Download Data Mining Classification: Basic Concepts, Decision Trees, and

Data Mining
Classification: Basic Concepts, Decision
Trees, and Model Evaluation
Examples of Classification Task
Lecture Notes for Chapter 4
Introduction to Data Mining
z
Predicting tumor cells as benign or malignant
z
Classifying credit card transactions
as legitimate or fraudulent
z
Classifying secondary structures of protein
as alpha-helix, beta-sheet, or random
coil
z
Categorizing news stories as finance,
weather, entertainment, sports, etc
by
Tan, Steinbach, Kumar
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
1
© Tan,Steinbach, Kumar
Classification: Definition
z
z
4/18/2004
4
Classification Techniques
Decision Tree based Methods
Rule-based Methods
z Memory based reasoning
z Neural Networks
z Naïve
Naï e Ba
Bayes
es and Ba
Bayesian
esian Belief Net
Networks
orks
z Support Vector Machines
z
Given a collection of records (training set )
z
– Each record contains a set of attributes, one of the
attributes is the class.
z
Introduction to Data Mining
Find a model for class attribute as a function
of the values of other attributes.
Goal: previously unseen records should be
assigned a class as accurately as possible.
– A test set is used to determine the accuracy of the
model. Usually, the given data set is divided into
training and test sets, with training set used to build
the model and test set used to validate it.
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
2
© Tan,Steinbach, Kumar
Illustrating Classification Task
Tid
Attrib1
1
Yes
Large
125K
No
2
No
Medium
Attrib2
100K
Attrib3
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
Introduction to Data Mining
4/18/2004
5
Example of a Decision Tree
Class
Learn
Model
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
No
7
Yes
Divorced 220K
No
8
No
Single
85K
Yes
9
No
Married
75K
No
10
No
Single
90K
Yes
Splitting Attributes
Refund
Yes
No
NO
MarSt
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
?
Apply
Model
60K
Single, Divorced
TaxInc
< 80K
NO
Married
NO
> 80K
YES
10
10
Model: Decision Tree
Training Data
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
3
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
6
Another Example of Decision Tree
Apply Model to Test Data
Test Data
MarSt
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
No
6
No
Married
7
Yes
Divorced 220K
No
8
No
Single
60K
85K
Yes
9
No
Married
75K
No
10
No
Single
90K
Yes
Married
NO
Single,
Divorced
Refund
Refund
Yes
No
Yes
NO
TaxInc
< 80K
Refund Marital
Status
Taxable
Income Cheat
No
80K
Married
?
10
No
NO
MarSt
> 80K
Single, Divorced
Married
YES
NO
TaxInc
< 80K
There could be more than one tree that
fits the same data!
NO
> 80K
YES
NO
10
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
7
Decision Tree Classification Task
Tid
Attrib1
1
Yes
Large
125K
No
2
No
Medium
Attrib2
100K
Attrib3
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
© Tan,Steinbach, Kumar
Introduction to Data Mining
Test Data
Class
Refund
Yes
Learn
Model
NO
Attrib1
11
No
Small
55K
?
12
Yes
Medium
80K
Attrib3
?
13
Yes
Large
110K
?
14
No
Small
95K
?
15
No
Large
67K
?
Apply
Model
Class
Refund Marital
Status
Taxable
Income Cheat
No
80K
Married
?
10
No
MarSt
Single, Divorced
Tid
10
Apply Model to Test Data
10
Attrib2
4/18/2004
Decision
Tree
TaxInc
< 80K
Married
NO
> 80K
YES
NO
10
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
8
Apply Model to Test Data
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
Apply Model to Test Data
Test Data
Start from the root of tree.
Refund
Yes
Test Data
Refund Marital
Status
Taxable
Income Cheat
No
80K
Married
?
Refund
10
No
NO
Yes
MarSt
Single, Divorced
TaxInc
< 80K
NO
© Tan,Steinbach, Kumar
TaxInc
< 80K
NO
4/18/2004
9
No
80K
Married
?
