CART – ID3, C4.5 – SLIQ,SPRINT © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›General Structure of Hunt’s Algorithm Tid Refund Marital Taxable  Let D be the set of training records Cheat Status Income t that reach a node t 1 Yes Single 125K No 2 No Married 100K No  General Procedure: 3 No Single 70K No – If D contains records that t 4 Yes Married 120K No belong the same class y , then t t 5 No Divorced 95K Yes is a leaf node labeled as y t 6 No Married 60K No – If D is an empty set, then t is a 7 Yes Divorced 220K No t leaf node labeled by the default 8 No Single 85K Yes 9 No Married 75K No class, y d 10 No Single 90K Yes 10 – If D contains records that t belong to more than one class, D t use an attribute test to split the data into smaller subsets. Recursively apply the procedure to each subset. © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Tid Refund Marital Taxable Cheat Hunt’s Algorithm Status Income 1 Yes Single 125K No 2 No Married 100K No Refund 3 No Single 70K No Don’t Yes No Cheat 4 Yes Married 120K No Don’t Don’t 5 No Divorced 95K Yes Cheat Cheat 6 No Married 60K No 7 Yes Divorced 220K No 8 No Single 85K Yes Refund 9 No Married 75K No Refund Yes No 10 No Single 90K Yes No Yes 10 Don’t Marital Don’t Marital Cheat Cheat Status Status Single, Single, Married Married Divorced Divorced Don’t Don’t Taxable Cheat Cheat Cheat Income 80K = 80K Don’t Cheat Cheat © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›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 © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›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 © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›How to Specify Test Condition  Depends on attribute types – Nominal – Ordinal – Continuous  Depends on number of ways to split – 2way split – Multiway split © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Splitting Based on Nominal Attributes  Multiway split: Use as many partitions as distinct values. CarType Family Luxury Sports  Binary split: Divides values into two subsets. Need to find optimal partitioning. CarType CarType Sports, Family, OR Family Sports Luxury Luxury © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Splitting Based on Ordinal Attributes  Multiway split: Use as many partitions as distinct values. Size Small Large Medium  Binary split: Divides values into two subsets. Need to find optimal partitioning. Size Size Small, Medium, OR Large Small Medium Large Size Small, Medium  What about this split Large © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Splitting Based on Continuous Attributes  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 © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Splitting Based on Continuous Attributes Taxable Taxable Income Income 80K 10K 80K Yes No 10K,25K) 25K,50K) 50K,80K) (i) Binary split (ii) Multiway split © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›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 © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›How to determine the Best Split Before Splitting: 10 records of class 0, 10 records of class 1 Own Car Student Car Type ID Family Luxury c c Yes No 1 20 c c 10 11 Sports C0: 6 C0: 4 C0: 1 C0: 8 C0: 1 C0: 1 C0: 1 C0: 0 C0: 0 ... ... C1: 4 C1: 6 C1: 3 C1: 0 C1: 7 C1: 0 C1: 0 C1: 1 C1: 1 Which test condition is the best © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›How to determine the Best Split  Greedy approach: – Nodes with homogeneous class distribution are preferred  Need a measure of node impurity: C0: 5 C0: 9 C1: 5 C1: 1 Nonhomogeneous, Homogeneous, High degree of impurity Low degree of impurity © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Measures of Node Impurity  Gini Index  Entropy  Misclassification error © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›How to Find the Best Split C0 N00 Before Splitting: M0 C1 N01 A B Yes No Yes No Node N1 Node N2 Node N3 Node N4 C0 N10 C0 N20 C0 N30 C0 N40 C1 N21 C1 N31 C1 N41 C1 N11 M2 M3 M4 M1 M12 M34 Gain = M0 – M12 vs M0 – M34 © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Measure of Impurity: GINI  Gini Index for a given node t : 2 GINI(t)1 p( j t)  j (NOTE: p( j t) is the relative frequency of class j at node t). – Maximum (1 1/n ) when records are equally c distributed