Data Mining Classification: Basic Concepts, Decision Trees, and Model Evaluation Lecture Notes for Chapter 4 Introduction to Data Mining by Tan, Steinbach, Kumar

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Introduction to Data Mining

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Classification: Definition
Given a collection of records (training set )
– Each record contains a set of attributes, one of the attributes is the class.

Find a model for class attribute as a function of the values of other attributes. Goal: previously unseen records should be assigned a class as accurately as possible.
– A test set is used to determine the accuracy of the model. Usually, the given data set is divided into training and test sets, with training set used to build the model and test set used to validate it.

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Illustrating Classification Task
Tid 1 2 3 4 5 6 7 8 9 10
10

Attrib1 Yes No No Yes No No Yes No No No

Attrib2 Large Medium Small Medium Large Medium Large Small Medium Small

Attrib3 125K 100K 70K 120K 95K 60K 220K 85K 75K 90K

Class No No No No Yes No No Yes No Yes

Learning algorithm Induction
Learn Model

Training Set
Tid 11 12 13 14 15
10

Model
Apply Model

Attrib1 No Yes Yes No No

Attrib2 Small Medium Large Small Large

Attrib3 55K 80K 110K 95K 67K

Class ? ? ? ? ?

Deduction

Test Set
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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 alpha-helix, beta-sheet, or random coil Categorizing news stories as finance, weather, entertainment, sports, etc
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Classification Techniques
Decision Tree based Methods Rule-based Methods Memory based reasoning Neural Networks Naïve Bayes and Bayesian Belief Networks Support Vector Machines

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Example of a Decision Tree al al us ric ric uo s in go go nt te te as cl ca ca co
Tid Refund Marital Status 1 2 3 4 5 6 7 8 9 10
10

Taxable Income Cheat 125K 100K 70K 120K No No No No Yes No No Yes No Yes

Splitting Attributes

Yes No No Yes No No Yes No No No

Single Married Single Married

Refund Yes NO No MarSt Single, Divorced TaxInc < 80K NO > 80K YES Married NO

Divorced 95K Married 60K

Divorced 220K Single Married Single 85K 75K 90K

Training Data
© Tan,Steinbach, Kumar Introduction to Data Mining

Model: Decision Tree
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Another Example of Decision Tree ric go al ric go al n co uo ti n us s as cl

te ca

te ca

Tid Refund Marital Status 1 2 3 4 5 6 7 8 9 10
10

Taxable Income Cheat 125K 100K 70K 120K No No No No Yes No No Yes No Yes

Married NO

MarSt

Single, Divorced Refund

Yes No No Yes No No Yes No No No

Single Married Single Married

Yes NO < 80K NO

No TaxInc > 80K YES

Divorced 95K Married 60K

Divorced 220K Single Married Single 85K 75K 90K

There could be more than one tree that fits the same data!

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Decision Tree Classification Task
Tid 1 2 3 4 5 6 7 8 9 10
10

Attrib1 Yes No No Yes No No Yes No No No

Attrib2 Large Medium Small Medium Large Medium Large Small Medium Small

Attrib3 125K 100K 70K 120K 95K 60K 220K 85K 75K 90K

Class No No No No Yes No No Yes No Yes

Tree Induction algorithm Induction
Learn Model

Training Set
Tid 11 12 13 14 15
10

Model
Apply Model
Decision Tree

Attrib1 No Yes Yes No No

Attrib2 Small Medium Large Small Large

Attrib3 55K 80K 110K 95K 67K

Class ? ? ? ? ?

Deduction

Test Set
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Apply Model to Test Data
Test Data Start from the root of tree.
Refund Marital Status No
1 0

Taxable Income Cheat 80K ?

Refund Yes NO No MarSt Single, Divorced TaxInc < 80K NO > 80K YES Married NO

Married

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Apply Model to Test Data
Test Data
Refund Marital Status Taxable Income Cheat 80K ?

Refund Yes NO No MarSt Single, Divorced TaxInc < 80K NO > 80K YES Married NO

No
1 0

Married

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Apply Model to Test Data
Test Data
Refund Marital Status Taxable Income Cheat 80K ?

Refund Yes NO No MarSt Single, Divorced TaxInc < 80K NO > 80K YES Married NO

No
1 0

Married

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Apply Model to Test Data
Test Data
Refund Marital Status Taxable Income Cheat 80K ?

Refund Yes NO No MarSt Single, Divorced TaxInc < 80K NO > 80K YES Married NO

No
1 0

Married

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Apply Model to Test Data
Test Data
Refund Marital Status Taxable Income Cheat 80K ?