MarSt
NO
YES
Taxable
Income Cheat
10
Single, Divorced
> 80K
Refund Marital
Status
No
NO
Married
Introduction to Data Mining
11
© Tan,Steinbach, Kumar
Married
NO
> 80K
YES
Introduction to Data Mining
4/18/2004
12
Apply Model to Test Data
Decision Tree Induction
Test Data
Refund
z
Refund Marital
Status
Taxable
Income Cheat
No
80K
Married
?
10
Yes
No
NO
MarSt
Many Algorithms:
– Hunt’s Algorithm (one of the earliest)
– CART
– ID3, C4.5
– SLIQ,SPRINT
SLIQ SPRINT
Married
Single, Divorced
TaxInc
NO
< 80K
> 80K
YES
NO
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
13
Apply Model to Test Data
Refund Marital
Status
Refund
Married
z
Taxable
Income Cheat
80K
z
?
10
Yes
No
NO
MarSt
Assign Cheat to “No”
Married
Single, Divorced
TaxInc
NO
< 80K
> 80K
YES
NO
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
Introduction to Data Mining
4/18/2004
14
Decision Tree Classification Task
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
l f node
leaf
d llabeled
b l db
by th
the d
default
f lt
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.
© Tan,Steinbach, Kumar
Attrib3
Don’t
Cheat
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
No
7
Yes
Divorced 220K
No
8
No
Single
85K
Yes
9
No
Married
75K
No
10
No
Single
90K
Yes
60K
Dt
?
Introduction to Data Mining
4/18/2004
Hunt’s Algorithm
Class
Tid Refund Marital
Status
10
Refund
Attrib2
16
General Structure of Hunt’s Algorithm
Test Data
No
© Tan,Steinbach, Kumar
17
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
No
Tid
Attrib1
1
Yes
Large
125K
No
2
No
Medium
100K
No
3
No
Small
70K
No
4
Yes
Medium
120K
No
6
No
Married
5
No
Large
95K
Yes
7
Yes
Divorced 220K
No
6
No
Medium
60K
No
8
No
Single
85K
Yes
9
No
Married
75K
No
10
No
Single
90K
Yes
7
Yes
Large
220K
No
8
No
Small
85K
Yes
9
No
Medium
75K
No
10
No
Small
90K
Yes
Yes
No
Don’t
Cheat
Don’t
Cheat
Learn
Model
Refund
Refund
Yes
Yes
No
No
60K
10
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
?
Apply
Model
Don’t
Cheat
Decision
Tree
Don’t
Cheat
Marital
Status
Single,
Divorced
Cheat
Married
Marital
Status
Single,
Divorced
Married
Don’t
Cheat
Taxable
Income
Don’t
Cheat
< 80K
>= 80K
Don’t
Cheat
Cheat
10
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
15
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
18
Tree Induction
z
Splitting Based on Nominal Attributes
Greedy strategy.
– Split the records based on an attribute test
that optimizes certain criterion.
z
Multi-way split: Use as many partitions as distinct
values.
CarType
Family
Luxury
Sports
z
Issues
– Determine how to split the records
z
‹How
to specify the attribute test condition?
‹How to determine the best split?
{Sports,
Luxury}
– Determine when to stop splitting
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
19
Tree Induction
z
z
CarType
{Family,
Luxury}
OR
{Family}
© Tan,Steinbach, Kumar
z
{Sports}
Introduction to Data Mining
4/18/2004
22
Multi-way split: Use as many partitions as distinct
values.
Size
Small
Medium
Issues
– Determine how to split the records
‹How
CarType
Splitting Based on Ordinal Attributes
Greedy strategy.
– Split the records based on an attribute test
that optimizes certain criterion.
‹How
Binary split: Divides values into two subsets.
Need to find optimal partitioning.
z
to specify the attribute test condition?
to determine the best split?
Large
Bi
Binary
split:
lit Di
Divides
id values
l
iinto
t ttwo subsets.
b t
Need to find optimal partitioning.
