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Transcript of AMCS/CS 340: Data Mining Classification I: Decision Tree Xiangliang Zhang King Abdullah University...
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 MaritalStatus
TaxableIncome 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 Yes10
categoric
al
categoric
al
continuous
class
Refund MaritalStatus
TaxableIncome Cheat
No Single 75K ?
Yes Married 50K ?
No Married 150K ?
Yes Divorced 90K ?
No Single 40K ?
No Married 80K ?10
TestSet
Training Set
ModelLearn
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
Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
6
Example of a Decision Tree
Tid Refund MaritalStatus
TaxableIncome 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 Yes10
categoric
al
categoric
al
continuous
class
Refund
MarSt
TaxInc
YESNO
NO
NO
Yes No
Married Single, Divorced
< 80K > 80K
Splitting Attributes
Training Data Model: Decision Tree
Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
7
Another Example of Decision Tree
Tid Refund MaritalStatus
TaxableIncome 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 Yes10
categoric
al
categoric
al
continuous
class 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!
Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
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
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
TreeInductionalgorithm
Training Set
Decision Tree
Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
9
Apply Model to Test Data
Refund
MarSt
TaxInc
YESNO
NO
NO
Yes No
Married Single, Divorced
< 80K > 80K
Refund Marital Status
Taxable Income Cheat
No Married 80K ? 10
Test DataStart from the root of tree.
10
Apply Model to Test Data
Refund
MarSt
TaxInc
YESNO
NO
NO
Yes No
Married Single, Divorced
< 80K > 80K
Refund Marital Status
Taxable Income Cheat
No Married 80K ? 10
Test Data
Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
11
Apply Model to Test Data
Refund
MarSt
TaxInc
YESNO
NO
NO
Yes No
Married Single, Divorced
< 80K > 80K
Refund Marital Status
Taxable Income Cheat
No Married 80K ? 10
Test Data
Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
12
Apply Model to Test Data
Refund
MarSt
TaxInc
YESNO
NO
NO
Yes No
Married Single, Divorced
< 80K > 80K
Refund Marital Status
Taxable Income Cheat
No Married 80K ? 10
Test Data
Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
13
Apply Model to Test Data
Refund
MarSt
TaxInc
YESNO
NO
NO
Yes No
Married Single, Divorced
< 80K > 80K
Refund Marital Status
Taxable Income Cheat
No Married 80K ? 10
Test Data
Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
14
Apply Model to Test Data
Refund
MarSt
TaxInc
YESNO
NO
NO
Yes No
Married Single, Divorced
< 80K > 80K
Refund Marital Status
Taxable Income Cheat
No Married 80K ? 10
Test Data
Assign Cheat to “No”
Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
15
Decision Tree Classification Task
Apply
Model
Induction
Deduction
Learn
Model
Model
Tid Attrib1 Attrib2 Attrib3 Class
1 Yes Large 125K No
2 No Medium 100K No
3 No Small 70K No
4 Yes Medium 120K No
5 No Large 95K Yes
6 No Medium 60K No
7 Yes Large 220K No
8 No Small 85K Yes
9 No Medium 75K No
10 No Small 90K Yes 10
Tid Attrib1 Attrib2 Attrib3 Class
11 No Small 55K ?
12 Yes Medium 80K ?
13 Yes Large 110K ?
14 No Small 95K ?
15 No Large 67K ? 10
Test Set
TreeInductionalgorithm
Training Set
Decision Tree
Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
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
Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
17
Decision Tree Induction
Many Algorithms: Hunt’s Algorithm (one of the earliest)
CART
ID3, C4.5
SLIQ,SPRINT
Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
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
1 Yes Single 125K No
2 No Married 100K No
3 No Single 70K No
4 Yes Married 120K No
5 No Divorced 95K Yes
6 No Married 60K No
7 Yes Divorced 220K No
8 No Single 85K Yes
9 No Married 75K No
10 No Single 90K Yes 10
Dt
?
Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
19
Hunt’s AlgorithmDon’t Cheat
Refund
Don’t Cheat
Don’t Cheat
Yes No
Refund
Don’t Cheat
Yes No
MaritalStatus
Don’t Cheat
Cheat
Single,Divorced
Married
TaxableIncome
Don’t Cheat
< 80K >= 80K
Refund
Don’t Cheat
Yes No
MaritalStatus
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 ?
Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
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
Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
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.
CarTypeFamily
Sports
Luxury
CarType{Family, Luxury} {Sports}
CarType{Sports, Luxury} {Family} OR
Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
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?
SizeSmall
Medium
Large
Size{Medium,
Large} {Small}
Size{Small,
Medium} {Large}OR
Size{Small, Large} {Medium}
Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
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.
Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
25
Splitting Based on Continuous Attributes
Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
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 ?
Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
27
How to determine the Best SplitBefore Splitting: 10 records of class 0,
10 records of class 1
Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
28
How to determine the Best Split
Greedy approach: Nodes with homogeneous class distribution are
preferred Need a measure of node impurity:
C0: 5C1: 5
C0: 9C1: 1
Non-homogeneous,
High degree of impurity
Homogeneous,
Low degree of impurity
Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
How to determine the Best SplitBefore Splitting: 10 records of class 0,
10 records of class 1
Non-homogeneous,
High degree of impurity
Homogeneous,
Low degree of impurity
29Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
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
Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
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)( tjptErrorj
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
31Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
Comparison among Measures of Node Impurity
j
tjptGINI 2)]|([1)(
j
tjptjptEntropy )|(log)|()(
)|(max1)( tjptErrorj
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
32Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
Comparison among Measures of Node Impurity
j
tjptGINI 2)]|([1)(
j
tjptjptEntropy )|(log)|()(
)|(max1)( tjptErrorj
For a 2-class problem:
33Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
When a node p is split into k partitions, the quality of split is computed as,
Or information gain
k
i
isplit iGINI
n
npGINIGAIN
1)()(
Quality of Split
p: n
child 1: n1
GINI(1)child k: nk
GINI(k)
child i: ni
GINI(i)
k
i
isplit iEntropy
n
npEntropyGAIN
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.
34Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
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
Decision Tree Induction
Many Algorithms: Hunt’s Algorithm (one of the earliest)
CART
ID3, C4.5
SLIQ,SPRINT
Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
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
37Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
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
Decision Tree Induction
Many Algorithms: Hunt’s Algorithm (one of the earliest)
CART
ID3, C4.5
SLIQ,SPRINT
Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
How to Find the Best Split
CarTypeFamily
Sports
LuxuryCarType
{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
40Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
Which Attribute to Split ?
Gain = 0.337Gain = 0.02 Gain = 0.5Best ?
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
41Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
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
GAINGainRATIO Split
split n
n
n
nSplitINFO ik
i
i log1
42Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
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 others43
Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
• 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
44Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
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
45Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
• 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
46Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
47
Decision Tree Induction
Many Algorithms: Hunt’s Algorithm (one of the earliest)
CART
ID3, C4.5
SLIQ,SPRINT
Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
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
48Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
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
49Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
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
Cbelow0 0
Cabove4 2
Ginisplit =0.44
HIGH
LOW
Cbelow3 0
Cabove1 2
Ginisplit =0.22
HIGH
LOW
Cbelow4 2
Cabove0 0
Ginisplit =0.44
50Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
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
Cbelow0 0
Cabove4 2
Ginisplit =0.44
HIGH
LOW
Cbelow3 0
Cabove1 2
Ginisplit =0.22
HIGH
LOW
Cbelow4 2
Cabove0 0
Ginisplit =0.44age
N1 N2
< 27.5 > 27.5
51
Numeric attributes splitting index
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
52Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
54
Issues: Underfitting and Overfitting
Missing values
Decision Tree Issues
Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
Underfitting and Overfitting
55Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
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”
56Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
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
DTerrDtTpruneerrt
tremove
57
58
Issues: Underfitting and Overfitting
Missing values
Decision Tree Issues
Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
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
59Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
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
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) = 0.3303Missing 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
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
RefundYes No
Class=Yes 0
Class=No 3
Cheat=Yes 2
Cheat=No 4
RefundYes
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
Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
65
Classification Techniques
• Decision Tree based Methods
• Rule-based Methods
• Learning from Neighbors
• Bayesian Classification
• Neural Networks
• Ensemble Methods
• Support Vector Machines
Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
Tree Rules
How tree can be used in other ways?
66Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
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
(Refund=No, Marital Status={Single,Divorced},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
(Refund=No, Marital Status={Single,Divorced},Taxable Income<80K) ==> No
(Refund=No, Marital Status={Single,Divorced},Taxable Income>80K) ==> Yes
(Refund=No, Marital Status={Married}) ==> No
• Rules can be simplified
(Marital Status = {Married}) ==> No
68Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
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 R69
Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
• 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
70Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
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
Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
Tid Refund Marital Status
Taxable Income Class
5 No Divorced 95K Yes
8 No Single 85K Yes
10 No Single 90K Yes 10
1 Yes Single 125K No
4 Yes Married 120K No
7 Yes Divorced 220K No
3 No Single 70K No
2 No Married 100K 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?
73Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
Metrics for Performance Evaluation
Focus on the predictive capability of a modelRather than how fast it takes to classify or build models, scalability, etc.
Confusion Matrix:
Most widely-used metric:
PREDICTED CLASS
ACTUALCLASS
Class=Yes Class=No
Class=Yes a (TP) b (FN)
Class=No c (FP) d (TN)
74
FNFPTNTP
TNTP
dcba
da
Accuracy
TP: True PositiveFP: False PositiveTN: True NegativeFN: False Negative
Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
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
75Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
Other Measures
cba
a
pr
rpba
aca
a
2
22(F) measure-F
(r) Recall
(p)Precision
dwcwbwawdwaw
4321
41Accuracy Weighted
76
PREDICTED CLASS
ACTUALCLASS
Class=Yes Class=No
Class=Yes a (TP) b (FN)
Class=No c (FP) d (TN)
ba
adc
c
rate positive True
rate positive False
Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
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?
77Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
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
78Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
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
79Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
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?
80Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
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 classXiangliang Zhang, KAUST AMCS/CS 340:
Data Mining
82
Using ROC for Model Comparison
Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
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 = Np = 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)?
85Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
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/ ZNpp
paccZP
Area = 1 -
Z/2 Z1- /2
)(2
4422
2/
222/
22/
ZN
accNaccNZZaccNp
86
Confidence Interval for Accuracy
Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
Confidence Interval for AccuracyConsider 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
Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
Test of SignificanceGiven 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
Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
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
2111
Ne
Ne
i
iii n
ee )1(ˆ 2
89Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
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 22
21
22
21
2
n
ee
n
eet
tt Zdd ˆ2/
90
Comparing Performance of 2 Models
),(~ 2ttdNd
Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
An Illustrative ExampleGiven: 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
0043.05000
)25.01(25.0
30
)15.01(15.0ˆ
d
128.0100.00043.096.1100.0 td
91Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining
Our Teaching Assistant
Name: Francisco Franco
Email: [email protected]
Responsibilities:
Homework
Project report
Final grades
Questions92
Xiangliang Zhang, KAUST AMCS/CS 340: Data Mining