Refund Yes NO No MarSt Single, Divorced TaxInc < 80K NO > 80K YES Married NO

No
1 0

Married

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Apply Model to Test Data
Test Data
Refund Marital Status Taxable Income Cheat 80K ?

Refund Yes NO No MarSt Single, Divorced TaxInc < 80K NO > 80K YES Married NO

No
1 0

Married

Assign Cheat to “No”

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Decision Tree Classification Task
Tid 1 2 3 4 5 6 7 8 9 10
10

Attrib1 Yes No No Yes No No Yes No No No

Attrib2 Large Medium Small Medium Large Medium Large Small Medium Small

Attrib3 125K 100K 70K 120K 95K 60K 220K 85K 75K 90K

Class No No No No Yes No No Yes No Yes

Tree Induction algorithm Induction
Learn Model

Training Set
Tid 11 12 13 14 15
10

Model
Apply Model
Decision Tree

Attrib1 No Yes Yes No No

Attrib2 Small Medium Large Small Large

Attrib3 55K 80K 110K 95K 67K

Class ? ? ? ? ?

Deduction

Test Set
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Decision Tree Induction
Many Algorithms: – Hunt’s Algorithm (one of the earliest) – CART – ID3, C4.5 – SLIQ,SPRINT

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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 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 1 2 3 4 5 6 7 8 9 10
10

Taxable Income Cheat 125K 100K 70K 120K No No No No Yes No No Yes No Yes

Yes No No Yes No No Yes No No No

Single Married Single Married

Divorced 95K Married 60K

Divorced 220K Single Married Single 85K 75K 90K

Dt

?

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Hunt’s Algorithm
Don’t Cheat

Tid Refund Marital Status 1 Yes No No Yes No No Yes No No No Single Married Single Married

Taxable Income Cheat 125K 100K 70K 120K No No No No Yes No No Yes No Yes

Refund
Yes Don’t Cheat No Don’t Cheat

2 3 4 5 6 7 8

Divorced 95K Married 60K

Divorced 220K Single Married Single 85K 75K 90K

Refund
Yes Don’t Cheat Single, Divorced No Yes Don’t Cheat Married Don’t Cheat < 80K Don’t Cheat
© Tan,Steinbach, Kumar

Refund
No
10

9 10

Marital Status

Single, Divorced

Marital Status
Married Don’t Cheat >= 80K

Cheat

Taxable Income

Cheat
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Introduction to Data Mining

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

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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

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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

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Splitting Based on Nominal Attributes
Multi-way split: Use as many partitions as distinct values.
CarType
Family Sports Luxury

Binary split: Divides values into two subsets. Need to find optimal partitioning.
{Sports, Luxury}

CarType
{Family}

OR

{Family, Luxury}

CarType
{Sports}

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Splitting Based on Ordinal Attributes
Multi-way split: Use as many partitions as distinct values.
Size
Small Medium Large

Binary split: Divides values into two subsets. Need to find optimal partitioning.
{Small, Medium}

Size
{Large}

OR

{Medium, Large}

Size
{Small}

What about this split?
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{Small, Large}

Size
{Medium}
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Introduction to Data Mining

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
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Splitting Based on Continuous Attributes

Taxable Income > 80K?
< 10K Yes No [10K,25K)

Taxable Income?
> 80K

[25K,50K)

[50K,80K)

(i) Binary split

(ii) Multi-way split

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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

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How to determine the Best Split
Before Splitting: 10 records of class 0, 10 records of class 1
Own Car? Yes
C0: 6 C1: 4

Car Type? No Family Sports Luxury c1
C0: 1 C1: 0

Student ID? c10
C0: 1 C1: 0

c11
C0: 0 C1: 1

c20

C0: 4 C1: 6

C0: 1 C1: 3

C0: 8 C1: 0

C0: 1 C1: 7

...

...

C0: 0 C1: 1

Which test condition is the best?

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How to determine the Best Split
Greedy approach: – Nodes with homogeneous class distribution are preferred Need a measure of node impurity:
C0: 5 C1: 5
Non-homogeneous, High degree of impurity

C0: 9 C1: 1
Homogeneous, Low degree of impurity

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Measures of Node Impurity
Gini Index Entropy Misclassification error

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How to Find the Best Split
Before Splitting:
C0 C1 N00 N01

M0 B?