Size
{Small,
Medium}
{Large}
OR
{Medium,
Large}
Size
{Small}
– Determine when to stop splitting
z
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
20
How to Specify Test Condition?
z
What about this split?
© Tan,Steinbach, Kumar
Depends on attribute types
– Nominal
– Ordinal
– Continuous
z
4/18/2004
23
Static – discretize once at the beginning
Dynamic
y
– ranges
g can be found by
y equal
q
interval
bucketing, equal frequency bucketing
(percentiles), or clustering.
– Binary Decision: (A < v) or (A ≥ v)
‹
‹
Introduction to Data Mining
{Medium}
Introduction to Data Mining
Different ways of handling
– Discretization to form an ordinal categorical
attribute
‹
Depends on number of ways to split
– 2-way split
– Multi-way split
© Tan,Steinbach, Kumar
Size
Splitting Based on Continuous Attributes
‹
z
{Small,
Large}
4/18/2004
21
consider all possible splits and finds the best cut
can be more compute intensive
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
24
Splitting Based on Continuous Attributes
How to determine the Best Split
Greedy approach:
– Nodes with homogeneous class distribution
are preferred
z Need a measure of node impurity:
z
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
25
Tree Induction
z
z
Non-homogeneous,
Homogeneous,
High degree of impurity
Low degree of impurity
© Tan,Steinbach, Kumar
Introduction to Data Mining
z
Gini Index
z
Entropy
Issues
– Determine how to split the records
z
Misclassification error
‹How
28
4/18/2004
29
Measures of Node Impurity
Greedy strategy.
– Split the records based on an attribute test
that optimizes certain criterion.
‹How
4/18/2004
to specify the attribute test condition?
to determine the best split?
– Determine when to stop splitting
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
26
How to determine the Best Split
© Tan,Steinbach, Kumar
Introduction to Data Mining
How to Find the Best Split
Before Splitting:
Before Splitting: 10 records of class 0,
10 records of class 1
C0
C1
N00
N01
M0
A?
B?
Yes
No
Node N1
C0
C1
Node N2
N10
N11
C0
C1
N20
N21
M2
M1
Yes
No
Node N3
C0
C1
Node N4
N30
N31
C0
C1
M3
N40
N41
M4
Which test condition is the best?
M12
M34
Gain = M0 – M12 vs M0 – M34
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
27
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
30
Measure of Impurity: GINI
z
Binary Attributes: Computing GINI Index
Gini Index for a given node t :
z
GINI (t ) = 1 − ∑ [ p ( j | t )]
z
2
j
Splits into two partitions
Effect of Weighing partitions:
– Larger and Purer Partitions are sought for.
Parent
(NOTE: p( j | t) is the relative frequency of class j at node t).
B?
– 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
Gini=0.000
© Tan,Steinbach, Kumar
C1
C2
1
5
C1
C2
Gini=0.278
2
4
Gini=0.444
C1
C2
Yes
Gini(N1)
= 1 – (5/7)2 – (2/7)2
= 0.408
3
3
4/18/2004
31
© Tan,Steinbach, Kumar
z
j
C1
C2
1
5
C1
C2
© Tan,Steinbach, Kumar
2
4
z
P(C1) = 0/6 = 0
z
P(C1) = 1/6
N1
5
2
N2
1
4
Gini(Children)
= 7/12 * 0.408 +
5/12 * 0.32
= 0.371
Gini=0.371
Introduction to Data Mining
Multi-way split
4/18/2004
34
P(C1) = 2/6
Two-way split
(find best partition of values)
CarType
P(C2) = 5/6
Gini = 1 – (1/6)2 – (5/6)2 = 0.278
C1
C2
Gini
P(C2) = 4/6
Family Sports Luxury
1
2
1
4
1
1
0.393
C1
C2
Gini
CarType
{Sports,
{Family}
Luxury}
3
1
2
4
0.400
CarType
{Family,
Luxury}
2
2
1
5
0.419
{Sports}
C1
C2
Gini
Gini = 1 – (2/6)2 – (4/6)2 = 0.444
Introduction to Data Mining
k
GINI split = ∑
i =1
4/18/2004
32
© Tan,Steinbach, Kumar
z
z
ni
GINI (i )
n
z
ni = number of records at child i,
n = number of records at node p.
