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Data Mining Lecture_10(a).pptx
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Data Mining Lecture_10(a).pptx

The document covers classification concepts in data mining, specifically focusing on decision trees used for predicting tax evasion based on attributes like marital status and taxable income. It discusses the process of building classification models, evaluating their effectiveness using confusion matrices, and outlines various classification techniques. Additionally, it delves into decision tree construction, including attribute selection and impurity measurements, emphasizing methods such as ID3, C4.5, and CART.

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DATA MINING LECTURE 10 Classification Basic Concepts Decision Trees Subrata Kumer Paul Assistant Professor, Dept. of CSE, BAUET sksubrata96@gmail.com
Catching tax-evasion Tid Refund Marital Status Taxable Income Cheat 1 Yes Single 125K No 2 No Married 100K No 3 No Single 70K No 4 Yes Married 120K No 5 No Divorced 95K Yes 6 No Married 60K No 7 Yes Divorced 220K No 8 No Single 85K Yes 9 No Married 75K No 10 No Single 90K Yes 10 Refund Marital Status Taxable Income Cheat No Married 80K ? 10 Tax-return data for year 2011 A new tax return for 2012 Is this a cheating tax return? An instance of the classification problem: learn a method for discriminating between records of different classes (cheaters vs non-cheaters)
What is classification? • Classification is the task of learning a target function f that maps attribute set x to one of the predefined class labels y Tid Refund Marital Status Taxable Income Cheat 1 Yes Single 125K No 2 No Married 100K No 3 No Single 70K No 4 Yes Married 120K No 5 No Divorced 95K Yes 6 No Married 60K No 7 Yes Divorced 220K No 8 No Single 85K Yes 9 No Married 75K No 10 No Single 90K Yes 10 One of the attributes is the class attribute In this case: Cheat Two class labels (or classes): Yes (1), No (0)
Why classification? • The target function f is known as a classification model • Descriptive modeling: Explanatory tool to distinguish between objects of different classes (e.g., understand why people cheat on their taxes) • Predictive modeling: Predict a class of a previously unseen record
Examples of Classification Tasks • Predicting tumor cells as benign or malignant • Classifying credit card transactions as legitimate or fraudulent • Categorizing news stories as finance, weather, entertainment, sports, etc • Identifying spam email, spam web pages, adult content • Understanding if a web query has commercial intent or not
General approach to classification • Training set consists of records with known class labels • Training set is used to build a classification model • A labeled test set of previously unseen data records is used to evaluate the quality of the model. • The classification model is applied to new records with unknown class labels
Illustrating Classification Task Apply Model Induction Deduction Learn Model Model Tid Attrib1 Attrib2 Attrib3 Class 1 Yes Large 125K No 2 No Medium 100K No 3 No Small 70K No 4 Yes Medium 120K No 5 No Large 95K Yes 6 No Medium 60K No 7 Yes Large 220K No 8 No Small 85K Yes 9 No Medium 75K No 10 No Small 90K Yes 10 Tid Attrib1 Attrib2 Attrib3 Class 11 No Small 55K ? 12 Yes Medium 80K ? 13 Yes Large 110K ? 14 No Small 95K ? 15 No Large 67K ? 10 Test Set Learning algorithm Training Set
Evaluation of classification models • Counts of test records that are correctly (or incorrectly) predicted by the classification model • Confusion matrix Class = 1 Class = 0 Class = 1 f11 f10 Class = 0 f01 f00 Predicted Class Actual Class 00 01 10 11 00 11 s prediction of # total s prediction correct # Accuracy f f f f f f       00 01 10 11 01 10 s prediction of # total s prediction wrong # rate Error f f f f f f      
Classification Techniques • Decision Tree based Methods • Rule-based Methods • Memory based reasoning • Neural Networks • Naïve Bayes and Bayesian Belief Networks • Support Vector Machines
Classification Techniques • Decision Tree based Methods • Rule-based Methods • Memory based reasoning • Neural Networks • Naïve Bayes and Bayesian Belief Networks • Support Vector Machines
Decision Trees • Decision tree • A flow-chart-like tree structure • Internal node denotes a test on an attribute • Branch represents an outcome of the test • Leaf nodes represent class labels or class distribution
Example of a Decision Tree Tid Refund Marital Status Taxable Income Cheat 1 Yes Single 125K No 2 No Married 100K No 3 No Single 70K No 4 Yes Married 120K No 5 No Divorced 95K Yes 6 No Married 60K No 7 Yes Divorced 220K No 8 No Single 85K Yes 9 No Married 75K No 10 No Single 90K Yes 10 Refund MarSt TaxInc YES NO NO NO Yes No Married Single, Divorced < 80K > 80K Splitting Attributes Training Data Model: Decision Tree Test outcome Class labels
Another Example of Decision Tree Tid Refund Marital Status Taxable Income Cheat 1 Yes Single 125K No 2 No Married 100K No 3 No Single 70K No 4 Yes Married 120K No 5 No Divorced 95K Yes 6 No Married 60K No 7 Yes Divorced 220K No 8 No Single 85K Yes 9 No Married 75K No 10 No Single 90K Yes 10 MarSt Refund TaxInc YES NO NO NO Yes No Married Single, Divorced < 80K > 80K There could be more than one tree that fits the same data!
Decision Tree Classification Task Apply Model Induction Deduction Learn Model Model Tid Attrib1 Attrib2 Attrib3 Class 1 Yes Large 125K No 2 No Medium 100K No 3 No Small 70K No 4 Yes Medium 120K No 5 No Large 95K Yes 6 No Medium 60K No 7 Yes Large 220K No 8 No Small 85K Yes 9 No Medium 75K No 10 No Small 90K Yes 10 Tid Attrib1 Attrib2 Attrib3 Class 11 No Small 55K ? 12 Yes Medium 80K ? 13 Yes Large 110K ? 14 No Small 95K ? 15 No Large 67K ? 10 Test Set Tree Induction algorithm Training Set Decision Tree
Apply Model to Test Data Refund MarSt TaxInc YES NO NO NO Yes No Married Single, Divorced < 80K > 80K Refund Marital Status Taxable Income Cheat No Married 80K ? 10 Test Data Start from the root of tree.