among all classes, implying least interesting information – Minimum (0.0) when all records belong to one class, implying most interesting information C1 0 C1 1 C1 2 C1 3 C2 6 C2 5 C2 4 C2 3 Gini=0.000 Gini=0.278 Gini=0.444 Gini=0.500 © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Examples for computing GINI 2 GINI(t)1 p( j t)  j C1 0 P(C1) = 0/6 = 0 P(C2) = 6/6 = 1 2 2 C2 6 Gini = 1 – P(C1) – P(C2) = 1 – 0 – 1 = 0 C1 1 P(C1) = 1/6 P(C2) = 5/6 C2 5 2 2 Gini = 1 – (1/6) – (5/6) = 0.278 P(C1) = 2/6 P(C2) = 4/6 C1 2 2 2 C2 4 Gini = 1 – (2/6) – (4/6) = 0.444 © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Splitting Based on GINI  Used in CART, SLIQ, SPRINT.  When a node p is split into k partitions (children), the quality of split is computed as, k n i GINI GINI(i) split n i1 where, n = number of records at child i, i n = number of records at node p. © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Binary Attributes: Computing GINI Index  Splits into two partitions  Effect of Weighing partitions: – Larger and Purer Partitions are sought for. Parent B C1 6 C2 6 Yes No Gini = 0.500 Node N1 Node N2 Gini(N1) 2 2 = 1 – (5/6) – (2/6) N1 N2 Gini(Children) = 0.194 C1 5 1 = 7/12 0.194 + Gini(N2) 5/12 0.528 C2 2 4 2 2 = 1 – (1/6) – (4/6) = 0.333 Gini=0.333 = 0.528 © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Categorical Attributes: Computing Gini Index  For each distinct value, gather counts for each class in the dataset  Use the count matrix to make decisions Multiway split Twoway split (find best partition of values) CarType CarType CarType Sports, Family, Family Sports Luxury Family Sports Luxury Luxury C1 1 2 1 C1 3 1 C1 2 2 C2 4 1 1 C2 2 4 C2 1 5 Gini 0.393 Gini 0.400 Gini 0.419 © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Continuous Attributes: Computing Gini Index Tid Refund Marital Taxable  Use Binary Decisions based on one Cheat Status Income value 1 Yes Single 125K No  Several Choices for the splitting value 2 No Married 100K No 3 No Single 70K No – Number of possible splitting values = Number of distinct values 4 Yes Married 120K No 5 No Divorced 95K Yes  Each splitting value has a count matrix 6 No Married 60K No associated with it 7 Yes Divorced 220K No – Class counts in each of the 8 No Single 85K Yes partitions, A v and A  v 9 No Married 75K No  Simple method to choose best v 10 No Single 90K Yes 10 – For each v, scan the database to Taxable gather count matrix and compute Income its Gini index 80K – Computationally Inefficient Repetition of work. Yes No © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Continuous Attributes: Computing Gini Index...  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 Cheat No No No Yes Yes Yes No No No No Taxable Income 60 70 75 85 90 95 100 120 125 220 Sorted Values 55 65 72 80 87 92 97 110 122 172 230 Split Positions = = = = = = = = = = = Yes 0 3 0 3 0 3 0 3 1 2 2 1 3 0 3 0 3 0 3 0 3 0 No 0 7 1 6 2 5 3 4 3 4 3 4 3 4 4 3 5 2 6 1 7 0 Gini 0.420 0.400 0.375 0.343 0.417 0.400 0.300 0.343 0.375 0.400 0.420 © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Alternative Splitting Criteria based on INFO  Entropy at a given node t:  Entropy(t) p( j t)log p( j t) j (NOTE: p( j t) is the relative frequency of class j at node t). – Measures homogeneity of a node. Maximum (log n ) when records are equally distributed c among all classes implying least information Minimum (0.0) when all records belong to one class, implying most information – Entropy based computations are similar to the GINI index computations © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Examples for computing Entropy  Entropy(t) p( j t)log p( j t) 2 j C1 0 P(C1) = 0/6 = 0 P(C2) = 6/6 = 1 C2 6 Entropy = – 0 log 0 – 1 log 1 = – 0 – 0 = 0 P(C1) = 1/6 P(C2) = 5/6 C1 1 C2 5 Entropy = – (1/6) log (1/6) – (5/6) log (1/6) = 0.65 2 2 P(C1) = 2/6 P(C2) = 4/6 C1 2 C2 4 Entropy = – (2/6) log (2/6) – (4/6) log (4/6) = 0.92 2 2 © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Splitting Based on INFO...  Information Gain: k n  i GAIN Entropy( p) Entropy(i)  split i1 n  Parent Node, p is split into k partitions; n is number of records in partition i 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. © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Splitting Based on INFO...  