A?
Yes Node N1
C0 C1 N10 N11

No Node N2
C0 C1 N20 N21

Yes Node N3
C0 C1 N30 N31

No Node N4
C0 C1 N40 N41

M1 M12

M2

M3 M34

M4

Gain = M0 – M12 vs M0 – M34
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Measure of Impurity: GINI
Gini Index for a given node t :

GINI (t ) = 1 − ∑ [ p ( j | t )]2 j (NOTE: p( j | t) is the relative frequency of class j at node t).

– Maximum (1 - 1/nc) when records are equally distributed among all classes, implying least interesting information – Minimum (0.0) when all records belong to one class, implying most interesting information
C1 C2 0 6 C1 C2 1 5 C1 C2 2 4 C1 C2 3 3

Gini=0.000

Gini=0.278

Gini=0.444

Gini=0.500

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Examples for computing GINI
GINI (t ) = 1 − ∑ [ p ( j | t )]2 j C1 C2

0 6

P(C1) = 0/6 = 0

P(C2) = 6/6 = 1

Gini = 1 – P(C1)2 – P(C2)2 = 1 – 0 – 1 = 0

C1 C2

1 5

P(C1) = 1/6

P(C2) = 5/6

Gini = 1 – (1/6)2 – (5/6)2 = 0.278 P(C1) = 2/6 P(C2) = 4/6

C1 C2
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2 4

Gini = 1 – (2/6)2 – (4/6)2 = 0.444
Introduction to Data Mining 4/18/2004 32

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,

GINI split = ∑ i =1

k

ni GINI (i ) n

where,

ni = number of records at child i, n = number of records at node p.

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Binary Attributes: Computing GINI Index
Splits into two partitions Effect of Weighing partitions: – Larger and Purer Partitions are sought for.
Parent

B?
Yes No Node N2

C1 C2

6 6

Gini = 0.500

Gini(N1) = 1 – (5/6)2 – (2/6)2 = 0.194 Gini(N2) = 1 – (1/6)2 – (4/6)2 = 0.528
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Node N1

C1 C2

N1 5 2

N2 1 4

Gini=0.333
Introduction to Data Mining

Gini(Children) = 7/12 * 0.194 + 5/12 * 0.528 = 0.333
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Categorical Attributes: Computing Gini Index
For each distinct value, gather counts for each class in the dataset Use the count matrix to make decisions
Multi-way split Two-way split (find best partition of values)
CarType {Sports, {Family} Luxury} 3 1 2 4 0.400 CarType {Family, {Sports} Luxury} 2 2 1 5 0.419

CarType C1 C2 Gini Family Sports Luxury 1 2 1 4 1 1 0.393
C1 C2 Gini

C1 C2 Gini

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Continuous Attributes: Computing Gini Index
Use Binary Decisions based on one value Several 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 Simple method to choose best v – For each v, scan the database to gather count matrix and compute its Gini index – Computationally Inefficient! Repetition of work.
© Tan,Steinbach, Kumar Introduction to Data Mining
Tid Refund Marital Status 1 2 3 4 5 6 7 8 9 10
10

Taxable Income Cheat 125K 100K 70K 120K No No No No Yes No No Yes No Yes

Yes No No Yes No No Yes No No No

Single Married Single Married

Divorced 95K Married 60K

Divorced 220K Single Married Single 85K 75K 90K

Taxable Income > 80K?
Yes No

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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

Sorted Values Split Positions
Yes No Gini 3 7 65 3 6 3 5 3 4 2 4 1 4 0 4 0 2

0.420

0.400

0.375

0.343

0.417

0.400

0.300

0.343

0.375

0.400

0.420

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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 nc) when records are equally distributed 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
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Examples for computing Entropy

Entropy (t ) = − ∑ p ( j | t ) log p ( j | t ) j 2

C1 C2

0 6

P(C1) = 0/6 = 0

P(C2) = 6/6 = 1

Entropy = – 0 log 0 – 1 log 1 = – 0 – 0 = 0

C1 C2

1 5

P(C1) = 1/6

P(C2) = 5/6

Entropy = – (1/6) log2 (1/6) – (5/6) log2 (1/6) = 0.65 P(C1) = 2/6 P(C2) = 4/6

C1 C2

2 4

Entropy = – (2/6) log2 (2/6) – (4/6) log2 (4/6) = 0.92
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© Tan,Steinbach, Kumar

Splitting Based on INFO...
Information Gain:

GAIN

split

 n = Entropy ( p ) −  ∑ Entropy (i )    n  k i i =1

Parent Node, p is split into k partitions; ni is number of records in partition i

