3830599

AMCS/CS 340: Data Mining
Classification I: Decision Tree
Xiangliang Zhang
King Abdullah University of Science and Technology

2
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.
Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining

3
Classification Example
Tid Refund Marital
Status
Taxable
Income Cheat
1 Yes Single 125K No
2 No Married 100K No
3 No Single 70K No
4 Yes Married 120K No
5 No Divorced 95K Yes
6 No Married 60K No
7 Yes Divorced 220K No
8 No Single 85K Yes
9 No Married 75K No
10 No Single 90K Yes
10
Refund Marital
Status
Taxable
Income Cheat
No Single 75K ?
Yes Married 50K ?
No Married 150K ?
Yes Divorced 90K ?
No Single 40K ?
No Married 80K ?
10
Test
Set
Training
Set
Model
Learn
Classifier
predicting borrowers who cheat on loan payments.

Issues: Evaluating Classification Methods
4
• Accuracy
– classifier accuracy: how well the class labels of test
data are predicted
• Speed
– time to construct the model (training time)
– time to use the model (classification/prediction time)
• Robustness: handling noise and missing values
• Scalability: efficiency in large-scale data
• Interpretability
– understanding and insight provided by the model
• Other measures, e.g., goodness of rules, such as decision tree
size or compactness of classification rules

5
Classification Techniques
• Decision Tree based Methods
• Rule-based Methods
• Learning from Neighbors
• Bayesian Classification
• Neural Networks
• Ensemble Methods
• Support Vector Machines

6
Example of a Decision Tree
Tid Refund Marital
Status
Taxable
Income Cheat
3 No Single 70K No
6 No Married 60K No
8 No Single 85K Yes
9 No Married 75K No
10
Refund
MarSt
TaxInc
YESNO
NO
NO
Yes No
MarriedSingle, Divorced
< 80K > 80K
Splitting Attributes
Training Data Model: Decision Tree

7
Another Example of Decision Tree
Tid Refund Marital
Status
Taxable
Income Cheat
3 No Single 70K No
6 No Married 60K No
8 No Single 85K Yes
9 No Married 75K No
10
MarSt
Refund
TaxInc
YESNO
NO
NO
Yes No
Married
Single,
Divorced
< 80K > 80K
There could be more than one tree that
fits the same data!

8
Decision Tree Classification Task
Apply
Model
Induction
Deduction
Learn
Model
Model
Tid Attrib1 Attrib2 Attrib3 Class
1 Yes Large 125K No
2 No Medium 100K No
3 No Small 70K No
4 Yes Medium 120K No
5 No Large 95K Yes
6 No Medium 60K No
7 Yes Large 220K No
8 No Small 85K Yes
9 No Medium 75K No
10 No Small 90K Yes
10
11 No Small 55K ?
12 Yes Medium 80K ?
13 Yes Large 110K ?
14 No Small 95K ?
15 No Large 67K ?
10
Test Set
Tree
Induction
algorithm
Training Set
Decision
Tree

9
Apply Model to Test Data
Refund
MarSt
TaxInc
YESNO
NO
NO
Yes No
< 80K > 80K
Refund Marital
Status
Taxable
Income Cheat
No Married 80K ?
10
Test Data
Start from the root of tree.

10
Refund
MarSt
TaxInc
YESNO
NO
NO
Yes No
< 80K > 80K
Refund Marital
Status
Taxable
Income Cheat
No Married 80K ?
10
Test Data

11
Refund
MarSt
TaxInc
YESNO
NO
NO
Yes No
< 80K > 80K
Refund Marital
Status
Taxable
Income Cheat
No Married 80K ?
10
Test Data

12
Refund
MarSt
TaxInc
YESNO
NO
NO
Yes No
< 80K > 80K
Refund Marital
Status
Taxable
Income Cheat
No Married 80K ?
10
Test Data

13
Refund
MarSt
TaxInc
YESNO
NO
NO
Yes No
< 80K > 80K
Refund Marital
Status
Taxable
Income Cheat
No Married 80K ?
10
Test Data