Introduction to Data Mining
Introduction to Data Mining
4/18/2004
35
Continuous Attributes: Computing Gini Index
z
© Tan,Steinbach, Kumar
C1
C2
For each distinct value, gather counts for each class in
the dataset
Use the count matrix to make decisions
Gini = 1 – P(C1)2 – P(C2)2 = 1 – 0 – 1 = 0
Used in CART, SLIQ, SPRINT.
When a node p is split into k partitions (children), the
quality of split is computed as,
where,
Node N2
P(C2) = 6/6 = 1
Splitting Based on GINI
z
6
Categorical Attributes: Computing Gini Index
GINI (t ) = 1 − ∑ [ p ( j | t )]2
0
6
Node N1
Gini(N2)
= 1 – (1/5)2 – (4/5)2
= 0.32
Examples for computing GINI
C1
C2
6
C2
Gini = 0.500
Gini=0.500
Introduction to Data Mining
No
C1
4/18/2004
33
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
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
No
7
Y
Yes
Di
Divorced
d 220K
N
No
8
No
Single
85K
Yes
9
No
Married
75K
No
10
No
Single
90K
Yes
60K
10
4/18/2004
36
Continuous Attributes: Computing Gini Index...
z
Splitting Based on INFO...
z
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
Information Gain:
GAIN
n
= Entropy ( p ) − ⎛⎜ ∑ Entropy (i ) ⎞⎟
⎝ n
⎠
k
split
i
i =1
Parent Node, p is split into k partitions;
Ch
Cheat
N
No
N
No
N
No
Y
Yes
Y
Yes
Y
Yes
N
No
N
No
N
No
ni is number of records in partition i
N
No
Taxable Income
60
Sorted Values
Split Positions
70
55
75
65
85
72
90
80
95
87
100
92
97
120
110
125
122
220
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
© Tan,Steinbach, Kumar
0.400
0.375
0.343
0.417
0.400
0.300
0.343
Introduction to Data Mining
0.375
0.400
0.420
4/18/2004
37
Alternative Splitting Criteria based on INFO
z
– Measures Reduction in Entropy achieved because of
the split. Choose the split that achieves most reduction
(maximizes GAIN)
– Used in ID3 and C4.5
– Disadvantage: Tends to prefer splits that result in large
number of partitions, each being small but pure.
© Tan,Steinbach, Kumar
Introduction to Data Mining
z
Entropy (t ) = − ∑ p ( j | t ) log p ( j | t )
Gain Ratio:
GainRATIO
j
split
(NOTE: p( j | t) is the relative frequency of class j at node t).
– Measures homogeneity of a node.
(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
38
Examples for computing Entropy
C1
C2
0
6
C1
C2
1
5
P(C1) = 1/6
C1
C2
2
4
P(C1) = 2/6
P(C1) = 0/6 = 0
© Tan,Steinbach, Kumar
z
2
i
i
i =1
Introduction to Data Mining
4/18/2004
41
Classification error at a node t :
Error (t ) = 1 − max P (i | t )
P(C2) = 6/6 = 1
i
Entropy = – 0 log 0 – 1 log 1 = – 0 – 0 = 0
z
© Tan,Steinbach, Kumar
k
Split
Splitting Criteria based on Classification Error
Entropy (t ) = − ∑ p ( j | t ) log p ( j | t )
j
GAIN
n
n
SplitINFO = − ∑ log
SplitINFO
n
n
– 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
– Entropy based computations are similar to the
GINI index computations
4/18/2004
=
Parent Node, p is split into k partitions
ni is the number of records in partition i
‹ Maximum
Introduction to Data Mining
40
Splitting Based on INFO...
Entropy at a given node t:
© Tan,Steinbach, Kumar
4/18/2004
Measures misclassification error made by a node
node.