Apply Model to Test Data Refund MarSt TaxInc YES NO NO NO Yes No Married Single, Divorced < 80K > 80K Refund Marital Status Taxable Income Cheat No Married 80K ? 10 Test Data
Apply Model to Test Data Refund MarSt TaxInc YES NO NO NO Yes No Married Single, Divorced < 80K > 80K Refund Marital Status Taxable Income Cheat No Married 80K ? 10 Test Data
Apply Model to Test Data Refund MarSt TaxInc YES NO NO NO Yes No Married Single, Divorced < 80K > 80K Refund Marital Status Taxable Income Cheat No Married 80K ? 10 Test Data
Apply Model to Test Data Refund MarSt TaxInc YES NO NO NO Yes No Married Single, Divorced < 80K > 80K Refund Marital Status Taxable Income Cheat No Married 80K ? 10 Test Data
Apply Model to Test Data Refund MarSt TaxInc YES NO NO NO Yes No Married Single, Divorced < 80K > 80K Refund Marital Status Taxable Income Cheat No Married 80K ? 10 Test Data Assign Cheat to “No”
Decision Tree Classification Task Apply Model Induction Deduction Learn Model Model Tid Attrib1 Attrib2 Attrib3 Class 1 Yes Large 125K No 2 No Medium 100K No 3 No Small 70K No 4 Yes Medium 120K No 5 No Large 95K Yes 6 No Medium 60K No 7 Yes Large 220K No 8 No Small 85K Yes 9 No Medium 75K No 10 No Small 90K Yes 10 Tid Attrib1 Attrib2 Attrib3 Class 11 No Small 55K ? 12 Yes Medium 80K ? 13 Yes Large 110K ? 14 No Small 95K ? 15 No Large 67K ? 10 Test Set Tree Induction algorithm Training Set Decision Tree
Tree Induction • Finding the best decision tree is NP-hard • Greedy strategy. • Split the records based on an attribute test that optimizes certain criterion. • Many Algorithms: • Hunt’s Algorithm (one of the earliest) • CART • ID3, C4.5 • SLIQ,SPRINT
General Structure of Hunt’s Algorithm • Let Dt be the set of training records that reach a node t • General Procedure: • If Dt contains records that belong the same class yt, then t is a leaf node labeled as yt • If Dt contains records with the same attribute values, then t is a leaf node labeled with the majority class yt • If Dt is an empty set, then t is a leaf node labeled by the default class, yd • If Dt 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. Tid Refund Marital Status Taxable Income Cheat 1 Yes Single 125K No 2 No Married 100K No 3 No Single 70K No 4 Yes Married 120K No 5 No Divorced 95K Yes 6 No Married 60K No 7 Yes Divorced 220K No 8 No Single 85K Yes 9 No Married 75K No 10 No Single 90K Yes 10 Dt ?
Hunt’s Algorithm Don’t Cheat Refund Don’t Cheat Don’t Cheat Yes No Refund Don’t Cheat Yes No Marital Status Don’t Cheat Cheat Single, Divorced Married Taxable Income Don’t Cheat < 80K >= 80K Refund Don’t Cheat Yes No Marital Status Don’t Cheat Cheat Single, Divorced Married Tid Refund Marital Status Taxable Income Cheat 1 Yes Single 125K No 2 No Married 100K No 3 No Single 70K No 4 Yes Married 120K No 5 No Divorced 95K Yes 6 No Married 60K No 7 Yes Divorced 220K No 8 No Single 85K Yes 9 No Married 75K No 10 No Single 90K Yes 10 Tid Refund Marital Status Taxable Income Cheat 1 Yes Single 125K No 4 Yes Married 120K No 7 Yes Divorced 220K No 2 No Married 100K No 3 No Single 70K No 5 No Divorced 95K Yes 6 No Married 60K No 8 No Single 85K Yes 9 No Married 75K No 10 No Single 90K Yes 10 Tid Refund Marital Status Taxable Income Cheat 1 Yes Single 125K No 4 Yes Married 120K No 7 Yes Divorced 220K No 2 No Married 100K No 6 No Married 60K No 9 No Married 75K No 3 No Single 70K No 5 No Divorced 95K Yes 8 No Single 85K Yes 10 No Single 90K Yes 10
Constructing decision-trees (pseudocode) GenDecTree(Sample S, Features F) 1. If stopping_condition(S,F) = true then a. leaf = createNode() b. leaf.label= Classify(S) c. return leaf 2. root = createNode() 3. root.test_condition = findBestSplit(S,F) 4. V = {v| v a possible outcome of root.test_condition} 5. for each value vєV: a. Sv: = {s | root.test_condition(s) = v and s є S}; b. child = GenDecTree(Sv ,F) ; c. Add child as a descent of root and label the edge (rootchild) as v 6. return root
Tree Induction • Issues • How to Classify a leaf node • Assign the majority class • If leaf is empty, assign the default class – the class that has the highest popularity. • Determine how to split the records • How to specify the attribute test condition? • How to determine the best split? • Determine when to stop splitting
How to Specify Test Condition? • Depends on attribute types • Nominal • Ordinal • Continuous • Depends on number of ways to split • 2-way split • Multi-way split
Splitting Based on Nominal Attributes • Multi-way split: Use as many partitions as distinct values. • 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
• Multi-way split: Use as many partitions as distinct values. • Binary split: Divides values into two subsets – respects the order. Need to find optimal partitioning. • 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}
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
Splitting Based on Continuous Attributes Taxable Income > 80K? Yes No Taxable Income? (i) Binary split (ii) Multi-way split < 10K [10K,25K) [25K,50K) [50K,80K) > 80K
How to determine the Best Split Own Car? C0: 6 C1: 4 C0: 4 C1: 6 C0: 1 C1: 3 C0: 8 C1: 0 C0: 1 C1: 7 Car Type? C0: 1 C1: 0 C0: 1 C1: 0 C0: 0 C1: 1 Student ID? ... Yes No Family Sports Luxury c1 c10 c20 C0: 0 C1: 1 ... c11 Before Splitting: 10 records of class 0, 10 records of class 1 Which test condition is the best?
