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

Lecture Notes for Chapter 4

Introduction to Data Mining

Tan, Steinbach, Kumar

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.

Illustrating Classification Task

�

Learning algorithm�

�

Induction�

Deduction�

Test Set�

Model�

Training Set�

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

Classification Techniques

Decision Tree based Methods
Rule-based Methods
Memory based reasoning
Neural Networks
Naïve Bayes and Bayesian Belief Networks
Support Vector Machines

Example of a Decision Tree

Refund

MarSt

TaxInc

YES

Yes

Married

Single, Divorced

< 80K

> 80K

Splitting Attributes

Training Data

Model: Decision Tree

categorical

continuous

class

Another Example of Decision Tree

categorical

continuous

class

MarSt

Refund

TaxInc

YES

Yes

Married

Single, Divorced

< 80K

> 80K

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

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

Decision Tree Classification Task

Decision Tree

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

�

Tree Induction algorithm�

�

�

Induction�

Deduction�

Test Set�

Model�

Training Set�

Apply Model to Test Data

Test Data

Start from the root of tree.

Refund

MarSt

TaxInc

YES

NO

NO

NO

Yes

No

Married

Single, Divorced

< 80K

> 80K

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

Refund

Marital

Status

Taxable

Income

Cheat

No

Married

80K

?

10

Apply Model to Test Data

Test Data

Refund

MarSt

TaxInc

YES

NO

NO

NO

Yes

No

Married

Single, Divorced

< 80K

> 80K

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

Refund

Marital

Status

Taxable

Income

Cheat

No

Married

80K

?

10

Apply Model to Test Data

Refund

MarSt

TaxInc

YES

NO

NO

NO

Yes

No

Married

Single, Divorced

< 80K

> 80K

Test Data

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

Refund

Marital

Status

Taxable

Income

Cheat

No

Married

80K

?

10

Apply Model to Test Data

Refund

MarSt

TaxInc

YES

NO

NO

NO

Yes

No

Married

Single, Divorced

< 80K

> 80K

Test Data

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

Refund

Marital

Status

Taxable

Income

Cheat

No

Married

80K

?

10

Apply Model to Test Data

Refund

MarSt

TaxInc

YES

NO

NO

NO

Yes

No

Married

Single, Divorced

< 80K

> 80K

Test Data

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

Refund

Marital

Status

Taxable

Income

Cheat

No

Married

80K

?

10

Apply Model to Test Data

Refund

MarSt

TaxInc

YES

NO

NO

NO

Yes

No

Married

Single, Divorced

< 80K

> 80K

Test Data

Assign Cheat to “No”

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

Refund

Marital

Status

Taxable

Income

Cheat

No

Married

80K

?

10

Decision Tree Classification Task

Decision Tree

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

�

Tree Induction algorithm�

�

�

Induction�

Deduction�

Test Set�

Model�

Training Set�

Decision Tree Induction

Many Algorithms:

Hunt’s Algorithm (one of the earliest)

CART

ID3, C4.5

SLIQ,SPRINT

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

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.

Dt

?

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

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

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

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

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

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

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

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

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

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

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

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.

OR

CarType

Family

Sports

Luxury

CarType

{Family,
Luxury}

{Sports}

CarType

{Sports, Luxury}

{Family}

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

Splitting Based on Ordinal Attributes

Multi-way split: Use as many partitions as distinct values.

Binary split: Divides values into two subsets.
Need to find optimal partitioning.

What about this split?

OR

Size

Small

Medium

Large

Size

{Medium,
Large}

{Small}

Size

{Small, Medium}

{Large}

Size

{Small, Large}

{Medium}

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

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

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

Splitting Based on Continuous Attributes

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

Taxable Income�> 80K?�

< 10K�

[10K,25K)�

Yes�

No�

[25K,50K)�

Taxable Income?�

[50K,80K)�

> 80K�

(i) Binary split�

(ii) Multi-way split�

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

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

How to determine the Best Split

Before Splitting: 10 records of class 0,
10 records of class 1

Which test condition is the best?