14
Refund
MarSt
TaxInc
YESNO
NO
NO
Yes No
< 80K > 80K
Refund Marital
Status
Taxable
Income Cheat
No Married 80K ?
10
Test Data
Assign Cheat to “No”

15
Decision Tree Classification Task
Apply
Model
Induction
Deduction
Learn
Model
Model
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
11 No Small 55K ?
12 Yes Medium 80K ?
13 Yes Large 110K ?
14 No Small 95K ?
15 No Large 67K ?
10
Test Set
Tree
Induction
algorithm
Training Set
Decision
Tree

16
Algorithm for Decision Tree Induction
• Basic algorithm (a greedy algorithm)
– Tree is constructed in a top-down recursive divide-and-conquer manner
– At start, all the training examples are at the root
– Examples are partitioned recursively based on selected attributes
– Attributes are categorical (if continuous-valued, they are discretized in
advance)
– Test attributes are selected on the basis of a heuristic or statistical
measure (e.g., information gain)
• Conditions for stopping partitioning
– All samples for a given node belong to the same class
– There are no remaining attributes for further partitioning – majority
voting is employed for classifying the leaf
– There are no samples left

17
Decision Tree Induction
 Many Algorithms:
 Hunt’s Algorithm (one of the earliest)
 CART
 ID3, C4.5
 SLIQ,SPRINT

18
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
Taxable
Income Cheat
3 No Single 70K No
6 No Married 60K No
8 No Single 85K Yes
9 No Married 75K No
10
Dt
?

19
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

20
Issues of Hunt's Algorithm
Issues: Determine how to split the records
– How to specify the attribute test condition?
 How many branches
 Partition threshold for splitting
– How to determine the best split?
 Choose which attribute ?

21
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

22
Splitting Based on Nominal Attributes
 Multi-way split: Use as many partitions as
distinct values.
 Binary split: Divides values into two subsets.
Need to find optimal partitioning.
CarType
Family
Sports
Luxury
CarType
{Family,
Luxury} {Sports}
CarType
{Sports,
Luxury} {Family} OR

23
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?
Size
Small
Medium
Large
Size
{Medium,
Large} {Small}
Size
{Small,
Medium} {Large}
OR
Size
{Small,
Large} {Medium}

24
Splitting Based on Continuous Attributes
 Different ways of handling
 Binary Decision: (A < v) or ( A >= v )
consider all possible splits and finds the best cut
 Discretization to form an ordinal categorical
attribute
ranges can be found by equal interval bucketing,
equal frequency bucketing (percentiles), or
clustering.

25
Splitting Based on Continuous Attributes

26
Issues of Hunt's Algorithm
Issues: Determine how to split the records
– How to specify the attribute test condition?
 How many branches
 Partition threshold for splitting
– How to determine the best split?
 Choose which attribute ?

27
How to determine the Best Split
Before Splitting: 10 records of class 0,
10 records of class 1

28
 Greedy approach:
 Nodes with homogeneous class distribution are
preferred
 Need a measure of node impurity:
C0: 5
C1: 5
C0: 9
C1: 1
Non-homogeneous,
High degree of impurity
Homogeneous,
Low degree of impurity

Before Splitting: 10 records of class 0,
10 records of class 1
Non-homogeneous,
High degree of impurity
Homogeneous,
Low degree of impurity
29

30
Measures of Node Impurity
• Gini Index
– Gini impurity is a measure of how frequently a
randomly chosen element from a set is incorrectly
labeled if it were labeled randomly according to the
distribution of labels in the subset.
• Entropy
– a measure of the uncertainty associated with a
random variable.
• Misclassification error
– the proportion of misclassified samples