‹ Maximum
P(C2) = 5/6
Entropy = – (1/6) log2 (1/6) – (5/6) log2 (5/6) = 0.65
(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
P(C2) = 4/6
Entropy = – (2/6) log2 (2/6) – (4/6) log2 (4/6) = 0.92
Introduction to Data Mining
4/18/2004
39
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
42
Examples for Computing Error
Stopping Criteria for Tree Induction
Error (t ) = 1 − max P (i | t )
z
Stop expanding a node when all the records
belong to the same class
z
Stop expanding a node when all the records have
similar attribute values
z
Early termination (to be discussed later)
i
C1
C2
0
6
C1
C2
1
5
P(C1) = 1/6
C1
C2
2
4
P(C1) = 2/6
© Tan,Steinbach, Kumar
P(C1) = 0/6 = 0
P(C2) = 6/6 = 1
Error = 1 – max (0, 1) = 1 – 1 = 0
P(C2) = 5/6
Error = 1 – max (1/6, 5/6) = 1 – 5/6 = 1/6
P(C2) = 4/6
Error = 1 – max (2/6, 4/6) = 1 – 4/6 = 1/3
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© Tan,Steinbach, Kumar
Introduction to Data Mining
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Comparison among Splitting Criteria
Decision Tree Based Classification
For a 2-class problem:
z
© Tan,Steinbach, Kumar
Introduction to Data Mining
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Tree Induction
z
z
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Simple depth-first construction.
Uses Information Gain
z Sorts Continuous Attributes at each node.
z Needs entire data to fit in memory.
z Unsuitable
Uns itable for Large Datasets
Datasets.
– Needs out-of-core sorting.
to specify the attribute test condition?
to determine the best split?
Introduction to Data Mining
Introduction to Data Mining
z
z
– Determine when to stop splitting
© Tan,Steinbach, Kumar
© Tan,Steinbach, Kumar
z
Issues
– Determine how to split the records
‹How
Advantages:
– Inexpensive to construct
– Extremely fast at classifying unknown records
– Easy to interpret for small-sized trees
– Accuracy
Acc rac is comparable to other classification
techniques for many simple data sets
Example: C4.5
Greedy strategy.
– Split the records based on an attribute test
that optimizes certain criterion.
‹How
47
You can download the software from:
http://www.cse.unsw.edu.au/~quinlan/c4.5r8.tar.gz
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Practical Issues of Classification
z
Underfitting and Overfitting
z
Missing Values
z
Costs of Classification
Overfitting due to Insufficient Examples
Lack of data points in the lower half of the diagram makes it difficult
to predict correctly the class labels of that region
- Insufficient number of training records in the region causes the
decision tree to predict the test examples using other training
records that are irrelevant to the classification task
© Tan,Steinbach, Kumar
Introduction to Data Mining
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Underfitting and Overfitting
© Tan,Steinbach, Kumar
Introduction to Data Mining
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How to Address Overfitting
z
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
M
restrictive
t i ti conditions:
diti
‹ 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).
Underfitting: when model is too simple, both training and test errors are large
© Tan,Steinbach, Kumar
Introduction to Data Mining
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Overfitting due to Noise
© Tan,Steinbach, Kumar
Introduction to Data Mining
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How to Address Overfitting…
z
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
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
Decision boundary is distorted by noise point
© Tan,Steinbach, Kumar
Introduction to Data Mining
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© Tan,Steinbach, Kumar
Introduction to Data Mining
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Handling Missing Attribute Values
z
Search Strategy
Missing values affect decision tree construction in
three different ways:
– Affects how impurity measures are computed
– Affects how to distribute instance with missing
value to child nodes
– Affects how a test instance with missing value
is classified
© Tan,Steinbach, Kumar
Introduction to Data Mining
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z
Finding an optimal decision tree is NP-hard
z
The algorithm presented so far uses a greedy,
top-down, recursive partitioning strategy to
induce a reasonable solution
z
Other strategies?