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

Own Car?�

C0: 6 C1: 4�

C0: 4 C1: 6�

Car Type?�

C0: 1 C1: 3�

C0: 8 C1: 0�

C0: 1 C1: 7�

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�

How to determine the Best Split

Greedy approach:

Nodes with homogeneous class distribution are preferred

Need a measure of node impurity:

Non-homogeneous,

High degree of impurity

Homogeneous,

Low degree of impurity

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

C0: 5 C1: 5�

C0: 9 C1: 1�

Measures of Node Impurity

Gini Index

Entropy

Misclassification error

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

How to Find the Best Split

B?

Yes

No

Node N3

Node N4

A?

Yes

No

Node N1

Node N2

Before Splitting:

Gain = M0 – M12 vs M0 – M34

M0

M1

M2

M3

M4

M12

M34

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

C0

N10

C1

N11

C0

N20

C1

N21

C0

N30

C1

N31

C0

N40

C1

N41

C0

N00

C1

N01

Measure of Impurity: GINI

Gini Index for a given node t :

(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

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

C1

0

C2

6

Gini=0.000

C1

2

C2

4

Gini=0.444

C1

3

C2

3

Gini=0.500

C1

1

C2

5

Gini=0.278

Examples for computing GINI

P(C1) = 0/6 = 0 P(C2) = 6/6 = 1

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

P(C1) = 1/6 P(C2) = 5/6

Gini = 1 – (1/6)2 – (5/6)2 = 0.278

P(C1) = 2/6 P(C2) = 4/6

Gini = 1 – (2/6)2 – (4/6)2 = 0.444

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

C1

0

C2

6

C1

2

C2

4

C1

1

C2

5

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,

where, ni = number of records at child i,

n = number of records at node p.

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

Binary Attributes: Computing GINI Index

Splits into two partitions

Effect of Weighing partitions:

Larger and Purer Partitions are sought for.

B?

Yes

No

Node N1

Node N2

Gini(N1)
= 1 – (5/6)2 – (2/6)2
= 0.194

Gini(N2)
= 1 – (1/6)2 – (4/6)2
= 0.528

Gini(Children)
= 7/12 * 0.194 +
5/12 * 0.528
= 0.333

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

Parent

C1

6

C2

6

Gini = 0.500

N1

N2

C1

5

1

C2

2

4

Gini=0.333

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)

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

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

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.

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

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�

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

Split Positions

Sorted Values

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

Alternative Splitting Criteria based on INFO

Entropy at a given node t:

(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

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

Examples for computing Entropy

P(C1) = 0/6 = 0 P(C2) = 6/6 = 1

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

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

Entropy = – (2/6) log2 (2/6) – (4/6) log2 (4/6) = 0.92

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

C1

0

C2

6

C1

2

C2

4

C1

1

C2

5

Splitting Based on INFO...

Information Gain:

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.

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

Splitting Based on INFO...

Gain Ratio:

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

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

Splitting Criteria based on Classification Error

Classification error at a node t :

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

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

Examples for Computing Error

P(C1) = 0/6 = 0 P(C2) = 6/6 = 1

Error = 1 – max (0, 1) = 1 – 1 = 0

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

Error = 1 – max (2/6, 4/6) = 1 – 4/6 = 1/3

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

C1

0

C2

6

C1

2

C2

4

C1

1

C2

5

Comparison among Splitting Criteria

For a 2-class problem:

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

Misclassification Error vs Gini

A?

Yes

No

Node N1

Node N2

Gini(N1)
= 1 – (3/3)2 – (0/3)2
= 0

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

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

Gini improves !!

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

Parent

C1

7

C2

3

Gini = 0.42

N1

N2

C1

3

4

C2

0

3

Gini=0.361

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

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

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)

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

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

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

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

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

Practical Issues of Classification

Underfitting and Overfitting

Missing Values

Costs of Classification

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

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

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

Underfitting and Overfitting

Overfitting

Underfitting: when model is too simple, both training and test errors are large

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

Overfitting due to Noise

Decision boundary is distorted by noise point

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

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

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

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

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

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)

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

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

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

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

Minimum Description Length (MDL)

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.

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

Sheet1

X y X y

1 ?

0 ?

0 ?

1 ?

… … … …

1 ?

Sheet2

Sheet3

Sheet1

X y X y

?