How to determine the Best Split • Greedy approach: • Nodes with homogeneous class distribution are preferred • Need a measure of node impurity: • Ideas? C0: 5 C1: 5 C0: 9 C1: 1 Non-homogeneous, High degree of impurity Homogeneous, Low degree of impurity
Measuring Node Impurity • p(i|t): fraction of records associated with node t belonging to class i • Used in ID3 and C4.5 • Used in CART, SLIQ, SPRINT.     c i t i p t i p t 1 ) | ( log ) | ( ) ( Entropy       c i t i p t 1 2 ) | ( 1 ) ( Gini   ) | ( max 1 ) ( error tion Classifica t i p t i  
Gain • Gain of an attribute split: compare the impurity of the parent node with the average impurity of the child nodes • Maximizing the gain  Minimizing the weighted average impurity measure of children nodes • If I() = Entropy(), then Δinfo is called information gain      k j j j v I N v N parent I 1 ) ( ) ( ) (
Example C1 0 C2 6 C1 2 C2 4 C1 1 C2 5 P(C1) = 0/6 = 0 P(C2) = 6/6 = 1 Gini = 1 – P(C1)2 – P(C2)2 = 1 – 0 – 1 = 0 Entropy = – 0 log 0 – 1 log 1 = – 0 – 0 = 0 Error = 1 – max (0, 1) = 1 – 1 = 0 P(C1) = 1/6 P(C2) = 5/6 Gini = 1 – (1/6)2 – (5/6)2 = 0.278 Entropy = – (1/6) log2 (1/6) – (5/6) log2 (1/6) = 0.65 Error = 1 – max (1/6, 5/6) = 1 – 5/6 = 1/6 P(C1) = 2/6 P(C2) = 4/6 Gini = 1 – (2/6)2 – (4/6)2 = 0.444 Entropy = – (2/6) log2 (2/6) – (4/6) log2 (4/6) = 0.92 Error = 1 – max (2/6, 4/6) = 1 – 4/6 = 1/3
Impurity measures • All of the impurity measures take value zero (minimum) for the case of a pure node where a single value has probability 1 • All of the impurity measures take maximum value when the class distribution in a node is uniform.
Comparison among Splitting Criteria For a 2-class problem: The different impurity measures are consistent
CategoricalAttributes • For binary values split in two • For multivalued attributes, for each distinct value, gather counts for each class in the dataset • Use the count matrix to make decisions CarType {Sports, Luxury} {Family} C1 3 1 C2 2 4 Gini 0.400 CarType {Sports} {Family, Luxury} C1 2 2 C2 1 5 Gini 0.419 CarType Family Sports Luxury C1 1 2 1 C2 4 1 1 Gini 0.393 Multi-way split Two-way split (find best partition of values)
ContinuousAttributes • Use Binary Decisions based on one value • Choices for the splitting value • Number of possible splitting values = Number of distinct values • Each splitting value has a count matrix associated with it • Class counts in each of the partitions, A < v and A  v • Exhaustive method to choose best v • For each v, scan the database to gather count matrix and compute the impurity index • Computationally Inefficient! Repetition of work. Tid Refund Marital Status Taxable Income Cheat 1 Yes Single 125K No 2 No Married 100K No 3 No Single 70K No 4 Yes Married 120K No 5 No Divorced 95K Yes 6 No Married 60K No 7 Yes Divorced 220K No 8 No Single 85K Yes 9 No Married 75K No 10 No Single 90K Yes 10 Taxable Income > 80K? Yes No
ContinuousAttributes • For efficient computation: for each attribute, • Sort the attribute on values • Linearly scan these values, each time updating the count matrix and computing impurity • Choose the split position that has the least impurity Cheat No No No Yes Yes Yes No No No No Taxable Income 60 70 75 85 90 95 100 120 125 220 55 65 72 80 87 92 97 110 122 172 230 <= > <= > <= > <= > <= > <= > <= > <= > <= > <= > <= > 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 Split Positions Sorted Values
Splitting based on impurity • Impurity measures favor attributes with large number of values • A test condition with large number of outcomes may not be desirable • # of records in each partition is too small to make predictions
Splitting based on INFO
Gain Ratio • Splitting using information gain Parent Node, p is split into k partitions ni is the number of records in partition 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 impurity SplitINFO GAIN GainRATIO Split split      k i i i n n n n SplitINFO 1 log
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)
Decision Tree Based Classification • Advantages: • Inexpensive to construct • Extremely fast at classifying unknown records • Easy to interpret for small-sized trees • Accuracy is comparable to other classification techniques for many simple data sets
Example: C4.5 • Simple depth-first construction. • Uses Information Gain • Sorts Continuous Attributes at each node. • Needs entire data to fit in memory. • Unsuitable for Large Datasets. • Needs out-of-core sorting. • You can download the software from: http://www.cse.unsw.edu.au/~quinlan/c4.5r8.tar.gz
Other Issues • Data Fragmentation • Expressiveness
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 • You can introduce a lower bound on the number of items per leaf node in the stopping criterion.
Expressiveness • A classifier defines a function that discriminates between two (or more) classes. • The expressiveness of a classifier is the class of functions that it can model, and the kind of data that it can separate • When we have discrete (or binary) values, we are interested in the class of boolean functions that can be modeled • If the data-points are real vectors we talk about the decision boundary that the classifier can model
Decision Boundary y < 0.33? : 0 : 3 : 4 : 0 y < 0.47? : 4 : 0 : 0 : 4 x < 0.43? Yes Yes No No Yes No 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 x y • 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 at-a-time
Expressiveness • Decision tree provides expressive representation for learning discrete-valued 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 • Less expressive for modeling continuous variables • Particularly when test condition involves only a single attribute at-a-time
Oblique Decision Trees x + y < 1 Class = + Class = • Test condition may involve multiple attributes • More expressive representation • Finding optimal test condition is computationally expensive
Practical Issues of Classification • Underfitting and Overfitting • Evaluation
Underfitting and Overfitting (Example) 500 circular and 500 triangular data points. Circular points: 0.5  sqrt(x1 2+x2 2)  1 Triangular points: sqrt(x1 2+x2 2) > 0.5 or sqrt(x1 2+x2 2) < 1
Underfitting and Overfitting Overfitting Underfitting: when model is too simple, both training and test errors are large Underfitting Overfitting: when model is too complex it models the details of the training set and fails on the test set
Overfitting due to Noise Decision boundary is distorted by noise point
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
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 • The model does not generalize well • Need new ways for estimating errors
Estimating Generalization Errors • Re-substitution 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) • Penalize large trees • 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% • Using validation set: • Split data into training, validation, test • Use validation dataset to estimate generalization error • Drawback: less data for training.
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
Minimum Description Length (MDL) • Cost(Model,Data) = Cost(Data|Model) + Cost(Model) • Search for the least costly model. • Cost(Data|Model) encodes the misclassification errors. • Cost(Model) encodes the decision tree • node encoding (number of children) plus splitting condition encoding. A B A? B? C? 1 0 0 1 Yes No B1 B2 C1 C2 X y X1 1 X2 0 X3 0 X4 1 … … Xn 1 X y X1 ? X2 ? X3 ? X4 ? … … Xn ?
