Classification: Basic Concepts and Decision Trees.
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Transcript of Classification: Basic Concepts and Decision Trees.
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Classification: Basic Concepts and Decision Trees
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A programming task
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Classification: Definition
• Given a collection of records (training set )– Each record contains a set of attributes, one of the
attributes is the class.• Find a model for class attribute as a function
of the values of other attributes.• Goal: previously unseen records should be
assigned a class as accurately as possible.– A test set is used to determine the accuracy of the
model. Usually, the given data set is divided into training and test sets, with training set used to build the model and test set used to validate it.
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Illustrating Classification Task
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
Learningalgorithm
Training Set
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Examples of Classification Task
• Predicting tumor cells as benign or malignant
• Classifying credit card transactions as legitimate or fraudulent
• Classifying secondary structures of protein as alpha-helix, beta-sheet, or random coil
• Categorizing news stories as finance, weather, entertainment, sports, etc
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Classification Using Distance• Place items in class to which they are
“closest”.• Must determine distance between an item
and a class.• Classes represented by
– Centroid: Central value.– Medoid: Representative point.– Individual points
• Algorithm: KNN
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Classification Techniques
• Decision Tree based Methods• Rule-based Methods• Memory based reasoning• Neural Networks• Naïve Bayes and Bayesian Belief Networks• Support Vector Machines
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A first exampleDatabase of 20,000 images of handwritten digits, each
labeled by a human
[28 x 28 greyscale; pixel values 0-255; labels 0-9]
Use these to learn a classifier which will label digit-images automatically…
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The learning problem
Input space X = {0,1,…,255}784
Output space Y = {0,1,…,9}
Training set (x1, y1), …, (xm, ym)
m = 20000
Classifier f: X ! Y
LearningAlgorithm
To measure how good f is: use a test set[Our test set: 100 instances of each digit.]
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A possible strategyInput space X = {0,1,…,255}784
Output space Y = {0,1,…,9}
Treat each image as a point in 784-dimensional Euclidean space
To classify a new image: find its nearest neighbor in the database (training set) and return that label
f = entire training set + search engine
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K Nearest Neighbor (KNN):
• Training set includes classes.• Examine K items near item to be classified.• New item placed in class with the most
number of close items.• O(q) for each tuple to be classified. (Here q
is the size of the training set.)
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KNN
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Nearest neighborImage to label Nearest neighbor
Overall:
error rate = 6%
(on test set)
Question: what is the error rate for random guessing?
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What does it get wrong?Who knows… but here’s a hypothesis:Each digit corresponds to some connected region of R784. Some of the
regions come close to each other; problems occur at these boundaries.
e.g. a random point in this ball has only a 70% chance of being in R2
R2R1
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Nearest neighbor: pros and consProsSimpleFlexibleExcellent performance on a wide range of tasks
ConsAlgorithmic: time consuming – with n training
points in Rd, time to label a newpoint is O(nd)
Statistical: memorization, not learning!no insight into the domainwould prefer a compact classifier
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Prototype selectionA possible fix: instead of the entire training set, just keep a
“representative sample”
Voronoicells
“Decision boundary”
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Prototype selectionA possible fix: instead of the entire training set, just keep a
“representative sample”
Voronoicells
“Decision boundary”
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How to pick prototypes?They needn’t be actual data points
Idea 2: one prototype per class: mean of training points
Examples:
Error = 23%
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Postscript: learning models
Training data
Classifier f
LearningAlgorithm
Batch learning On-line learning
See a new point x
predict label
test
See y
Updateclassifer
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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
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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
classMarSt
Refund
TaxInc
YESNO
NO
NO
Yes No
Married Single,
Divorced
< 80K > 80K
There could be more than one tree that fits the same data!
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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 SetDecision Tree
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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.
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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
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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
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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
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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
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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”
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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
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Decision Tree Induction
• Many Algorithms:– Hunt’s Algorithm (one of the earliest)– CART– ID3, C4.5– SLIQ,SPRINT
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General Structure of Hunt’s Algorithm
• Let Dt be the set of training records that reach a node t
• General Procedure:– If Dt contains records that belong
the same class yt, then t is a leaf node labeled as yt
– If Dt is an empty set, then t is a leaf node labeled by the default class, yd
– If Dt contains records that belong to more than one class, use an attribute test to split the data into smaller subsets. Recursively apply the procedure to each subset.
Tid Refund Marital Status
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
?
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Hunt’s Algorithm
Don’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
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
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Tree Induction
• Greedy strategy.– Split the records based on an attribute test that
optimizes certain criterion.
• Issues– Determine how to split the records
• How to specify the attribute test condition?• How to determine the best split?