– Measures Reduction in Entropy achieved because of the split. Choose the split that achieves most reduction (maximizes GAIN) – Used in ID3 and C4.5 – Disadvantage: Tends to prefer splits that result in large number of partitions, each being small but pure.
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Splitting Based on INFO...
Gain Ratio:

GainRATIO

split

=

GAIN n n SplitINFO = − ∑ log SplitINFO n n
Split

k

i

i

i =1

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 Information Gain
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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/nc) when records are equally distributed among all classes, implying least interesting information Minimum (0.0) when all records belong to one class, implying most interesting information

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Examples for Computing Error

Error (t ) = 1 − max P (i | t ) i C1 C2

0 6

P(C1) = 0/6 = 0

P(C2) = 6/6 = 1

Error = 1 – max (0, 1) = 1 – 1 = 0

C1 C2

1 5

P(C1) = 1/6

P(C2) = 5/6

Error = 1 – max (1/6, 5/6) = 1 – 5/6 = 1/6 P(C1) = 2/6 P(C2) = 4/6

C1 C2

2 4

Error = 1 – max (2/6, 4/6) = 1 – 4/6 = 1/3
Introduction to Data Mining 4/18/2004 43

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Comparison among Splitting Criteria
For a 2-class problem:

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Misclassification Error vs Gini
A?
Yes Node N1 No Node N2

Parent C1 C2 7 3

Gini = 0.42

Gini(N1) = 1 – (3/3)2 – (0/3)2 =0 Gini(N2) = 1 – (4/7)2 – (3/7)2 = 0.489

C1 C2

N1 3 0

N2 4 3

Gini=0.361

Gini(Children) = 3/10 * 0 + 7/10 * 0.489 = 0.342 Gini improves !!

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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

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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)

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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

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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 © Tan,Steinbach, Kumar

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Practical Issues of Classification
Underfitting and Overfitting Missing Values Costs of Classification

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Underfitting and Overfitting (Example)

500 circular and 500 triangular data points. Circular points: 0.5 ≤ sqrt(x12+x22) ≤ 1 Triangular points: sqrt(x12+x22) > 0.5 or sqrt(x12+x22) < 1

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Underfitting and Overfitting
Overfitting

Underfitting: when model is too simple, both training and test errors are large
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Overfitting due to Noise

Decision boundary is distorted by noise point
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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
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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

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Estimating Generalization Errors
Re-substitution errors: error on training (Σ e(t) ) Generalization errors: error on testing (Σ 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
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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

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Minimum Description Length (MDL)
X X1 X2 X3 X4 y 1 0 0 1
A? Yes 0 B1 No B? B2 1 C2 1

A

C? C1 0

B

…
Xn

…
1

X X1 X2 X3 X4

y ? ? ? ?

…
Xn

…
?

Cost(Model,Data) = Cost(Data|Model) + Cost(Model) – Cost is the number of bits needed for encoding. – Search for the least costly model. Cost(Data|Model) encodes the misclassification errors. Cost(Model) uses node encoding (number of children) plus splitting condition encoding.
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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 χ 2 test) Stop if expanding the current node does not improve impurity measures (e.g., Gini or information gain).

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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

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Example of Post-Pruning
Training Error (Before splitting) = 10/30 Class = Yes Class = No 20 10 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!

Error = 10/30

A?
A1 A2
Class = Yes Class = No 8 4 Class = Yes Class = No 3 4

A4 A3
Class = Yes Class = No 4 1 Class = Yes Class = No 5 1
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Examples of Post-pruning
– Optimistic error?
Don’t prune for both cases

Case 1:

– Pessimistic error?
Don’t prune case 1, prune case 2

C0: 11 C1: 3

C0: 2 C1: 4

– Reduced error pruning?
Case 2:
Depends on validation set

C0: 14 C1: 3

C0: 2 C1: 2
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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

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Computing Impurity Measure
Tid Refund Marital Status 1 2 3 4 5 6 7 8 9 10
10

Taxable Income Class 125K 100K 70K 120K No No No No Yes No No Yes No Yes

Yes No No Yes No No Yes No No ?

Single Married Single Married

Before Splitting: Entropy(Parent) = -0.3 log(0.3)-(0.7)log(0.7) = 0.8813
Class Class = Yes = No 0 3 2 4 1 0

Divorced 95K Married 60K

Refund=Yes Refund=No Refund=?