?

?

?

… … … …

?

Sheet2

Sheet3

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).

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

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

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

Example of Post-Pruning

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 20

Class = No 10

Error = 10/30

Class = Yes 8

Class = No 4

Class = Yes 3

Class = No 4

Class = Yes 4

Class = No 1

Class = Yes 5

Class = No 1

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

Examples of Post-pruning

Optimistic error?

Pessimistic error?

Reduced error pruning?

Don’t prune for both cases

Don’t prune case 1, prune case 2

Case 1:

Case 2:

Depends on validation set

C0: 11

C1: 3

C0: 2

C1: 4

C0: 14

C1: 3

C0: 2

C1: 2

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

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

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

Computing Impurity Measure

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

Before Splitting:
Entropy(Parent)
= -0.3 log(0.3)-(0.7)log(0.7) = 0.8813

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

Tid

Refund

Marital

Status

Taxable

Income

Class

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

?

Single

90K

Yes

10

Class

= Yes

Class = No

Refund=Yes

0

3

Refund=No

2

4

Refund=?

1

0

Distribute Instances

Refund

Yes

No

Refund

Yes

No

Probability that Refund=Yes is 3/9

Probability that Refund=No is 6/9

Assign record to the left child with weight = 3/9 and to the right child with weight = 6/9

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

Tid

Refund

Marital

Status

Taxable

Income

Class

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

Class=Yes

0

Class=No

3

Cheat=Yes

2

Cheat=No

4

Tid

Refund

Marital

Status

Taxable

Income

Class

10

?

Single

90K

Yes

10

Class=Yes

2 + 6/9

Class=No

4

Class=Yes

0 + 3/9

Class=No

3

Classify Instances

Refund

MarSt

TaxInc

YES

NO

NO

NO

Yes

No

Married

Single,
Divorced

< 80K

> 80K

New record:

Probability that Marital Status
= Married is 3.67/6.67

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

Married Single Divorced Total

Class=No 3 1 0 4

Class=Yes 6/9 1 1 2.67

Total 3.67 2 1 6.67

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

Tid

Refund

Marital

Status

Taxable

Income

Class

11

No

?

85K

?

10

Other Issues

Data Fragmentation

Search Strategy

Expressiveness

Tree Replication

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

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

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

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

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

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

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

Decision Boundary

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

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

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�

Oblique Decision Trees

Test condition may involve multiple attributes

More expressive representation

Finding optimal test condition is computationally expensive

x + y < 1

Class = +

Class =

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

Tree Replication

Same subtree appears in multiple branches

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

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?

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

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?

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

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:

a: TP (true positive)

b: FN (false negative)

c: FP (false positive)

d: TN (true negative)

PREDICTED CLASS

ACTUAL CLASS Class=Yes Class=No

Class=Yes a b

Class=No c d

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

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)

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

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

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

Cost Matrix

C(i|j): Cost of misclassifying class j example as class i

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) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

Computing Cost of Classification

Accuracy = 80%

Cost = 3910

Accuracy = 90%

Cost = 4255

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

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

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

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

Cost-Sensitive Measures

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)

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

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?

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

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

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

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

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

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

k-fold: train on k-1 partitions, test on the remaining one

Leave-one-out: k=n

Stratified sampling

oversampling vs undersampling

Bootstrap

Sampling with replacement

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

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?

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

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 TP (on the y-axis) against FP (on the x-axis)

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

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

ROC Curve

- 1-dimensional 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

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

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

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

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

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

How to Construct an ROC curve

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)

Instance P(+|A) True Class

1 0.95 +

2 0.93 +

3 0.87 -

4 0.85 -

5 0.85 -

6 0.85 +

7 0.76 -

8 0.53 +

9 0.43 -

10 0.25 +

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

How to construct an ROC curve

Threshold >=

ROC Curve:

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

Class

+

-

+

-

-

-

+

-

+

+

P

0.25

0.43

0.53

0.76

0.85

0.85

0.85

0.87

0.93

0.95

1.00

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

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?

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

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)?