How to Address Overfitting • Pre-Pruning (Early Stopping Rule) • Stop the algorithm before it becomes a fully-grown 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 user-specified threshold • Stop if class distribution of instances are independent of the available features (e.g., using  2 test) • Stop if expanding the current node does not improve impurity measures (e.g., Gini or information gain).
How to Address Overfitting… • Post-pruning • Grow decision tree to its entirety • Trim the nodes of the decision tree in a bottom-up fashion • If generalization error improves after trimming, replace sub-tree by a leaf node. • Class label of leaf node is determined from majority class of instances in the sub-tree • Can use MDL for post-pruning
Example of Post-Pruning A? A1 A2 A3 A4 Class = Yes 20 Class = No 10 Error = 10/30 Training Error (Before splitting) = 10/30 Pessimistic error = (10 + 0.5)/30 = 10.5/30 Training Error (After splitting) = 9/30 Pessimistic error (After splitting) = (9 + 4  0.5)/30 = 11/30 PRUNE! Class = Yes 8 Class = No 4 Class = Yes 3 Class = No 4 Class = Yes 4 Class = No 1 Class = Yes 5 Class = No 1
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?
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?
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 ACTUAL CLASS Class=Yes Class=No Class=Yes a b Class=No c d a: TP (true positive) b: FN (false negative) c: FP (false positive) d: TN (true negative)
Metrics for Performance Evaluation… • Most widely-used metric: PREDICTED CLASS ACTUAL CLASS Class=Yes Class=No Class=Yes a (TP) b (FN) Class=No c (FP) d (TN) FN FP TN TP TN TP d c b a d a           Accuracy
Limitation of Accuracy • Consider a 2-class 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
Cost Matrix PREDICTED CLASS ACTUAL CLASS C(i|j) Class=Yes Class=No Class=Yes C(Yes|Yes) C(No|Yes) Class=No C(Yes|No) C(No|No) C(i|j): Cost of classifying class j example as class i d w c w b w a w d w a w 4 3 2 1 4 1 Accuracy Weighted     
Computing Cost of Classification Cost Matrix PREDICTED CLASS ACTUAL CLASS C(i|j) + - + -1 100 - 1 0 Model M1 PREDICTED CLASS ACTUAL CLASS + - + 150 40 - 60 250 Model M2 PREDICTED CLASS ACTUAL CLASS + - + 250 45 - 5 200 Accuracy = 80% Cost = 3910 Accuracy = 90% Cost = 4255
Cost vs Accuracy Count PREDICTED CLASS ACTUAL CLASS Class=Yes Class=No Class=Yes a b Class=No c d Cost PREDICTED CLASS ACTUAL CLASS Class=Yes Class=No Class=Yes p q Class=No q p N = a + b + c + d Accuracy = (a + d)/N Cost = p (a + d) + q (b + c) = p (a + d) + q (N – a – d) = q N – (q – p)(a + d) = N [q – (q-p)  Accuracy] Accuracy is proportional to cost if 1. C(Yes|No)=C(No|Yes) = q 2. C(Yes|Yes)=C(No|No) = p
Precision-Recall FN FP TP TP c b a a p r rp p r FN TP TP b a a FP TP TP c a a                         2 2 2 2 2 2 / 1 / 1 1 (F) measure - F (r) Recall (p) Precision  Precision is biased towards C(Yes|Yes) & C(Yes|No)  Recall is biased towards C(Yes|Yes) & C(No|Yes)  F-measure is biased towards all except C(No|No) Count PREDICTED CLASS ACTUAL CLASS Class=Yes Class=No Class=Yes a b Class=No c d
Precision-Recall plot • Usually for parameterized models, it controls the precision/recall tradeoff
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?
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
Methods of Estimation • Holdout • Reserve 2/3 for training and 1/3 for testing • Random subsampling • One sample may be biased -- Repeated holdout • Cross validation • Partition data into k disjoint subsets • k-fold: train on k-1 partitions, test on the remaining one • Leave-one-out: k=n • Guarantees that each record is used the same number of times for training and testing • Bootstrap • Sampling with replacement • ~63% of records used for training, ~27% for testing
Dealing with class Imbalance • If the class we are interested in is very rare, then the classifier will ignore it. • The class imbalance problem • Solution • We can modify the optimization criterion by using a cost sensitive metric • We can balance the class distribution • Sample from the larger class so that the size of the two classes is the same • Replicate the data of the class of interest so that the classes are balanced • Over-fitting issues
Learning Curve  Learning curve shows how accuracy changes with varying sample size  Requires a sampling schedule for creating learning curve Effect of small sample size: - Bias in the estimate - Variance of estimate
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?
ROC (Receiver Operating Characteristic) • Developed in 1950s for signal detection theory to analyze noisy signals • Characterize the trade-off between positive hits and false alarms • ROC curve plots TPR (on the y-axis) against FPR (on the x-axis) FN TP TP TPR   TN FP FP FPR   PREDICTED CLASS Actual Yes No Yes a (TP) b (FN) No c (FP) d (TN) Fraction of positive instances predicted correctly Fraction of negative instances predicted incorrectly
ROC (Receiver Operating Characteristic) • Performance of a classifier represented as a point on the ROC curve • Changing some parameter of the algorithm, sample distribution or cost matrix changes the location of the point
ROC Curve At threshold t: TP=0.5, FN=0.5, FP=0.12, FN=0.88 - 1-dimensional data set containing 2 classes (positive and negative) - any points located at x > t is classified as positive
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 PREDICTED CLASS Actual Yes No Yes a (TP) b (FN) No c (FP) d (TN)
Using ROC for Model Comparison  No model consistently outperform the other  M1 is better for small FPR  M2 is better for large FPR  Area Under the ROC curve (AUC)  Ideal: Area = 1  Random guess:  Area = 0.5
ROC curve vs Precision-Recall curve Area Under the Curve (AUC) as a single number for evaluation

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Data Mining Lecture_10(a).pptx

  • 1.
    DATA MINING LECTURE 10 Classification BasicConcepts Decision Trees Subrata Kumer Paul Assistant Professor, Dept. of CSE, BAUET sksubrata96@gmail.com
  • 2.
    Catching tax-evasion Tid RefundMarital Status Taxable Income Cheat 1 Yes Single 125K No 2 No Married 100K No 3 No Single 70K No 4 Yes Married 120K No 5 No Divorced 95K Yes 6 No Married 60K No 7 Yes Divorced 220K No 8 No Single 85K Yes 9 No Married 75K No 10 No Single 90K Yes 10 Refund Marital Status Taxable Income Cheat No Married 80K ? 10 Tax-return data for year 2011 A new tax return for 2012 Is this a cheating tax return? An instance of the classification problem: learn a method for discriminating between records of different classes (cheaters vs non-cheaters)
  • 3.