Divorced 220K Single Married Single 85K 75K 90K

Split on Refund: Entropy(Refund=Yes) = 0 Entropy(Refund=No) = -(2/6)log(2/6) – (4/6)log(4/6) = 0.9183 Entropy(Children) = 0.3 (0) + 0.6 (0.9183) = 0.551 Gain = 0.9 × (0.8813 – 0.551) = 0.3303

Missing value
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Distribute Instances
Tid Refund Marital Status 1 2 3 4 5 6 7 8 9
10

Taxable Income Class 125K 100K 70K 120K No No No No Yes No No Yes No
10

Yes No No Yes No No Yes No No

Single Married Single Married

Tid Refund Marital Status 10 ? Single

Taxable Income Class 90K Yes

Divorced 95K Married 60K

Yes
Class=Yes Class=No

Refund
0 + 3/9 3

No
2 + 6/9 4

Divorced 220K Single Married 85K 75K

Class=Yes Class=No

Probability that Refund=Yes is 3/9 Probability that Refund=No is 6/9
2 4

Yes
Class=Yes Class=No 0 3

Refund

No

Cheat=Yes Cheat=No

Assign record to the left child with weight = 3/9 and to the right child with weight = 6/9
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© Tan,Steinbach, Kumar

Introduction to Data Mining

Classify Instances
New record:
Tid Refund Marital Status 11
1 0

Married
Taxable Income Class 85K ?

Single 1 1 2

Divorced Total 0 1 1 4 2.67 6.67

Class=No Class=Yes Total

3 6/9 3.67

No

?

Refund Yes NO Single, Divorced TaxInc < 80K NO > 80K YES No MarSt Married NO

Probability that Marital Status = Married is 3.67/6.67 Probability that Marital Status ={Single,Divorced} is 3/6.67

© Tan,Steinbach, Kumar

Introduction to Data Mining

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Other Issues
Data Fragmentation Search Strategy Expressiveness Tree Replication

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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

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Search Strategy
Finding an optimal decision tree is NP-hard The algorithm presented so far uses a greedy, top-down, recursive partitioning strategy to induce a reasonable solution Other strategies? – Bottom-up – Bi-directional

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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

Not expressive enough for modeling continuous variables – Particularly when test condition involves only a single attribute at-a-time
© Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 70

Decision Boundary
1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0

x < 0.43? Yes y < 0.47? Yes :4 :0
0 0.1 0.2 0.3 0.4 0.5

No y < 0.33? Yes :0 :3 No :4 :0

y

No :0 :4

x

0.6

0.7

0.8

0.9

1

• 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
© Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 71

Oblique Decision Trees

x+y 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 89

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 90

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
Ideal: Area = 1 Random guess: Area = 0.5
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How to Construct an ROC curve
Instance 1 2 3 4 5 6 7 8 9 10 P(+|A) 0.95 0.93 0.87 0.85 0.85 0.85 0.76 0.53 0.43 0.25 True Class + + + + +

• Use classifier that produces posterior probability for each test instance P(+|A) • Sort the instances according to P(+|A) in decreasing order • Apply threshold at each unique value of P(+|A) • Count the number of TP, FP, TN, FN at each threshold • TP rate, TPR = TP/(TP+FN) • FP rate, FPR = FP/(FP + TN)

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Introduction to Data Mining

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How to construct an ROC curve
Class

+
0.25 5 5 0 0 1 1

0.43 4 5 0 1 0.8 1

+
0.53 4 4 1 1 0.8 0.8

0.76 3 4 1 2 0.6 0.8

0.85 3 3 2 2 0.6 0.6

0.85 3 2 3 2 0.6 0.4

+
0.85 3 1 4 2 0.6 0.2

0.87 2 1 4 3 0.4 0.2

+
0.93 2 0 5 3 0.4 0

+
0.95 1 0 5 4 0.2 0 1.00 0 0 5 5 0 0

Threshold >=
TP FP TN FN TPR FPR

ROC Curve:

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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?

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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 95

Confidence Interval for Accuracy
For large test sets (N > 30),
– acc has a normal distribution with mean p and variance p(1-p)/N
Area = 1 - α

P( Z < α /2

acc − p Interval contains 0 => difference may not be statistically significant
© Tan,Steinbach, Kumar Introduction to Data Mining 4/18/2004 100

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 cross-validation)
– For each set: compute dj = e1j – e2j – dj has mean dt and variance σt k 2 – Estimate: ∑ (d − d )

σ = ˆ
2 t t

j =1

j

k (k − 1) d = d ±t σ ˆ
1−α , k −1 t

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