– Determine when to stop splitting
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Tree Induction
• Greedy strategy.– Split the records based on an attribute test that
optimizes certain criterion.
• Issues– Determine how to split the records
• How to specify the attribute test condition?• How to determine the best split?
– Determine when to stop splitting
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How to Specify Test Condition?
• Depends on attribute types– Nominal– Ordinal– Continuous
• Depends on number of ways to split– 2-way split– Multi-way split
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Splitting Based on Nominal Attributes• Multi-way split: Use as many partitions as distinct values.
• Binary split: Divides values into two subsets. Need to find optimal partitioning.
CarTypeFamily
Sports
Luxury
CarType{Family, Luxury} {Sports}
CarType{Sports, Luxury} {Family}
OR
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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}
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Splitting Based on Continuous Attributes
• Different ways of handling– Discretization to form an ordinal categorical
attribute• Static – discretize once at the beginning• Dynamic – ranges can be found by equal interval
bucketing, equal frequency bucketing(percentiles), or clustering.
– Binary Decision: (A < v) or (A v)• consider all possible splits and finds the best cut• can be more compute intensive
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Splitting Based on Continuous Attributes
TaxableIncome> 80K?
Yes No
TaxableIncome?
(i) Binary split (ii) Multi-way split
< 10K
[10K,25K) [25K,50K) [50K,80K)
> 80K
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Tree Induction
• Greedy strategy.– Split the records based on an attribute test that
optimizes certain criterion.
• Issues– Determine how to split the records
• How to specify the attribute test condition?• How to determine the best split?
– Determine when to stop splitting
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How to determine the Best Split
OwnCar?
C0: 6C1: 4
C0: 4C1: 6
C0: 1C1: 3
C0: 8C1: 0
C0: 1C1: 7
CarType?
C0: 1C1: 0
C0: 1C1: 0
C0: 0C1: 1
StudentID?
...
Yes No Family
Sports
Luxury c1c10
c20
C0: 0C1: 1
...
c11
Before Splitting: 10 records of class 0,10 records of class 1
Which test condition is the best?
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How to determine the Best Split
C0: 5C1: 5
• Greedy approach: – Nodes with homogeneous class distribution are
preferred• Need a measure of node impurity:
C0: 9C1: 1
Non-homogeneous,
High degree of impurity
Homogeneous,
Low degree of impurity
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Measures of Node Impurity
• Gini Index
• Entropy
• Misclassification error
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How to Find the Best Split
C0 N10 C1 N11
B?
Yes No
Node N3 Node N4
A?
Yes No
Node N1 Node N2
Before Splitting:
C0 N20 C1 N21
C0 N30 C1 N31
C0 N40 C1 N41
C0 N00 C1 N01
M0
M1 M2 M3 M4
M12 M34Gain = M0 – M12 vs M0 – M34
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Measure of Impurity: GINI• Gini Index for a given node t :
(NOTE: p( j | t) is the relative frequency of class j at node t).
– Maximum (1 - 1/nc) when records are equally distributed among all classes, implying least interesting information
– Minimum (0.0) when all records belong to one class, implying most interesting information
j
tjptGINI 2)]|([1)(
C1 0C2 6
Gini=0.000
C1 2C2 4
Gini=0.444
C1 3C2 3
Gini=0.500
C1 1C2 5
Gini=0.278
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Examples for computing GINI
C1 0 C2 6
C1 2 C2 4
C1 1 C2 5
P(C1) = 0/6 = 0 P(C2) = 6/6 = 1
Gini = 1 – P(C1)2 – P(C2)2 = 1 – 0 – 1 = 0
j
tjptGINI 2)]|([1)(
P(C1) = 1/6 P(C2) = 5/6
Gini = 1 – (1/6)2 – (5/6)2 = 0.278
P(C1) = 2/6 P(C2) = 4/6
Gini = 1 – (2/6)2 – (4/6)2 = 0.444
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Splitting Based on GINI• Used in CART, SLIQ, SPRINT.• When a node p is split into k partitions (children), the quality of
split is computed as,
where, ni = number of records at child i,
n = number of records at node p.
k
i
isplit iGINI
n
nGINI
1
)(
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Binary Attributes: Computing GINI Index Splits into two partitions Effect of Weighing partitions:
Larger and Purer Partitions are sought for.
B?