    What is classification? •Classification is the task of learning a target function f that maps attribute set x to one of the predefined class labels y Tid Refund Marital Status Taxable Income Cheat 1 Yes Single 125K No 2 No Married 100K No 3 No Single 70K No 4 Yes Married 120K No 5 No Divorced 95K Yes 6 No Married 60K No 7 Yes Divorced 220K No 8 No Single 85K Yes 9 No Married 75K No 10 No Single 90K Yes 10 One of the attributes is the class attribute In this case: Cheat Two class labels (or classes): Yes (1), No (0)
  • 4.
    Why classification? • Thetarget function f is known as a classification model • Descriptive modeling: Explanatory tool to distinguish between objects of different classes (e.g., understand why people cheat on their taxes) • Predictive modeling: Predict a class of a previously unseen record
  • 5.
    Examples of ClassificationTasks • Predicting tumor cells as benign or malignant • Classifying credit card transactions as legitimate or fraudulent • Categorizing news stories as finance, weather, entertainment, sports, etc • Identifying spam email, spam web pages, adult content • Understanding if a web query has commercial intent or not
  • 6.
    General approach toclassification • Training set consists of records with known class labels • Training set is used to build a classification model • A labeled test set of previously unseen data records is used to evaluate the quality of the model. • The classification model is applied to new records with unknown class labels
  • 7.
    Illustrating Classification Task Apply Model Induction Deduction Learn Model Model TidAttrib1 Attrib2 Attrib3 Class 1 Yes Large 125K No 2 No Medium 100K No 3 No Small 70K No 4 Yes Medium 120K No 5 No Large 95K Yes 6 No Medium 60K No 7 Yes Large 220K No 8 No Small 85K Yes 9 No Medium 75K No 10 No Small 90K Yes 10 Tid Attrib1 Attrib2 Attrib3 Class 11 No Small 55K ? 12 Yes Medium 80K ? 13 Yes Large 110K ? 14 No Small 95K ? 15 No Large 67K ? 10 Test Set Learning algorithm Training Set
  • 8.
    Evaluation of classificationmodels • Counts of test records that are correctly (or incorrectly) predicted by the classification model • Confusion matrix Class = 1 Class = 0 Class = 1 f11 f10 Class = 0 f01 f00 Predicted Class Actual Class 00 01 10 11 00 11 s prediction of # total s prediction correct # Accuracy f f f f f f       00 01 10 11 01 10 s prediction of # total s prediction wrong # rate Error f f f f f f      
  • 9.
    Classification Techniques • DecisionTree based Methods • Rule-based Methods • Memory based reasoning • Neural Networks • Naïve Bayes and Bayesian Belief Networks • Support Vector Machines
  • 10.
    Classification Techniques • DecisionTree based Methods • Rule-based Methods • Memory based reasoning • Neural Networks • Naïve Bayes and Bayesian Belief Networks • Support Vector Machines
  • 11.
    Decision Trees • Decisiontree • A flow-chart-like tree structure • Internal node denotes a test on an attribute • Branch represents an outcome of the test • Leaf nodes represent class labels or class distribution
  • 12.
    Example of aDecision Tree Tid Refund Marital Status Taxable Income Cheat 1 Yes Single 125K No 2 No Married 100K No 3 No Single 70K No 4 Yes Married 120K No 5 No Divorced 95K Yes 6 No Married 60K No 7 Yes Divorced 220K No 8 No Single 85K Yes 9 No Married 75K No 10 No Single 90K Yes 10 Refund MarSt TaxInc YES NO NO NO Yes No Married Single, Divorced < 80K > 80K Splitting Attributes Training Data Model: Decision Tree Test outcome Class labels
  • 13.
    Another Example ofDecision Tree Tid Refund Marital Status Taxable Income Cheat 1 Yes Single 125K No 2 No Married 100K No 3 No Single 70K No 4 Yes Married 120K No 5 No Divorced 95K Yes 6 No Married 60K No 7 Yes Divorced 220K No 8 No Single 85K Yes 9 No Married 75K No 10 No Single 90K Yes 10 MarSt Refund TaxInc YES NO NO NO Yes No Married Single, Divorced < 80K > 80K There could be more than one tree that fits the same data!
  • 14.
    Decision Tree ClassificationTask Apply Model Induction Deduction Learn Model Model Tid Attrib1 Attrib2 Attrib3 Class 1 Yes Large 125K No 2 No Medium 100K No 3 No Small 70K No 4 Yes Medium 120K No 5 No Large 95K Yes 6 No Medium 60K No 7 Yes Large 220K No 8 No Small 85K Yes 9 No Medium 75K No 10 No Small 90K Yes 10 Tid Attrib1 Attrib2 Attrib3 Class 11 No Small 55K ? 12 Yes Medium 80K ? 13 Yes Large 110K ? 14 No Small 95K ? 15 No Large 67K ? 10 Test Set Tree Induction algorithm Training Set Decision Tree
  • 15.
    Apply Model toTest Data Refund MarSt TaxInc YES NO NO NO Yes No Married Single, Divorced < 80K > 80K Refund Marital Status Taxable Income Cheat No Married 80K ? 10 Test Data Start from the root of tree.
  • 16.
    Apply Model toTest Data Refund MarSt TaxInc YES NO NO NO Yes No Married Single, Divorced < 80K > 80K Refund Marital Status Taxable Income Cheat No Married 80K ? 10 Test Data
  • 17.
    Apply Model toTest Data Refund MarSt TaxInc YES NO NO NO Yes No Married Single, Divorced < 80K > 80K Refund Marital Status Taxable Income Cheat No Married 80K ? 10 Test Data
  • 18.
    Apply Model toTest Data Refund MarSt TaxInc YES NO NO NO Yes No Married Single, Divorced < 80K > 80K Refund Marital Status Taxable Income Cheat No Married 80K ? 10 Test Data
  • 19.
    Apply Model toTest Data Refund MarSt TaxInc YES NO NO NO Yes No Married Single, Divorced < 80K > 80K Refund Marital Status Taxable Income Cheat No Married 80K ? 10 Test Data
  • 20.
    Apply Model toTest Data Refund MarSt TaxInc YES NO NO NO Yes No Married Single, Divorced < 80K > 80K Refund Marital Status Taxable Income Cheat No Married 80K ? 10 Test Data Assign Cheat to “No”
  • 21.