Measures of Node Impurity
• Gini Index
p( j | t) is the relative frequency of class j at node t
• Entropy
• Misclassification error

j
tjptGINI 2
)]|([1)(

j
tjptjptEntropy )|(log)|()(
)|(max1)( tjptError
j

Node t
Class j count
C1 0
C2 6
60
6
)|2(
60
0
)|1(




tCp
tCp
0)66,60max(1)(
0)66(log)66()60(log)60()(
0)66()60(1)(
22
22



tError
tEntropy
tGINI
31

Comparison among Measures of Node Impurity

j
tjptGINI 2)]|([1)(

j
)|(max1)( tjptError
j

Node t1
Class j count
C1 0
C2 6
0)1(
0)1(
0)1(



tError
tEntropy
tGINIFor a 2-class problem:
Node t2
Class j count
C1 1
C2 5
167.0)2(
650.0)2(
278.0)2(



tError
tEntropy
tGINI
Node t3
Class j count
C1 3
C2 3
5.0)3(
1)3(
5.0)3(



tError
tEntropy
tGINI
32

Comparison among Measures of Node Impurity

j
tjptGINI 2)]|([1)(

j
)|(max1)( tjptError
j

For a 2-class problem:
33

When a node p is split into k partitions,
the quality of split is computed as,
Or information gain


k
i
i
split iGINI
n
n
pGINIGAIN
1
)()(
Quality of Split
p: n
child 1: n1
GINI(1)
child k: nk
GINI(k)
child i: ni
GINI(i)



k
i
i
split
iEntropy
n
n
pEntropyGAIN
1
)()(
• Measures reduction in GINI/Entropy achieved because of the split.
• Choose the split that achieves most reduction (maximizes GAIN)
where, ni = number of records at child i, n = number of records at node p.
34

Quality of Split: binary attributes
• Splits into two partitions (binary attributes)
P
Yes No
Node N1 Node N2
Parent
C1 6
C2 6
Gini = 0.500
N1 N2
C1 4 2
C2 3 3
Gini=0.486
Gini(N1)
= 1 – (4/7)2 – (3/7)2
= 0.4898
Gini(N2)
= 1 – (2/5)2 – (3/5)2
= 0.480
Ginisplit(Children)
= 7/12 * 0.4898 +
5/12 * 0.480
= 0.486
Gainsplit = Gini (parent) – Ginisplit (children) = 0.014

36
 CART
 ID3, C4.5
 SLIQ,SPRINT

CART
CART: Classification and Regression Trees
• constructs trees with only binary splits (simplifies
the splitting criterion)
• use Gini Index as a splitting criterion
• split the attribute who provides the smallest
Ginisplit(p) or the largest GAINsplit(p)
• need to enumerate all the possible splitting points
for each attribute
37

Continuous Attributes: Computing Gini Index
For efficient computation: for each attribute,
• Sort the attribute on values
• Set candidate split positions as the midpoints between two
adjacent sorted values.
• Linearly scan these values, each time updating the count matrix
and computing Gini index
• Choose the split position that has the least Gini index
Cheat No No No Yes Yes Yes No No No No
Taxable Income
60 70 75 85 90 95 100 120 125 220
55 65 72 80 87 92 97 110 122 172 230
<= > <= > <= > <= > <= > <= > <= > <= > <= > <= > <= >
Yes 0 3 0 3 0 3 0 3 1 2 2 1 3 0 3 0 3 0 3 0 3 0
No 0 7 1 6 2 5 3 4 3 4 3 4 3 4 4 3 5 2 6 1 7 0
Gini 0.420 0.400 0.375 0.343 0.417 0.400 0.300 0.343 0.375 0.400 0.420
Split Positions
Sorted Values

39
 CART
 ID3, C4.5
 SLIQ,SPRINT

How to Find the Best Split
CarType
Family
Sports
Luxury
CarType
{Sports,
Luxury} {Family}
CarType
{Family,
Luxury} {Sports}
CarType
{Sports,
Luxury}
{Family}
C1 9 1
C2 7 3
Gini 0.468
CarType
{Family,
Luxury}
{Sports}
C1 2 8
C2 10 0
Gini 0.167
CarType
Family Sports Luxury
C1 1 8 1
C2 3 0 7
Gini 0.163
Multi-way splitTwo-way split
(find best partition of values)
Parent
C1 10
C2 10
Gini = 0.500
largest Gain = 0.337
40