Yes No
Node N1 Node N2
Parent
C1 6
C2 6
Gini = 0.500
N1 N2 C1 5 1
C2 2 4
Gini=0.333
Gini(N1) = 1 – (5/6)2 – (2/6)2 = 0.194
Gini(N2) = 1 – (1/6)2 – (4/6)2 = 0.528
Gini(Children) = 7/12 * 0.194 + 5/12 * 0.528= 0.333
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Categorical Attributes: Computing Gini Index• For each distinct value, gather counts for each class in the
dataset• Use the count matrix to make decisions
CarType{Sports,Luxury}
{Family}
C1 3 1
C2 2 4
Gini 0.400
CarType
{Sports}{Family,Luxury}
C1 2 2
C2 1 5
Gini 0.419
CarType
Family Sports Luxury
C1 1 2 1
C2 4 1 1
Gini 0.393
Multi-way split Two-way split (find best partition of values)
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Continuous Attributes: Computing Gini Index
• Use Binary Decisions based on one value• Several Choices for the splitting value
– Number of possible splitting values = Number of distinct values
• Each splitting value has a count matrix associated with it– Class counts in each of the partitions, A < v
and A v• Simple method to choose best v
– For each v, scan the database to gather count matrix and compute its Gini index
– Computationally Inefficient! Repetition of work.
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
TaxableIncome> 80K?
Yes No
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Continuous Attributes: Computing Gini Index...
• For efficient computation: for each attribute,– Sort the attribute on values– Linearly scan these values, each time updating the count matrix and
computing gini index– Choose the split position that has the least gini index
Cheat No No No Yes Yes Yes No No No No
Taxable Income
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
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Alternative Splitting Criteria based on INFO• Entropy at a given node t:
(NOTE: p( j | t) is the relative frequency of class j at node t).
– Measures homogeneity of a node. • Maximum (log nc) when records are equally distributed
among all classes implying least information• Minimum (0.0) when all records belong to one class,
implying most information
– Entropy based computations are similar to the GINI index computations
j
tjptjptEntropy )|(log)|()(
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Examples for computing Entropy
C1 0 C2 6
C1 2 C2 4
C1 1 C2 5
P(C1) = 0/6 = 0 P(C2) = 6/6 = 1
Entropy = – 0 log 0 – 1 log 1 = – 0 – 0 = 0
P(C1) = 1/6 P(C2) = 5/6
Entropy = – (1/6) log2 (1/6) – (5/6) log2 (1/6) = 0.65
P(C1) = 2/6 P(C2) = 4/6
Entropy = – (2/6) log2 (2/6) – (4/6) log2 (4/6) = 0.92
j
tjptjptEntropy )|(log)|()(2
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Splitting Based on INFO...• Information Gain:
Parent Node, p is split into k partitions;ni is number of records in partition i
– Measures Reduction in Entropy achieved because of the split. Choose the split that achieves most reduction (maximizes GAIN)
– Used in ID3 and C4.5– Disadvantage: Tends to prefer splits that result in large number of
partitions, each being small but pure.
k
i
i
splitiEntropy
nn
pEntropyGAIN1
)()(
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Splitting Based on INFO...• Gain Ratio:
Parent Node, p is split into k partitionsni is the number of records in partition i
– Adjusts Information Gain by the entropy of the partitioning (SplitINFO). Higher entropy partitioning (large number of small partitions) is penalized!
– Used in C4.5– Designed to overcome the disadvantage of Information Gain
SplitINFO
GAINGainRATIO Split
split
k
i
ii
nn
nn
SplitINFO1
log
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Splitting Criteria based on Classification Error
• Classification error at a node t :
• Measures misclassification error made by a node. • Maximum (1 - 1/nc) when records are equally distributed among all
classes, implying least interesting information• Minimum (0.0) when all records belong to one class, implying most
interesting information
)|(max1)( tiPtErrori
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Examples for Computing Error
C1 0 C2 6
C1 2 C2 4
C1 1 C2 5
P(C1) = 0/6 = 0 P(C2) = 6/6 = 1
Error = 1 – max (0, 1) = 1 – 1 = 0
P(C1) = 1/6 P(C2) = 5/6
Error = 1 – max (1/6, 5/6) = 1 – 5/6 = 1/6
P(C1) = 2/6 P(C2) = 4/6
Error = 1 – max (2/6, 4/6) = 1 – 4/6 = 1/3
)|(max1)( tiPtErrori
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Comparison among Splitting Criteria
For a 2-class problem:
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Misclassification Error vs Gini
A?
Yes No
Node N1 Node N2
Parent
C1 7
C2 3
Gini = 0.42
N1 N2 C1 3 4
C2 0 3
Gini(N1) = 1 – (3/3)2 – (0/3)2 = 0
Gini(N2) = 1 – (4/7)2 – (3/7)2 = 0.489
Gini(Children) = 3/10 * 0 + 7/10 * 0.489= 0.342
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Tree Induction
• Greedy strategy.– Split the records based on an attribute test that
optimizes certain criterion.