Gain Ratio: GAIN k n n Split i i GainRATIO SplitINFO log split i1 SplitINFO n n Parent Node, p is split into k partitions n is the number of records in partition i i – Adjusts Information Gain by the entropy of the partitioning (SplitINFO). Higher entropy partitioning (large number of small partitions) is penalized – Used in C4.5 – Designed to overcome the disadvantage of Information Gain © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Splitting Criteria based on Classification Error  Classification error at a node t : Error(t)1 max P(i t) i  Measures misclassification error made by a node.  Maximum (1 1/n ) when records are equally distributed c among all classes, implying least interesting information  Minimum (0.0) when all records belong to one class, implying most interesting information © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Examples for Computing Error Error(t)1 max P(i t) i C1 0 P(C1) = 0/6 = 0 P(C2) = 6/6 = 1 C2 6 Error = 1 – max (0, 1) = 1 – 1 = 0 P(C1) = 1/6 P(C2) = 5/6 C1 1 C2 5 Error = 1 – max (1/6, 5/6) = 1 – 5/6 = 1/6 P(C1) = 2/6 P(C2) = 4/6 C1 2 C2 4 Error = 1 – max (2/6, 4/6) = 1 – 4/6 = 1/3 © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Comparison among Splitting Criteria For a 2class problem: © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Misclassification Error vs Gini Parent A C1 7 Yes No C2 3 Gini = 0.42 Node N1 Node N2 Gini(N1) N1 N2 2 2 Gini(Children) = 1 – (3/3) – (0/3) C1 3 4 = 3/10 0 = 0 C2 0 3 + 7/10 0.489 Gini(N2) Gini=0.361 = 0.342 2 2 = 1 – (4/7) – (3/7) Gini improves = 0.489 © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›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 © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Stopping Criteria for Tree Induction  Stop expanding a node when all the records belong to the same class  Stop expanding a node when all the records have similar attribute values  Early termination (to be discussed later) © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Decision Tree Based Classification  Advantages: – Inexpensive to construct – Extremely fast at classifying unknown records – Easy to interpret for smallsized trees – Accuracy is comparable to other classification techniques for many simple data sets © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Example: C4.5  Simple depthfirst construction.  Uses Information Gain  Sorts Continuous Attributes at each node.  Needs entire data to fit in memory.  Unsuitable for Large Datasets. – Needs outofcore sorting.  You can download the software from: http://www.cse.unsw.edu.au/quinlan/c4.5r8.tar.gz © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Practical Issues of Classification  Underfitting and Overfitting  Missing Values  Costs of Classification © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Underfitting and Overfitting (Example) 500 circular and 500 triangular data points. Circular points: 2 2 0.5  sqrt(x +x )  1 1 2 Triangular points: 2 2 sqrt(x +x ) 0.5 or 1 2 2 2 sqrt(x +x ) 1 1 2 © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Underfitting and Overfitting Overfitting Underfitting: when model is too simple, both training and test errors are large © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Overfitting due to Noise Decision boundary is distorted by noise point © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Overfitting due to Insufficient Examples Lack of data points in the lower half of the diagram makes it difficult to predict correctly the class labels of that region Insufficient number of training records in the region causes the decision tree to predict the test examples using other training records that are irrelevant to the classification task © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Notes on Overfitting  Overfitting results in decision trees that are more complex than necessary  Training error no longer provides a good estimate of how well the tree will perform on previously unseen records  Need new ways for estimating errors © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Estimating Generalization Errors  Resubstitution errors: error on training ( e(t) )  Generalization errors: error on testing ( e’(t))  Methods for estimating generalization errors: – Optimistic approach: e’(t) = e(t) – Pessimistic approach:  For each