Which Attribute to Split ?
Gain = 0.337Gain = 0.02 Gain = 0.5
Best ?
Disadvantage: Tends to prefer splits that result in large number of
partitions, each being small but pure.
• Unique value for each record  not predictable
• Small number of recodes in each node  not reliable prediction
41

Splitting Based on Gain Ratio
Gain Ratio:
Parent node p is split into k partitions
where ni is the number of records in partition i
• designed to overcome the disadvantage of Information Gain
• adjusts Information Gain by the entropy of the partitioning
(SplitINFO).
• higher entropy partitioning (large number of small partitions)
is penalized!
SplitINFO
GAIN
GainRATIO Split
split  n
n
n
n
SplitINFO ik
i
i log
1



42

September 14, 2010 Data Mining: Concepts and Techniques 43
Comparing Attribute Selection Measures
• The three measures, in general, return good results but
– Gini gain:
• biased to multivalued attributes
• has difficulty when # of classes is large
• tends to favor tests that result in equal-sized
partitions and purity in both partitions
– Information gain:
• biased towards multivalued attributes
– Gain ratio:
• tends to prefer unbalanced splits in which one
partition is much smaller than the others
43

• ID3 (Ross Quinlan1986) is the precursor to the C4.5
algorithm (Ross Quinlan 1993)
• C4.5 is an extension of earlier ID3 algorithm
- For each unused attribute Ai, count the information GAIN
(ID3) or GAINRatio (C4.5) from splitting on Ai
- Find the best splitting attribute Abest with the highest GAIN
or GAINRatio
- Create a decision node that splits on Abest
- Recur on the sublists obtained by splitting on Abest , and add
those nodes as children of node
ID3 and C4.5
44

 Handling both continuous and discrete attributes
o In order to handle continuous attributes, C4.5 creates a threshold
and then splits the list into those whose attribute value is above
the threshold and those that are less than or equal to it.
 Handling training data with missing attribute values
o C4.5 allows attribute values to be marked as ? for missing.
Missing attribute values are simply not used in gain and entropy
calculations.
 Pruning trees after creation
o C4.5 goes back through the tree once it's been created and
attempts to remove branches that do not help by replacing them
with leaf nodes.
Improvements of C4.5 from ID3 algorithm
45

• Issues:
o Needs entire data to fit in memory.
o Unsuitable for Large Datasets.
– Needs a lot of computation at every stage of
construction of decision tree.
• You can download the software from:
http://www2.cs.uregina.ca/~dbd/cs831/notes/ml/dtrees/c4.5/c4.5r8.tar.gz
• More information
http://www2.cs.uregina.ca/~dbd/cs831/notes/ml/dtrees/c4.5/tutorial.html
C4.5
46

47
 CART
 ID3, C4.5
 SLIQ,SPRINT

 SLIQ, Supervised Learning In Quest (EDBT’96 — Mehta et al.)
o Uses a pre-sorting technique in the tree growing phase
(eliminates the need to sort data at each node)
o creates a separate list for each attribute of the training data
o A separate list, called class list, is created for the class labels
attached to the examples.
o SLIQ requires that the class list and (only) one attribute list
could be kept in the memory at any time
o Suitable for classification of large disk-resident datasets
o Applies to both numerical and categorical attributes
SLIQ – a decision tree classifier
48

Generate
attribute list
for each
attribute
Sort attribute
lists for
NUMERIC
Attributes
Create
decision tree
by
partitioning
records
Start End
SLIQ Methodology
Drivers
Age
CarType Class (risk)
23 Family HIGH
17 Sports HIGH
43 Sports HIGH
68 Family LOW
32 Truck LOW
20 Family HIGH
Only NUMERIC attributes sorted
Example
Age Class RecId
17 HIGH 2
20 HIGH 6
23 HIGH 1
32 LOW 5
43 HIGH 3
68 LOW 4
49