• Issues– Determine how to split the records
• How to specify the attribute test condition?• How to determine the best split?
– Determine when to stop splitting
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Stopping Criteria for Tree Induction
• Stop expanding a node when all the records belong to the same class
• Stop expanding a node when all the records have similar attribute values
• Early termination (to be discussed later)
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Decision Tree Based Classification
• Advantages:– Inexpensive to construct– Extremely fast at classifying unknown records– Easy to interpret for small-sized trees– Accuracy is comparable to other classification
techniques for many simple data sets
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Example: C4.5
• Simple depth-first construction.• Uses Information Gain• Sorts Continuous Attributes at each node.• Needs entire data to fit in memory.• Unsuitable for Large Datasets.
– Needs out-of-core sorting.
• You can download the software from:http://www.cse.unsw.edu.au/~quinlan/c4.5r8.tar.gz
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Practical Issues of Classification
• Underfitting and Overfitting
• Missing Values
• Costs of Classification
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Underfitting and Overfitting (Example)
500 circular and 500 triangular data points.
Circular points:
0.5 sqrt(x12+x2
2) 1
Triangular points:
sqrt(x12+x2
2) > 0.5 or
sqrt(x12+x2
2) < 1
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Underfitting and Overfitting
Overfitting
Underfitting: when model is too simple, both training and test errors are large
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Overfitting due to Noise
Decision boundary is distorted by noise point
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Overfitting due to Insufficient Examples
Lack of data points in the lower half of the diagram makes it difficult to predict correctly the class labels of that region
- Insufficient number of training records in the region causes the decision tree to predict the test examples using other training records that are irrelevant to the classification task
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Notes on Overfitting
• Overfitting results in decision trees that are more complex than necessary
• Training error no longer provides a good estimate of how well the tree will perform on previously unseen records
• Need new ways for estimating errors
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Estimating Generalization Errors• Re-substitution errors: error on training ( e(t) )• Generalization errors: error on testing ( e’(t))• Methods for estimating generalization errors:
– Optimistic approach: e’(t) = e(t)– Pessimistic approach:
• For each leaf node: e’(t) = (e(t)+0.5) • Total errors: e’(T) = e(T) + N 0.5 (N: number of leaf nodes)• For a tree with 30 leaf nodes and 10 errors on training
(out of 1000 instances): Training error = 10/1000 = 1%
Generalization error = (10 + 300.5)/1000 = 2.5%– Reduced error pruning (REP):
• uses validation data set to estimate generalization error
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Occam’s Razor
• Given two models of similar generalization errors, one should prefer the simpler model over the more complex model
• For complex models, there is a greater chance that it was fitted accidentally by errors in data
• Therefore, one should include model complexity when evaluating a model
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Minimum Description Length (MDL)
• Cost(Model,Data) = Cost(Data|Model) + Cost(Model)– Cost is the number of bits needed for encoding.– Search for the least costly model.
• Cost(Data|Model) encodes the misclassification errors.• Cost(Model) uses node encoding (number of children) plus splitting
condition encoding.
A B
A?
B?
C?
10
0
1
Yes No
B1 B2
C1 C2
X yX1 1X2 0X3 0X4 1
… …Xn 1
X yX1 ?X2 ?X3 ?X4 ?
… …Xn ?
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How to Address Overfitting• Pre-Pruning (Early Stopping Rule)
– Stop the algorithm before it becomes a fully-grown tree– Typical stopping conditions for a node:
• Stop if all instances belong to the same class• Stop if all the attribute values are the same
– More restrictive conditions:• Stop if number of instances is less than some user-specified threshold• Stop if class distribution of instances are independent of the available features
(e.g., using 2 test)• Stop if expanding the current node does not improve impurity
measures (e.g., Gini or information gain).
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How to Address Overfitting…
• Post-pruning– Grow decision tree to its entirety– Trim the nodes of the decision tree in a bottom-up
fashion– If generalization error improves after trimming,
replace sub-tree by a leaf node.– Class label of leaf node is determined from
majority class of instances in the sub-tree– Can use MDL for post-pruning
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Example of Post-Pruning
A?
A1
A2 A3
A4
Class = Yes 20
Class = No 10
Error = 10/30
Training Error (Before splitting) = 10/30
Pessimistic error = (10 + 0.5)/30 = 10.5/30
Training Error (After splitting) = 9/30
Pessimistic error (After splitting)
= (9 + 4 0.5)/30 = 11/30
PRUNE!