leaf node: e’(t) = (e(t)+0.5)  Total errors: e’(T) = e(T) + N  0.5 (N: number of leaf nodes)  For a tree with 30 leaf nodes and 10 errors on training (out of 1000 instances): Training error = 10/1000 = 1 Generalization error = (10 + 300.5)/1000 = 2.5 – Reduced error pruning (REP):  uses validation data set to estimate generalization error © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Occam’s Razor  Given two models of similar generalization errors, one should prefer the simpler model over the more complex model  For complex models, there is a greater chance that it was fitted accidentally by errors in data  Therefore, one should include model complexity when evaluating a model © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Minimum Description Length (MDL) A X y Yes No X y X 1 1 0 B X 1 B B X 1 2 0 2 X 2 C X 1 0 3 A B X C C 3 1 2 X 1 4 X 0 1 4 … … … … X 1 n X n  Cost(Model,Data) = Cost(DataModel) + Cost(Model) – Cost is the number of bits needed for encoding. – Search for the least costly model.  Cost(DataModel) encodes the misclassification errors.  Cost(Model) uses node encoding (number of children) plus splitting condition encoding. © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›How to Address Overfitting  PrePruning (Early Stopping Rule) – Stop the algorithm before it becomes a fullygrown tree – Typical stopping conditions for a node:  Stop if all instances belong to the same class  Stop if all the attribute values are the same – More restrictive conditions:  Stop if number of instances is less than some userspecified threshold  Stop if class distribution of instances are independent of the 2 available features (e.g., using  test)  Stop if expanding the current node does not improve impurity measures (e.g., Gini or information gain). © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›How to Address Overfitting…  Postpruning – Grow decision tree to its entirety – Trim the nodes of the decision tree in a bottomup fashion – If generalization error improves after trimming, replace subtree by a leaf node. – Class label of leaf node is determined from majority class of instances in the subtree – Can use MDL for postpruning © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Example of PostPruning Training Error (Before splitting) = 10/30 Pessimistic error = (10 + 0.5)/30 = 10.5/30 Class = Yes 20 Training Error (After splitting) = 9/30 Class = No 10 Pessimistic error (After splitting) Error = 10/30 = (9 + 4  0.5)/30 = 11/30 PRUNE A A1 A4 A2 A3 Class = Yes 8 Class = Yes 3 Class = Yes 4 Class = Yes 5 Class = No 4 Class = No 4 Class = No 1 Class = No 1 © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Examples of Postpruning Case 1: – Optimistic error Don’t prune for both cases C0: 11 C0: 2 C1: 3 C1: 4 – Pessimistic error Don’t prune case 1, prune case 2 – Reduced error pruning Case 2: Depends on validation set C0: 14 C0: 2 C1: 3 C1: 2 © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Handling Missing Attribute Values  Missing values affect decision tree construction in three different ways: – Affects how impurity measures are computed – Affects how to distribute instance with missing value to child nodes – Affects how a test instance with missing value is classified © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Computing Impurity Measure Before Splitting: Tid Refund Marital Taxable Class Status Income Entropy(Parent) = 0.3 log(0.3)(0.7)log(0.7) = 0.8813 1 Yes Single 125K No 2 No Married 100K No Class Class = Yes = No 3 No Single 70K No Refund=Yes 0 3 4 Yes Married 120K No Refund=No 2 4 5 No Divorced 95K Yes Refund= 1 0 6 No Married 60K No Split on Refund: 7 Yes Divorced 220K No Entropy(Refund=Yes) = 0 8 No Single 85K Yes 9 No Married 75K No Entropy(Refund=No) = (2/6)log(2/6) – (4/6)log(4/6) = 0.9183 10 Single 90K Yes 10 Entropy(Children) Missing = 0.3 (0) + 0.6 (0.9183) = 0.551 value Gain = 0.9  (0.8813 – 0.551) = 0.3303 © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Distribute Instances Tid Refund Marital Taxable Class Status Income Tid Refund Marital Taxable 1 Yes Single 125K No Class Status Income 2 No Married 100K No 10 Single 90K Yes 10 3 No Single 70K No 4 Yes Married 120K No Refund 5 No Divorced 95K Yes Yes No 6 No Married 60K No Class=Yes 0 + 3/9 Class=Yes 2 + 6/9 7 Yes Divorced 