– Bottom-up
– Bi-directional
© Tan,Steinbach, Kumar
Introduction to Data Mining
Other Issues
Expressiveness
Data Fragmentation
z Search Strategy
z Expressiveness
z Tree Replication
z
z
‹
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
z
Introduction to Data Mining
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Decision tree provides expressive representation for
learning discrete-valued function
– But they do not generalize well to certain types of
Boolean functions
‹
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Data Fragmentation
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
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Decision Boundary
z
Number of instances gets smaller as you traverse
down the tree
z
Number of instances at the leaf nodes could be
too ssmall
a to make
a ea
any
y stat
statistically
st ca y ssignificant
g ca t
decision
• 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
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© Tan,Steinbach, Kumar
Introduction to Data Mining
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Oblique Decision Trees
Model Evaluation
z
Metrics for Performance Evaluation
– How to evaluate the performance of a model?
z
Methods for Performance Evaluation
– How
Ho to obtain reliable estimates?
z
Methods for Model Comparison
– How to compare the relative performance
among competing models?
x+y<1
Class = +
Class =
• Test condition may involve multiple attributes
• More expressive representation
• Finding optimal test condition is computationally expensive
© Tan,Steinbach, Kumar
Introduction to Data Mining
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Tree Replication
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Metrics for Performance Evaluation
P
Focus on the predictive capability of a model
– Rather than how fast it takes to classify or
build models, scalability, etc.
z Confusion Matrix:
z
Q
S
© Tan,Steinbach, Kumar
R
0
Q
1
PREDICTED CLASS
0
1
S
0
Class=Yes
0
1
Class=Yes
ACTUAL
CLASS Class=No
Class=No
a
b
c
d
a: TP (true positive)
b: FN (false negative)
c: FP (false positive)
d: TN (true negative)
• Same subtree appears in multiple branches
© Tan,Steinbach, Kumar
Introduction to Data Mining
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Model Evaluation
z
z
z
© Tan,Steinbach, Kumar
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PREDICTED CLASS
Class=Yes
Class=Yes
ACTUAL
CLASS Class=No
Class No
Methods for Performance Evaluation
– How
Ho to obtain reliable estimates?
Methods for Model Comparison
– How to compare the relative performance
among competing models?
Introduction to Data Mining
4/18/2004
Metrics for Performance Evaluation…
Metrics for Performance Evaluation
– How to evaluate the performance of a model?
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
z
a
(TP)
b
(FN)
c
(FP)
d
(TN)
Most widely-used metric:
Accuracy =
74
Class=No
© Tan,Steinbach, Kumar
a+d
TP + TN
=
a + b + c + d TP + TN + FP + FN
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Limitation of Accuracy
z
z
Cost-Sensitive Measures
Precision (p) =
Consider a 2-class problem
– Number of Class 0 examples = 9990
– Number of Class 1 examples = 10
Recall (r) =
a
a+c
a
a+b
F - measure (F) =
If model predicts e
everything
er thing to be class 0
0,
accuracy is 9990/10000 = 99.9 %
– Accuracy is misleading because model does
not detect any class 1 example
z
z
z
2rp
2a
=
r + p 2a + b + c
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)
Weighted Accuracy =
wa + w d
wa + wb+ wc+ w d
1
© Tan,Steinbach, Kumar
Introduction to Data Mining
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Cost Matrix
© Tan,Steinbach, Kumar
C(i|j)
Class=Yes
Class=Yes
C(Yes|Yes)
C(No|Yes)
C(Yes|No)
C(No|No)
ACTUAL
CLASS Class=No
© Tan,Steinbach, Kumar
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Metrics for Performance Evaluation
– How to evaluate the performance of a model?
z
Methods for Performance Evaluation
– How
Ho to obtain reliable estimates?
z
Methods for Model Comparison
– How to compare the relative performance
among competing models?