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

Confidence Interval for Accuracy

For large test sets (N > 30),

acc has a normal distribution
with mean p and variance
p(1-p)/N

Confidence Interval for p:

Area = 1 - 

Z/2

Z1-  /2

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

Confidence Interval for Accuracy

Consider a model that produces an accuracy of 80% when evaluated on 100 test instances:

N=100, acc = 0.8

Let 1- = 0.95 (95% confidence)

From probability table, Z/2=1.96

1- Z

0.99 2.58

0.98 2.33

0.95 1.96

0.90 1.65

N 50 100 500 1000 5000

p(lower) 0.670 0.711 0.763 0.774 0.789

p(upper) 0.888 0.866 0.833 0.824 0.811

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

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 = e1

M2 is tested on D2 (size=n2), found error rate = e2

Assume D1 and D2 are independent

If n1 and n2 are sufficiently large, then

Approximate:

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

Comparing Performance of 2 Models

To test if performance difference is statistically significant: d = e1 – e2

d ~ N(dt,t) where dt is the true difference

Since D1 and D2 are independent, their variance adds up:

At (1-) confidence level,

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

An Illustrative Example

Given: M1: n1 = 30, e1 = 0.15
M2: n2 = 5000, e2 = 0.25

d = |e2 – e1| = 0.1 (2-sided test)

At 95% confidence level, Z/2=1.96

=> Interval contains 0 => difference may not be
statistically significant

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

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 t

Estimate:

(C) Vipin Kumar, Parallel Issues in Data Mining, VECPAR 2002

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

C0: 9

C1: 1

C0: 5

C1: 5

)

|

(

max

1

)

(

t

i

P

t

Error

i

-

=

N1

N2

C1

5

1

C2

2

4

Gini=0.3

33

Parent

C1

7

C2

3

Gini = 0.

42

N1

N2

C1

3

4

C2

0

3

Gini=0.361

å

=

-

=

k

i

i

i

n

n

n

n

SplitINFO

1

log

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

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

å

-

=

j

t

j

p

t

GINI

2

)]

|

(

[

1

)

(

C1

0

C2

6

Gini=0.000

C1

2

C2

4

Gini=0.444

C1

3

C2

3

Gini=0.500

C1

1

C2

5

Gini=0.278

å

=

=

k

i

i

split

i

GINI

n

n

GINI

1

)

(

SplitINFO

GAIN

GainRATIO

Split

split

=

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

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

Taxable

Income

> 80K?

YesNo

Taxable

Income?

(i) Binary split(ii) Multi-way split

< 10K

[10K,25K)[25K,50K)[50K,80K)

> 80K

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

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

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

å

-

=

j

t

j

p

t

j

p

t

Entropy

)

|

(

log

)

|

(

)

(

÷

ø

ö

ç

è

æ

-

=

å

=

k

i

i

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i

Entropy

n

n

p

Entropy

GAIN

1

)

(

)

(

Refund

Marital

Status

Taxable

Income

Cheat

No

Married

80K

?

10

Refund

Marital

Status

Taxable

Income

Cheat

No

Married

80K

?

10

Refund

Marital

Status

Taxable

Income

Cheat

No

Married

80K

?

10

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

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

Luxuryc

1

c

10

c

20

C0: 0

C1: 1

...

c

11

C0 N10

C1 N11

C0 N00

C1 N01

C0 N20

C1 N21

C0 N30

C1 N31

C0 N40

C1 N41

Taxable

Income

> 80K?

YesNo

C1

1

C2

5

C1

0

C2

6

C1

2

C2

4

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Status

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Income

Class

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

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No

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Status

Taxable

Income

Class

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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

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Yes

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No

10

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Tid Refund Marital

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00.10.20.30.40.50.60.70.80.91

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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

PREDICTED CLASS
ACTUAL CLASS	Class=Yes	Class=No
Class=Yes	a	b
Class=No	c	d

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)

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

Instance	P(+\|A)	True Class
1	0.95	+
2	0.93	+
3	0.87	-
4	0.85	-
5	0.85	-
6	0.85	+
7	0.76	-
8	0.53	+
9	0.43	-
10	0.25	+

N	50	100	500	1000	5000
p(lower)	0.670	0.711	0.763	0.774	0.789
p(upper)	0.888	0.866	0.833	0.824	0.811