220K No Class=No 3 Class=No 4 8 No Single 85K Yes 9 No Married 75K No 10 Probability that Refund=Yes is 3/9 Refund Probability that Refund=No is 6/9 Yes No Assign record to the left child with Cheat=Yes 2 Class=Yes 0 weight = 3/9 and to the right child Cheat=No 4 Class=No 3 with weight = 6/9 © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Classify Instances New record: Married Single Divorced Total Tid Refund Marital Taxable Class Status Income Class=No 3 1 0 4 11 No 85K 10 Class=Yes 6/9 1 1 2.67 Total 3.67 2 1 6.67 Refund Yes No NO MarSt Single, Married Probability that Marital Status Divorced = Married is 3.67/6.67 TaxInc NO Probability that Marital Status 80K 80K =Single,Divorced is 3/6.67 YES NO © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Other Issues  Data Fragmentation  Search Strategy  Expressiveness  Tree Replication © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Data Fragmentation  Number of instances gets smaller as you traverse down the tree  Number of instances at the leaf nodes could be too small to make any statistically significant decision © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Search Strategy  Finding an optimal decision tree is NPhard  The algorithm presented so far uses a greedy, topdown, recursive partitioning strategy to induce a reasonable solution  Other strategies – Bottomup – Bidirectional © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Expressiveness  Decision tree provides expressive representation for learning discretevalued function – But they do not generalize well to certain types of Boolean functions  Example: parity function: – Class = 1 if there is an even number of Boolean attributes with truth value = True – Class = 0 if there is an odd number of Boolean attributes with truth value = True  For accurate modeling, must have a complete tree  Not expressive enough for modeling continuous variables – Particularly when test condition involves only a single attribute atatime © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Decision Boundary 1 0.9 x 0.43 0.8 0.7 Yes No 0.6 0.5 y 0.33 y 0.47 0.4 Yes No Yes No 0.3 0.2 : 4 : 0 : 0 : 4 0.1 : 0 : 4 : 3 : 0 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 x • Border line between two neighboring regions of different classes is known as decision boundary • Decision boundary is parallel to axes because test condition involves a single attribute atatime © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹› yOblique Decision Trees x + y 1 Class = Class = + • Test condition may involve multiple attributes • More expressive representation • Finding optimal test condition is computationally expensive © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Tree Replication P Q R S 0 Q 1 0 1 S 0 0 1 • Same subtree appears in multiple branches © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Model Evaluation  Metrics for Performance Evaluation – How to evaluate the performance of a model  Methods for Performance Evaluation – How to obtain reliable estimates  Methods for Model Comparison – How to compare the relative performance among competing models © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Model Evaluation  Metrics for Performance Evaluation – How to evaluate the performance of a model  Methods for Performance Evaluation – How to obtain reliable estimates  Methods for Model Comparison – How to compare the relative performance among competing models © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Metrics for Performance Evaluation  Focus on the predictive capability of a model – Rather than how fast it takes to classify or build models, scalability, etc.  Confusion Matrix: PREDICTED CLASS Class=Yes Class=No a: TP (true positive) b: FN (false negative) Class=Yes a b c: FP (false positive) ACTUAL d: TN (true negative) CLASS Class=No c d © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Metrics for Performance Evaluation… PREDICTED CLASS Class=Yes Class=No Class=Yes a b ACTUAL (TP) (FN) CLASS Class=No c d (FP) (TN)  Most widelyused metric: a d TPTN Accuracy  a b c d TPTN FP FN © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Limitation of Accuracy  Consider a 2class problem – Number of Class 0 examples = 9990 – Number of Class 1 examples = 10  If model predicts everything to be class 