Numeric attributes splitting index
Numeric attributes sorted
Age Class RecId
17 HIGH 2
20 HIGH 6
23 HIGH 1
32 LOW 5
43 HIGH 3
68 LOW 4
Partition
position
Position 0
Position 3
Position 6
States of class histograms
HIGH LOW
Cbelow
0 0
Cabove
4 2
Ginisplit =0.44
HIGH LOW
Cbelow
3 0
Cabove
1 2
Ginisplit =0.22
HIGH LOW
Cbelow
4 2
Cabove
0 0
Ginisplit =0.44
50

Numeric attributes sorted
Age Class RecId
17 HIGH 2
20 HIGH 6
23 HIGH 1
32 LOW 5
43 HIGH 3
68 LOW 4
Partition
position
Position 0
Position 3
Position 6
States of class histograms
HIGH LOW
Cbelow
0 0
Cabove
4 2
Ginisplit =0.44
HIGH LOW
Cbelow
3 0
Cabove
1 2
Ginisplit =0.22
HIGH LOW
Cbelow
4 2
Cabove
0 0
Ginisplit =0.44
age
N1 N2
< 27.5 > 27.5
51

age
N1 N2
< 27.5 > 27.5
Age Class Car type RecId
32 LOW truck 5
43 HIGH sport 3
68 LOW family 4
Age Class Car type RecId
17 HIGH sport 2
20 HIGH family 6
23 HIGH family 1
HIGH
Car type
52

54
 Issues:
 Underfitting and Overfitting
 Missing values
Decision Tree Issues

Underfitting and Overfitting
55

September 14, 2010 Data Mining: Concepts and Techniques 56
Overfitting and Tree Pruning
• Overfitting: An induced tree may overfit the training data
– Too many branches, some may reflect anomalies due to noise or outliers
– Poor accuracy for unseen samples
• Two approaches to avoid overfitting
– Pre-pruning: Stop tree construction early—do not split a node if this
would result in the gain measure falling below a threshold
• Difficult to choose an appropriate threshold
– Post-pruning: Remove branches from a “fully grown” tree
• Use a set of data different from the training data to decide which is
the “best pruned tree”
56

Data Mining: Concepts and Techniques 57
Tree Post-Pruning
• Reduced error pruning
– Start from leaves
– Replace each node with its most popular class
– Keep the change, if the prediction accuracy is not affected
• Cost complexity pruning
– Generate a series of trees T0,…,Tm, (T0 is the initial tree, Tm is the root alone)
– Construct tree Ti by
• Replacing a subtree of tree Ti-1 with a leaf node.
• the class label of leaf node is determined from majority class of
instances in the sub-tree
• Which subtree to remove ?
)),(()(
),()),,((
minarg
tTpruneleavesTleaves
DTerrDtTpruneerr
t
t
remove



57

58
 Issues:
 Underfitting and Overfitting
 Missing values
Decision Tree Issues

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
Handling Missing Values of Attribute
59

Tid Refund Marital
Status
Taxable
Income Class
3 No Single 70K No
6 No Married 60K 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
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
Computing Impurity Measure
60

Tid Refund Marital
Status
Taxable
Income Class
3 No Single 70K No
6 No Married 60K No
8 No Single 85K Yes
9 No Married 75K No
10
Refund
Yes No
Class=Yes 0
Class=No 3
Cheat=Yes 2
Cheat=No 4
Refund
Yes
Tid Refund Marital
Status
Taxable
Income Class
10 ? Single 90K Yes
10
No
Class=Yes 2 + 6/9
Class=No 4
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
Class=Yes 0 + 3/9
Class=No 3
Distribute Instances
61

Classify Instances
Refund
MarSt
TaxInc
YESNO
NO
NO
Yes
No
Married
Single,
Divorced
< 80K > 80K
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
Tid Refund Marital
Status
Taxable
Income Class
11 No ? 85K ?
10
New record:
Probability that Marital Status
= Married is 3.67/6.67 = 0.55
Probability that Marital Status
={Single,Divorced} is 3/6.67 = 0.45
62