    Decision Tree ClassificationTask Apply Model Induction Deduction Learn Model Model Tid Attrib1 Attrib2 Attrib3 Class 1 Yes Large 125K No 2 No Medium 100K No 3 No Small 70K No 4 Yes Medium 120K No 5 No Large 95K Yes 6 No Medium 60K No 7 Yes Large 220K No 8 No Small 85K Yes 9 No Medium 75K No 10 No Small 90K Yes 10 Tid Attrib1 Attrib2 Attrib3 Class 11 No Small 55K ? 12 Yes Medium 80K ? 13 Yes Large 110K ? 14 No Small 95K ? 15 No Large 67K ? 10 Test Set Tree Induction algorithm Training Set Decision Tree
  • 22.
    Tree Induction • Findingthe best decision tree is NP-hard • Greedy strategy. • Split the records based on an attribute test that optimizes certain criterion. • Many Algorithms: • Hunt’s Algorithm (one of the earliest) • CART • ID3, C4.5 • SLIQ,SPRINT
  • 23.
    General Structure ofHunt’s Algorithm • Let Dt be the set of training records that reach a node t • General Procedure: • If Dt contains records that belong the same class yt, then t is a leaf node labeled as yt • If Dt contains records with the same attribute values, then t is a leaf node labeled with the majority class yt • If Dt is an empty set, then t is a leaf node labeled by the default class, yd • If Dt 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. Tid Refund Marital Status Taxable Income Cheat 1 Yes Single 125K No 2 No Married 100K No 3 No Single 70K No 4 Yes Married 120K No 5 No Divorced 95K Yes 6 No Married 60K No 7 Yes Divorced 220K No 8 No Single 85K Yes 9 No Married 75K No 10 No Single 90K Yes 10 Dt ?
  • 24.
    Hunt’s Algorithm Don’t Cheat Refund Don’t Cheat Don’t Cheat Yes No Refund Don’t Cheat YesNo Marital Status Don’t Cheat Cheat Single, Divorced Married Taxable Income Don’t Cheat < 80K >= 80K Refund Don’t Cheat Yes No Marital Status Don’t Cheat Cheat Single, Divorced Married Tid Refund Marital Status Taxable Income Cheat 1 Yes Single 125K No 2 No Married 100K No 3 No Single 70K No 4 Yes Married 120K No 5 No Divorced 95K Yes 6 No Married 60K No 7 Yes Divorced 220K No 8 No Single 85K Yes 9 No Married 75K No 10 No Single 90K Yes 10 Tid Refund Marital Status Taxable Income Cheat 1 Yes Single 125K No 4 Yes Married 120K No 7 Yes Divorced 220K No 2 No Married 100K No 3 No Single 70K No 5 No Divorced 95K Yes 6 No Married 60K No 8 No Single 85K Yes 9 No Married 75K No 10 No Single 90K Yes 10 Tid Refund Marital Status Taxable Income Cheat 1 Yes Single 125K No 4 Yes Married 120K No 7 Yes Divorced 220K No 2 No Married 100K No 6 No Married 60K No 9 No Married 75K No 3 No Single 70K No 5 No Divorced 95K Yes 8 No Single 85K Yes 10 No Single 90K Yes 10
  • 25.
    Constructing decision-trees (pseudocode) GenDecTree(SampleS, Features F) 1. If stopping_condition(S,F) = true then a. leaf = createNode() b. leaf.label= Classify(S) c. return leaf 2. root = createNode() 3. root.test_condition = findBestSplit(S,F) 4. V = {v| v a possible outcome of root.test_condition} 5. for each value vєV: a. Sv: = {s | root.test_condition(s) = v and s є S}; b. child = GenDecTree(Sv ,F) ; c. Add child as a descent of root and label the edge (rootchild) as v 6. return root
  • 26.
    Tree Induction • Issues •How to Classify a leaf node • Assign the majority class • If leaf is empty, assign the default class – the class that has the highest popularity. • Determine how to split the records • How to specify the attribute test condition? • How to determine the best split? • Determine when to stop splitting
  • 27.
    How to SpecifyTest Condition? • Depends on attribute types • Nominal • Ordinal • Continuous • Depends on number of ways to split • 2-way split • Multi-way split
  • 28.
    Splitting Based onNominal Attributes • Multi-way split: Use as many partitions as distinct values. • 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
  • 29.
    • Multi-way split:Use as many partitions as distinct values. • Binary split: Divides values into two subsets – respects the order. Need to find optimal partitioning. • 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}
  • 30.
    Splitting Based onContinuous 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
  • 31.
    Splitting Based onContinuous Attributes Taxable Income > 80K? Yes No Taxable Income? (i) Binary split (ii) Multi-way split < 10K [10K,25K) [25K,50K) [50K,80K) > 80K
  • 32.
    How to determinethe Best Split Own Car? C0: 6 C1: 4 C0: 4 C1: 6 C0: 1 C1: 3 C0: 8 C1: 0 C0: 1 C1: 7 Car Type? C0: 1 C1: 0 C0: 1 C1: 0 C0: 0 C1: 1 Student ID? ... Yes No Family Sports Luxury c1 c10 c20 C0: 0 C1: 1 ... c11 Before Splitting: 10 records of class 0, 10 records of class 1 Which test condition is the best?
  • 33.
    How to determinethe Best Split • Greedy approach: • Nodes with homogeneous class distribution are preferred • Need a measure of node impurity: • Ideas? C0: 5 C1: 5 C0: 9 C1: 1 Non-homogeneous, High degree of impurity Homogeneous, Low degree of impurity
  • 34.
    Measuring Node Impurity •p(i|t): fraction of records associated with node t belonging to class i • Used in ID3 and C4.5 • Used in CART, SLIQ, SPRINT.     c i t i p t i p t 1 ) | ( log ) | ( ) ( Entropy       c i t i p t 1 2 ) | ( 1 ) ( Gini   ) | ( max 1 ) ( error tion Classifica t i p t i  
  • 35.
    Gain • Gain ofan attribute split: compare the impurity of the parent node with the average impurity of the child nodes • Maximizing the gain  Minimizing the weighted average impurity measure of children nodes • If I() = Entropy(), then Δinfo is called information gain      k j j j v I N v N parent I 1 ) ( ) ( ) (
  • 36.