Class=No
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Computing Cost of Classification
Cost
Matrix
C(i|j)
+
-
+
-1
100
-
1
0
PREDICTED CLASS
+
-
+
150
40
-
60
250
Accuracy = 80%
Cost = 3910
Model
M2
ACTUAL
CLASS
© Tan,Steinbach, Kumar
Introduction to Data Mining
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Methods for Performance Evaluation
PREDICTED CLASS
ACTUAL
CLASS
© Tan,Steinbach, Kumar
3
Introduction to Data Mining
z
C(i|j): Cost of misclassifying class j example as class i
ACTUAL
CLASS
4
2
Model Evaluation
PREDICTED CLASS
Model
M1
1
PREDICTED CLASS
+
-
+
250
45
-
5
200
z
How to obtain a reliable estimate of
performance?
z
Performance of a model may depend on other
factors
acto s bes
besides
des tthe
e learning
ea
ga
algorithm:
go t
– Class distribution
– Cost of misclassification
– Size of training and test sets
Accuracy = 90%
Cost = 4255
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© Tan,Steinbach, Kumar
Introduction to Data Mining
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Learning Curve
ROC (Receiver Operating Characteristic)
z
Learning curve shows
how accuracy changes
with varying sample size
z
Requires a sampling
schedule for creating
learning curve:
z
Arithmetic sampling
(Langley, et al)
z
Geometric sampling
(Provost et al)
Effect of small sample size:
© Tan,Steinbach, Kumar
-
Bias in the estimate
-
Variance of estimate
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Methods of Estimation
z
z
z
z
z
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Model Evaluation
z
z
z
Methods for Performance Evaluation
– How
Ho to obtain reliable estimates?
Methods for Model Comparison
– How to compare the relative performance
among competing models?
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Instance
P(+|A)
True Class
1
0.95
+
2
0.93
+
3
0.87
-
4
0.85
-
5
0 85
0.85
-
6
0.85
+
7
0.76
-
8
0.53
+
9
0.43
-
10
0.25
+
• Use classifier that produces
posterior probability for each
test instance P(+|A)
• Sort the instances according
to P(+|A) in decreasing order
• Apply
A l th
threshold
h ld att each
h
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)
© Tan,Steinbach, Kumar
Introduction to Data Mining
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How to construct a ROC curve
Metrics for Performance Evaluation
– How to evaluate the performance of a model?
© Tan,Steinbach, Kumar
© Tan,Steinbach, Kumar
How to construct a ROC curve
Holdout
– Reserve 2/3 for training and 1/3 for testing
Random subsampling
– Repeated holdout
Cross validation
– Partition data into k disjoint
j
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
© Tan,Steinbach, Kumar
Developed in 1950s for signal detection theory to
analyze noisy signals
– Characterize the trade-off between positive
hits and false alarms
z ROC curve plots TPR (on the y-axis) against FPR
(on the xx-axis)
axis)
z 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
z
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+
-
+
-
-
-
+
-
+
+
0.25
0.43
0.53
0.76
0.85
0.85
0.85
0.87
0.93
0.95
TP
5
4
4
3
3
3
3
2
2
1
FP
5
5
4
4
3
2
1
1
0
0
0
TN
0
0
1
1
2
3
4
4
5
5
5
5
Class
Threshold >=
1.00
0
FN
0
1
1
2
2
2
2
3
3
4
TPR
1
0.8
0.8
0.6
0.6
0.6
0.6
0.4
0.4
0.2
0
FPR
1
1
0.8
0.8
0.6
0.4
0.2
0.2
0
0
0
ROC Curve:
87
© Tan,Steinbach, Kumar
Introduction to Data Mining
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ROC Curve
(TPR,FPR):
z (0,0): declare everything
to be negative class
z (1,1): declare everything
to be positive class
z (1,0):
(1 0): ideal
z
Diagonal line:
– Random guessing
– Below diagonal line:
‹ prediction is opposite of
the true class
© Tan,Steinbach, Kumar
Introduction to Data Mining
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Using ROC for Model Comparison
z
No model consistently
outperform the other
z M1 is better for
small FPR
z M2 is better for
large FPR
z
Area Under the ROC
curve
z
Ideal:
z
Random guess:
ƒ Area
ƒ Area
© Tan,Steinbach, Kumar
Introduction to Data Mining
=1
= 0.5
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