Classify Instances
Refund
MarSt
TaxInc
YESNO
NO
NO
Yes
No
Married
Single,
Divorced
< 80K > 80K
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
Tid Refund Marital
Status
Taxable
Income Class
11 No ? 85K ?
10
New record:
Probability that Class = YES is 0.45
Probability that Class = NO is 0.55
63

 What is Decision Tree?
 How to choose the best attribute to split?
 How to decide the branches when splitting?
 What are CART, ID3 and C4.5? What are their differences?
 How Decision Tree can be used to learn from large data?
 How to avoid overfitting in Decision Tree learning?
 How missing values can be handled by Decision Tree?
64
Questions

65
Classification Techniques
• Decision Tree based Methods
• Rule-based Methods
• Learning from Neighbors
• Bayesian Classification
• Neural Networks
• Ensemble Methods
• Support Vector Machines

Tree  Rules
How tree can be used in other ways?
66

Rule Extraction from a Decision Tree
YESYESNONO
NONO
NONO
Yes No
{Married}
{Single,
Divorced}
< 80K > 80K
Taxable
Income
Marital
Status
Refund Classification Rules
(Refund=Yes) ==> No
(Refund=No, Marital Status={Single,Divorced},
Taxable Income<80K) ==> No
Taxable Income>80K) ==> Yes
(Refund=No, Marital Status={Married}) ==> No
• One rule is created for each path from the root to a leaf
• Each attribute-value pair along a path forms a conjunction: the leaf holds
the class prediction
• Rules are mutually exclusive (rules are independent and each record is covered by at most
one rule) and exhaustive (each record is covered by at least one rule)
• Rule set contains as much information as the tree 67

Rule Extraction from a Decision Tree
YESYESNONO
NONO
NONO
Yes No
{Married}
{Single,
Divorced}
< 80K > 80K
Taxable
Income
Marital
Status
Refund Classification Rules
(Refund=Yes) ==> No
Taxable Income<80K) ==> No
Taxable Income>80K) ==> Yes
(Refund=No, Marital Status={Married}) ==> No
• Rules can be simplified
(Marital Status = {Married}) ==> No
68

Rule-Based Classifier
• Classify records by using a collection of “if…then…”
rules
• Rule: Represent the knowledge in the form of IF-THEN rules
 (Condition)  y, where
‾ Condition is a conjunctions of attributes
‾ y is the class label
 Examples of classification rules:
‾ (Blood Type=Warm)  (Lay Eggs=Yes)  Birds
‾ (Taxable Income < 50K)  (Refund=Yes)  Cheat=No
‾ (rule antecedent or condition)  (rule consequent)
• Classifier: A rule R covers an instance x if the attributes of the instance
satisfy the condition of the rule, instance x labeled by the consequent of R
69

• Extract rules directly from training data
e.g., FOIL (Quinlan 1990), AQ (Michalski et al 1986), CN2(Clark & Niblett89), RIPPER
(Cohen95)
• Steps:
1. Start from an empty rule
2. Grow a rule for one class Ci using the Learn-One-Rule function
3. Remove training records covered by the rule
4. Repeat Step (2) and (3) until stopping criterion is met, e.g., when no more
training examples or when the quality of a rule returned is below a user-specified threshold
Compare with decision-tree induction: DT learns a set of
rules simultaneously
Sequential covering algorithm
70

Rules for 2-class and multi-class
71
 For 2-class problem
 choose one of the classes as positive class, and
the other as negative class
 Learn rules for positive class
 Negative class will be default class
 For multi-class problem
 Order the classes according to increasing class
prevalence (fraction of instances that belong to a
particular class)
 Learn the rule set for smallest class first, treat the
rest as negative class
 Repeat with next smallest class as positive class
Tid Refund Marital
Status
Taxable
Income Class
8 No Single 85K Yes
10
3 No Single 70K No
6 No Married 60K No
9 No Married 75K No