0, accuracy is 9990/10000 = 99.9 – Accuracy is misleading because model does not detect any class 1 example © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Cost Matrix PREDICTED CLASS Class=Yes Class=No C(ij) Class=Yes C(YesYes) C(NoYes) ACTUAL CLASS Class=No C(YesNo) C(NoNo) C(ij): Cost of misclassifying class j example as class i © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Computing Cost of Classification Cost PREDICTED CLASS Matrix C(ij) + ACTUAL + 1 100 CLASS 1 0 Model PREDICTED CLASS Model PREDICTED CLASS M M 1 2 + + ACTUAL ACTUAL + 150 40 + 250 45 CLASS CLASS 60 250 5 200 Accuracy = 80 Accuracy = 90 Cost = 3910 Cost = 4255 © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Cost vs Accuracy PREDICTED CLASS Accuracy is proportional to cost if Count 1. C(YesNo)=C(NoYes) = q Class=Yes Class=No 2. C(YesYes)=C(NoNo) = p Class=Yes a b ACTUAL N = a + b + c + d CLASS Class=No c d Accuracy = (a + d)/N PREDICTED CLASS Cost Cost = p (a + d) + q (b + c) Class=Yes Class=No = p (a + d) + q (N – a – d) Class=Yes = q N – (q – p)(a + d) p q ACTUAL = N q – (qp)  Accuracy CLASS Class=No q p © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›CostSensitive Measures a Precision (p)  a c a Recall (r) a b 2rp 2a F measure (F) r p 2a b c  Precision is biased towards C(YesYes) C(YesNo)  Recall is biased towards C(YesYes) C(NoYes)  Fmeasure is biased towards all except C(NoNo) w a w d 1 4 Weighted Accuracy  w a w b w c w d 1 2 3 4 © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Model Evaluation  Metrics for Performance Evaluation – How to evaluate the performance of a model  Methods for Performance Evaluation – How to obtain reliable estimates  Methods for Model Comparison – How to compare the relative performance among competing models © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Methods for Performance Evaluation  How to obtain a reliable estimate of performance  Performance of a model may depend on other factors besides the learning algorithm: – Class distribution – Cost of misclassification – Size of training and test sets © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Learning Curve  Learning curve shows how accuracy changes with varying sample size  Requires a sampling schedule for creating learning curve:  Arithmetic sampling (Langley, et al)  Geometric sampling (Provost et al) Effect of small sample size: Bias in the estimate Variance of estimate © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Methods of Estimation  Holdout – Reserve 2/3 for training and 1/3 for testing  Random subsampling – Repeated holdout  Cross validation – Partition data into k disjoint subsets – kfold: train on k1 partitions, test on the remaining one – Leaveoneout: k=n  Stratified sampling – oversampling vs undersampling  Bootstrap – Sampling with replacement © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Model Evaluation  Metrics for Performance Evaluation – How to evaluate the performance of a model  Methods for Performance Evaluation – How to obtain reliable estimates  Methods for Model Comparison – How to compare the relative performance among competing models © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›ROC (Receiver Operating Characteristic)  Developed in 1950s for signal detection theory to analyze noisy signals – Characterize the tradeoff between positive hits and false alarms  ROC curve plots TP (on the yaxis) against FP (on the xaxis)  Performance of each classifier represented as a point on the ROC curve – changing the threshold of algorithm, sample distribution or cost matrix changes the location of the point © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›ROC Curve 1dimensional data set containing 2 classes (positive and negative) any points located at x t is classified as positive At threshold t: TP=0.5, FN=0.5, FP=0.12, FN=0.88 © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›ROC Curve (TP,FP):  (0,0): declare everything to be negative class  (1,1): declare everything to be positive class  (1,0): ideal  Diagonal line: – Random guessing – Below diagonal line:  prediction is opposite of the true class © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Using ROC for Model Comparison  No model consistently outperform the other  M is better for 1 small FPR  M is better for 2 large