    Example C1 0 C2 6 C12 C2 4 C1 1 C2 5 P(C1) = 0/6 = 0 P(C2) = 6/6 = 1 Gini = 1 – P(C1)2 – P(C2)2 = 1 – 0 – 1 = 0 Entropy = – 0 log 0 – 1 log 1 = – 0 – 0 = 0 Error = 1 – max (0, 1) = 1 – 1 = 0 P(C1) = 1/6 P(C2) = 5/6 Gini = 1 – (1/6)2 – (5/6)2 = 0.278 Entropy = – (1/6) log2 (1/6) – (5/6) log2 (1/6) = 0.65 Error = 1 – max (1/6, 5/6) = 1 – 5/6 = 1/6 P(C1) = 2/6 P(C2) = 4/6 Gini = 1 – (2/6)2 – (4/6)2 = 0.444 Entropy = – (2/6) log2 (2/6) – (4/6) log2 (4/6) = 0.92 Error = 1 – max (2/6, 4/6) = 1 – 4/6 = 1/3
  • 37.
    Impurity measures • Allof the impurity measures take value zero (minimum) for the case of a pure node where a single value has probability 1 • All of the impurity measures take maximum value when the class distribution in a node is uniform.
  • 38.
    Comparison among SplittingCriteria For a 2-class problem: The different impurity measures are consistent
  • 39.
    CategoricalAttributes • For binaryvalues split in two • For multivalued attributes, for each distinct value, gather counts for each class in the dataset • Use the count matrix to make decisions CarType {Sports, Luxury} {Family} C1 3 1 C2 2 4 Gini 0.400 CarType {Sports} {Family, Luxury} C1 2 2 C2 1 5 Gini 0.419 CarType Family Sports Luxury C1 1 2 1 C2 4 1 1 Gini 0.393 Multi-way split Two-way split (find best partition of values)
  • 40.
    ContinuousAttributes • Use BinaryDecisions based on one value • Choices for the splitting value • Number of possible splitting values = Number of distinct values • Each splitting value has a count matrix associated with it • Class counts in each of the partitions, A < v and A  v • Exhaustive method to choose best v • For each v, scan the database to gather count matrix and compute the impurity index • Computationally Inefficient! Repetition of work. Tid Refund Marital Status Taxable Income Cheat 1 Yes Single 125K No 2 No Married 100K No 3 No Single 70K No 4 Yes Married 120K No 5 No Divorced 95K Yes 6 No Married 60K No 7 Yes Divorced 220K No 8 No Single 85K Yes 9 No Married 75K No 10 No Single 90K Yes 10 Taxable Income > 80K? Yes No
  • 41.
    ContinuousAttributes • For efficientcomputation: for each attribute, • Sort the attribute on values • Linearly scan these values, each time updating the count matrix and computing impurity • Choose the split position that has the least impurity Cheat No No No Yes Yes Yes No No No No Taxable Income 60 70 75 85 90 95 100 120 125 220 55 65 72 80 87 92 97 110 122 172 230 <= > <= > <= > <= > <= > <= > <= > <= > <= > <= > <= > 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 Split Positions Sorted Values
  • 42.
    Splitting based onimpurity • Impurity measures favor attributes with large number of values • A test condition with large number of outcomes may not be desirable • # of records in each partition is too small to make predictions
  • 43.
  • 44.
    Gain Ratio • Splittingusing information gain Parent Node, p is split into k partitions ni is the number of records in partition 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 impurity SplitINFO GAIN GainRATIO Split split      k i i i n n n n SplitINFO 1 log
  • 45.
    Stopping Criteria forTree 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)
  • 46.
    Decision Tree BasedClassification • Advantages: • Inexpensive to construct • Extremely fast at classifying unknown records • Easy to interpret for small-sized trees • Accuracy is comparable to other classification techniques for many simple data sets
  • 47.
    Example: C4.5 • Simpledepth-first construction. • Uses Information Gain • Sorts Continuous Attributes at each node. • Needs entire data to fit in memory. • Unsuitable for Large Datasets. • Needs out-of-core sorting. • You can download the software from: http://www.cse.unsw.edu.au/~quinlan/c4.5r8.tar.gz
  • 48.
    Other Issues • DataFragmentation • Expressiveness
  • 49.
    Data Fragmentation • Numberof 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 • You can introduce a lower bound on the number of items per leaf node in the stopping criterion.
  • 50.
    Expressiveness • A classifierdefines a function that discriminates between two (or more) classes. • The expressiveness of a classifier is the class of functions that it can model, and the kind of data that it can separate • When we have discrete (or binary) values, we are interested in the class of boolean functions that can be modeled • If the data-points are real vectors we talk about the decision boundary that the classifier can model
  • 51.
    Decision Boundary y <0.33? : 0 : 3 : 4 : 0 y < 0.47? : 4 : 0 : 0 : 4 x < 0.43? Yes Yes No No Yes No 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 x y • 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 at-a-time
  • 52.
    Expressiveness • Decision treeprovides expressive representation for learning discrete-valued 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 • Less expressive for modeling continuous variables • Particularly when test condition involves only a single attribute at-a-time
  • 53.
    Oblique Decision Trees x+ y < 1 Class = + Class = • Test condition may involve multiple attributes • More expressive representation • Finding optimal test condition is computationally expensive
  • 54.
    Practical Issues ofClassification • Underfitting and Overfitting • Evaluation
  • 55.
    Underfitting and Overfitting(Example) 500 circular and 500 triangular data points. Circular points: 0.5  sqrt(x1 2+x2 2)  1 Triangular points: sqrt(x1 2+x2 2) > 0.5 or sqrt(x1 2+x2 2) < 1
  • 56.
    Underfitting and Overfitting Overfitting Underfitting:when model is too simple, both training and test errors are large Underfitting Overfitting: when model is too complex it models the details of the training set and fails on the test set
  • 57.
    Overfitting due toNoise Decision boundary is distorted by noise point
  • 58.
    Overfitting due toInsufficient 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
  • 59.
    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 • The model does not generalize well • Need new ways for estimating errors
  • 60.
    Estimating Generalization Errors •Re-substitution 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) • Penalize large trees • 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% • Using validation set: • Split data into training, validation, test • Use validation dataset to estimate generalization error • Drawback: less data for training.
  • 61.
    Occam’s Razor • Giventwo 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
  • 62.
    Minimum Description Length(MDL) • Cost(Model,Data) = Cost(Data|Model) + Cost(Model) • Search for the least costly model. • Cost(Data|Model) encodes the misclassification errors. • Cost(Model) encodes the decision tree • node encoding (number of children) plus splitting condition encoding. A B A? B? C? 1 0 0 1 Yes No B1 B2 C1 C2 X y X1 1 X2 0 X3 0 X4 1 … … Xn 1 X y X1 ? X2 ? X3 ? X4 ? … … Xn ?
  • 63.