AMCS/CS 340: Data Mining
Classification: Evaluation
Xiangliang Zhang
King Abdullah University of Science and Technology

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?
73

Focus on the predictive capability of a model
Rather than how fast it takes to classify or build models, scalability, etc.
Confusion Matrix:
Most widely-used metric:
PREDICTED CLASS
ACTUAL
CLASS
Class=Yes Class=No
Class=Yes a (TP) b (FN)
Class=No c (FP) d (TN)
74
FNFPTNTP
TNTP
dcba
da





Accuracy
TP: True Positive
FP: False Positive
TN: True Negative
FN: False Negative

Limitation of Accuracy
Consider a 2-class problem
Number of Class 0 examples = 9990
Number of Class 1 examples = 10
Unbalanced classes
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
75

Other Measures
cba
a
pr
rp
ba
a
ca
a








2
22
(F)measure-F
(r)Recall
(p)Precision
dwcwbwaw
dwaw
4321
41
AccuracyWeighted



76
PREDICTED CLASS
ACTUAL
CLASS
Class=Yes Class=No
Class=Yes a (TP) b (FN)
Class=No c (FP) d (TN)
ba
a
dc
c




ratepositiveTrue
ratepositiveFalse

Model Evaluation
77

How to obtain a reliable estimate of performance?
Performance of a model may depend on other
factors besides the learning algorithm:
Class distribution
Cost of misclassification
Size of training and test sets
78

Methods of Estimation
Holdout
Reserve 2/3 for training and 1/3 for testing
Random subsampling
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
Bootstrap
Sampling with replacement
79

Model Evaluation
80

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 rate (y-axis) against FP rate (x-axis)
Performance of each classifier represented as a point on ROC curve
changing the threshold of algorithm, or sample distribution changes
the location of the point
At threshold t:
TPR=0.5, FPR=0.12 81
1-dimensional data set containing 2 classes (positive and negative)
- any points located at x > t is classified as positive

ROC Curve
(TPR,FPR):
(0,0): declare everything
to be negative class
(1,1): declare everything
to be positive class
(0,1): ideal
Diagonal line:
‾ Random guessing
‾ Below diagonal line:
prediction is opposite of the true class
Xiangliang Zhang, KAUST AMCS/CS 340:
Data Mining
82

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
83

How to construct an ROC curve
P(+|x) 0.95 0.93 0.87 0.85 0.85 0.85 0.76 0.53 0.43 0.25
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
Threshold: t
ROC Curve:
84
# of + >= t
# of - >= t
Posterior probability of
test instance x

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)?
85

For large test sets (N > 30),
acc has a normal distribution
with mean p and variance
p(1-p)/N
Confidence Interval for p:






 
1
)
/)1(
( 2/12/ Z
Npp
pacc
ZP
Area = 1 - 
Z/2 Z1-  /2
)(2
442
2
2/
22
2/
2
2/


ZN
accNaccNZZaccN
p



86

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
87
Standard Normal distribution

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?
88

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 of variance (Binomial distribution):
),(~
),(~
2
222
2
111


Ne
Ne
i
ii
i
n
ee )1(
ˆ 2 

89

To test if performance difference is statistically significant:
d = e1 – e2
where dt is the true difference
Since D1 and D2 are independent, their variance adds up:
At (1-) confidence level,
2
)21(2
1
)11(1
ˆˆ 2
2
2
1
2
2
2
1
2
n
ee
n
ee
t




 
tt Zdd  ˆ2/
90
Comparing Performance of 2 Models
),(~
2
ttdNd 

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
0043.0
5000
)25.01(25.0
30
)15.01(15.0
ˆ 



d
128.0100.00043.096.1100.0 td
91

Our Teaching Assistant
Name: Francisco Franco
Email: francisco.franco@kaust.edu.sa
Responsibilities:
Homework
Project report
Final grades
Questions
92

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