FPR  Area Under the ROC curve  Ideal:  Area = 1  Random guess:  Area = 0.5 © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›How to Construct an ROC curve • Use classifier that produces Instance P(+A) True Class posterior probability for each 1 0.95 + test instance P(+A) 2 0.93 + • Sort the instances according 3 0.87 to P(+A) in decreasing order 4 0.85 5 0.85 • Apply threshold at each 6 0.85 + unique value of P(+A) 7 0.76 • Count the number of TP, FP, 8 0.53 + TN, FN at each threshold 9 0.43 10 0.25 + • TP rate, TPR = TP/(TP+FN) • FP rate, FPR = FP/(FP + TN) © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›How to construct an ROC curve Class + + + + + P 0.25 0.43 0.53 0.76 0.85 0.85 0.85 0.87 0.93 0.95 1.00 Threshold = TP 5 4 4 3 3 3 3 2 2 1 0 FP 5 5 4 4 3 2 1 1 0 0 0 TN 0 0 1 1 2 3 4 4 5 5 5 FN 0 1 1 2 2 2 2 3 3 4 5 TPR 1 0.8 0.8 0.6 0.6 0.6 0.6 0.4 0.4 0.2 0 FPR 1 1 0.8 0.8 0.6 0.4 0.2 0.2 0 0 0 ROC Curve: © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Test of Significance  Given two models: – Model M1: accuracy = 85, tested on 30 instances – Model M2: accuracy = 75, tested on 5000 instances  Can we say M1 is better than M2 – How much confidence can we place on accuracy of M1 and M2 – Can the difference in performance measure be explained as a result of random fluctuations in the test set © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Confidence Interval for Accuracy  Prediction can be regarded as a Bernoulli trial – A Bernoulli trial has 2 possible outcomes – Possible outcomes for prediction: correct or wrong – Collection of Bernoulli trials has a Binomial distribution:  x  Bin(N, p) x: number of correct predictions  e.g: Toss a fair coin 50 times, how many heads would turn up Expected number of heads = Np = 50  0.5 = 25  Given x ( of correct predictions) or equivalently, acc=x/N, and N ( of test instances), Can we predict p (true accuracy of model) © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Confidence Interval for Accuracy Area = 1   For large test sets (N 30), – acc has a normal distribution with mean p and variance p(1p)/N acc p P(Z Z )  / 2 1 / 2 p(1 p) / N 1 Z Z /2 1 /2  Confidence Interval for p: 2 2 2 2  N  acc  Z  Z  4  N  acc  4  N  acc  / 2  / 2 p  2 2(N  Z )  / 2 © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Confidence Interval for Accuracy  Consider a model that produces an accuracy of 80 when evaluated on 100 test instances: – N=100, acc = 0.8 1 Z – Let 1 = 0.95 (95 confidence) 0.99 2.58 – From probability table, Z =1.96 /2 0.98 2.33 N 50 100 500 1000 5000 0.95 1.96 0.90 1.65 p(lower) 0.670 0.711 0.763 0.774 0.789 p(upper) 0.888 0.866 0.833 0.824 0.811 © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Comparing Performance of 2 Models  Given two models, say M1 and M2, which is better – M1 is tested on D1 (size=n1), found error rate = e 1 – M2 is tested on D2 (size=n2), found error rate = e 2 – Assume D1 and D2 are independent – If n1 and n2 are sufficiently large, then e N , 1 1 1 e N , 2 2 2 e (1 e ) i i  ˆ – Approximate: i n i © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Comparing Performance of 2 Models  To test if performance difference is statistically significant: d = e1 – e2 – d N(d , ) where d is the true difference t t t – Since D1 and D2 are independent, their variance adds up: 2 2 2 2 2  ˆ ˆ t 1 2 1 2 e1(1 e1) e2(1 e2)  n1 n2 – At (1) confidence level, d d Z ˆ t / 2 t © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›An Illustrative Example  Given: M1: n1 = 30, e1 = 0.15 M2: n2 = 5000, e2 = 0.25  d = e2 – e1 = 0.1 (2sided test) 0.15(1 0.15) 0.25(1 0.25)  0.0043 ˆ d 30 5000  At 95 confidence level, Z =1.96 /2 d 0.1001.96 0.0043 0.100 0.128 t = Interval contains 0 = difference may not be statistically significant © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›Comparing Performance of 2 Algorithms  Each learning algorithm may produce k models: – L1 may produce M11 , M12, …, M1k – L2 may produce M21 , M22, …, M2k  If models are generated on the same test sets D1,D2, …, Dk (e.g., via crossvalidation) – For each set: compute d = e – e j 1j 2j – d has mean d and variance  j t t k 2 – Estimate: (d d) j 2 j1  ˆ t k(k1) d d t ˆ t 1 ,k1 t © Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 ‹›
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