    How to AddressOverfitting • Pre-Pruning (Early Stopping Rule) • Stop the algorithm before it becomes a fully-grown 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 user-specified threshold • Stop if class distribution of instances are independent of the available features (e.g., using  2 test) • Stop if expanding the current node does not improve impurity measures (e.g., Gini or information gain).
  • 64.
    How to AddressOverfitting… • Post-pruning • Grow decision tree to its entirety • Trim the nodes of the decision tree in a bottom-up fashion • If generalization error improves after trimming, replace sub-tree by a leaf node. • Class label of leaf node is determined from majority class of instances in the sub-tree • Can use MDL for post-pruning
  • 65.
    Example of Post-Pruning A? A1 A2A3 A4 Class = Yes 20 Class = No 10 Error = 10/30 Training Error (Before splitting) = 10/30 Pessimistic error = (10 + 0.5)/30 = 10.5/30 Training Error (After splitting) = 9/30 Pessimistic error (After splitting) = (9 + 4  0.5)/30 = 11/30 PRUNE! Class = Yes 8 Class = No 4 Class = Yes 3 Class = No 4 Class = Yes 4 Class = No 1 Class = Yes 5 Class = No 1
  • 66.
    Model Evaluation • Metricsfor 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?
  • 67.
    Model Evaluation • Metricsfor 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?
  • 68.
    Metrics for PerformanceEvaluation • 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 ACTUAL CLASS Class=Yes Class=No Class=Yes a b Class=No c d a: TP (true positive) b: FN (false negative) c: FP (false positive) d: TN (true negative)
  • 69.
    Metrics for PerformanceEvaluation… • Most widely-used metric: PREDICTED CLASS ACTUAL CLASS Class=Yes Class=No Class=Yes a (TP) b (FN) Class=No c (FP) d (TN) FN FP TN TP TN TP d c b a d a           Accuracy
  • 70.
    Limitation of Accuracy •Consider a 2-class 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
  • 71.
    Cost Matrix PREDICTED CLASS ACTUAL CLASS C(i|j)Class=Yes Class=No Class=Yes C(Yes|Yes) C(No|Yes) Class=No C(Yes|No) C(No|No) C(i|j): Cost of classifying class j example as class i d w c w b w a w d w a w 4 3 2 1 4 1 Accuracy Weighted     
  • 72.
    Computing Cost ofClassification Cost Matrix PREDICTED CLASS ACTUAL CLASS C(i|j) + - + -1 100 - 1 0 Model M1 PREDICTED CLASS ACTUAL CLASS + - + 150 40 - 60 250 Model M2 PREDICTED CLASS ACTUAL CLASS + - + 250 45 - 5 200 Accuracy = 80% Cost = 3910 Accuracy = 90% Cost = 4255
  • 73.
    Cost vs Accuracy CountPREDICTED CLASS ACTUAL CLASS Class=Yes Class=No Class=Yes a b Class=No c d Cost PREDICTED CLASS ACTUAL CLASS Class=Yes Class=No Class=Yes p q Class=No q p N = a + b + c + d Accuracy = (a + d)/N Cost = p (a + d) + q (b + c) = p (a + d) + q (N – a – d) = q N – (q – p)(a + d) = N [q – (q-p)  Accuracy] Accuracy is proportional to cost if 1. C(Yes|No)=C(No|Yes) = q 2. C(Yes|Yes)=C(No|No) = p
  • 74.
    Precision-Recall FN FP TP TP c b a a p r rp p r FN TP TP b a a FP TP TP c a a                         2 2 2 2 2 2 / 1 / 1 1 (F) measure - F (r) Recall (p) Precision  Precisionis biased towards C(Yes|Yes) & C(Yes|No)  Recall is biased towards C(Yes|Yes) & C(No|Yes)  F-measure is biased towards all except C(No|No) Count PREDICTED CLASS ACTUAL CLASS Class=Yes Class=No Class=Yes a b Class=No c d
  • 75.
    Precision-Recall plot • Usuallyfor parameterized models, it controls the precision/recall tradeoff
  • 76.
    Model Evaluation • Metricsfor 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?
  • 77.
    Methods for PerformanceEvaluation • 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
  • 78.
    Methods of Estimation •Holdout • Reserve 2/3 for training and 1/3 for testing • Random subsampling • One sample may be biased -- Repeated holdout • Cross validation • Partition data into k disjoint subsets • k-fold: train on k-1 partitions, test on the remaining one • Leave-one-out: k=n • Guarantees that each record is used the same number of times for training and testing • Bootstrap • Sampling with replacement • ~63% of records used for training, ~27% for testing
  • 79.
    Dealing with classImbalance • If the class we are interested in is very rare, then the classifier will ignore it. • The class imbalance problem • Solution • We can modify the optimization criterion by using a cost sensitive metric • We can balance the class distribution • Sample from the larger class so that the size of the two classes is the same • Replicate the data of the class of interest so that the classes are balanced • Over-fitting issues
  • 80.
    Learning Curve  Learningcurve shows how accuracy changes with varying sample size  Requires a sampling schedule for creating learning curve Effect of small sample size: - Bias in the estimate - Variance of estimate
  • 81.
    Model Evaluation • Metricsfor 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?
  • 82.
    ROC (Receiver OperatingCharacteristic) • Developed in 1950s for signal detection theory to analyze noisy signals • Characterize the trade-off between positive hits and false alarms • ROC curve plots TPR (on the y-axis) against FPR (on the x-axis) FN TP TP TPR   TN FP FP FPR   PREDICTED CLASS Actual Yes No Yes a (TP) b (FN) No c (FP) d (TN) Fraction of positive instances predicted correctly Fraction of negative instances predicted incorrectly
  • 83.
    ROC (Receiver OperatingCharacteristic) • Performance of a classifier represented as a point on the ROC curve • Changing some parameter of the algorithm, sample distribution or cost matrix changes the location of the point
  • 84.
    ROC Curve At thresholdt: TP=0.5, FN=0.5, FP=0.12, FN=0.88 - 1-dimensional data set containing 2 classes (positive and negative) - any points located at x > t is classified as positive
  • 85.
    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 PREDICTED CLASS Actual Yes No Yes a (TP) b (FN) No c (FP) d (TN)
  • 86.
    Using ROC forModel Comparison  No model consistently outperform the other  M1 is better for small FPR  M2 is better for large FPR  Area Under the ROC curve (AUC)  Ideal: Area = 1  Random guess:  Area = 0.5
  • 87.
    ROC curve vsPrecision-Recall curve Area Under the Curve (AUC) as a single number for evaluation