Class = Yes
8
Class = No 4
Class = Yes
3
Class = No 4
Class = Yes
4
Class = No 1
Class = Yes
5
Class = No 1
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Examples of Post-pruning
– Optimistic error?
– Pessimistic error?
– Reduced error pruning?
C0: 11C1: 3
C0: 2C1: 4
C0: 14C1: 3
C0: 2C1: 2
Don’t prune for both cases
Don’t prune case 1, prune case 2
Case 1:
Case 2:
Depends on validation set
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Handling Missing Attribute Values
• Missing values affect decision tree construction in three different ways:– Affects how impurity measures are computed– Affects how to distribute instance with missing
value to child nodes– Affects how a test instance with missing value is
classified
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Computing Impurity MeasureTid 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
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Distribute Instances
Class=Yes 0 + 3/ 9
Class=No 3
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
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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
Probability that Marital Status ={Single,Divorced} is 3/6.67
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Scalable Decision Tree Induction Methods
• SLIQ (EDBT’96 — Mehta et al.)– Builds an index for each attribute and only class list and the current
attribute list reside in memory
• SPRINT (VLDB’96 — J. Shafer et al.)– Constructs an attribute list data structure
• PUBLIC (VLDB’98 — Rastogi & Shim)– Integrates tree splitting and tree pruning: stop growing the tree earlier
• RainForest (VLDB’98 — Gehrke, Ramakrishnan & Ganti)– Builds an AVC-list (attribute, value, class label)
• BOAT (PODS’99 — Gehrke, Ganti, Ramakrishnan & Loh)– Uses bootstrapping to create several small samples
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Postscript: learning models
Training data
Classifier f
LearningAlgorithm
Batch learning On-line learning
See a new point x
predict label
test
See y
Updateclassifer
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Preprocessing step
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Generative and Discriminative Classifiers
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Generative vs. Discriminative Classifiers
Training classifiers involves estimating f: X Y, or P(Y|X)
Discriminative classifiers (also called ‘informative’ by Rubinstein&Hastie):
1. Assume some functional form for P(Y|X)
2. Estimate parameters of P(Y|X) directly from training data
Generative classifiers
3. Assume some functional form for P(X|Y), P(X)
4. Estimate parameters of P(X|Y), P(X) directly from training data
5. Use Bayes rule to calculate P(Y|X= xi)
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Bayes Formula
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Generative Model
• Color• Size• Texture• Weight• …
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Discriminative Model
• Logistic Regression
• Color• Size• Texture• Weight• …
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Comparison
• Generative models– Assume some functional form for P(X|Y), P(Y)– Estimate parameters of P(X|Y), P(Y) directly from
training data– Use Bayes rule to calculate P(Y|X= x)
• Discriminative models– Directly assume some functional form for P(Y|X)– Estimate parameters of P(Y|X) directly from
training data
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Probability Basics
111
• Prior, conditional and joint probability for random variables– Prior probability:
– Conditional probability: – Joint probability: – Relationship:– Independence:
• Bayesian Rule
)| ,)( 121 XP(XX|XP 2
)()()(
)(X
XX
PCPC|P
|CP
)(XP
) )( ),,( 22 ,XP(XPXX 11 XX
)()|()()|() 2211122 XPXXPXPXXP,XP(X1
)()() ),()|( ),()|( 212121212 XPXP,XP(XXPXXPXPXXP 1
EvidencePriorLikelihood
Posterior
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Probability Basics
112
• Quiz: We have two six-sided dice. When they are tolled, it could end up with the following occurance: (A) dice 1 lands on side “3”, (B) dice 2 lands on side “1”, and (C) Two dice sum to eight. Answer the following questions:
? equals ),( Is 8)
?),( 7)
?),( 6)
?)|( 5)
?)|( 4)
? 3)
? 2)
? )( )1
P(C)P(A)CAP
CAP
BAP
ACP
BAP
P(C)
P(B)
AP
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Probabilistic Classification
113
• Establishing a probabilistic model for classification– Discriminative model
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Probabilistic Classification
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• Establishing a probabilistic model for classification (cont.)– Generative model
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Probabilistic Classification
115
• MAP classification rule– MAP: Maximum A Posterior– Assign x to c* if
• Generative classification with the MAP rule– Apply Bayesian rule to convert them into posterior
probabilities
– Then apply the MAP rule
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Naïve Bayes
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• Bayes classification
Difficulty: learning the joint probability
• Naïve Bayes classification– Assumption that all input attributes are conditionally
independent!
– MAP classification rule: for
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Naïve Bayes
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• Naïve Bayes Algorithm (for discrete input attributes)– Learning Phase: Given a training set S,
Output: conditional probability tables; for
elements
– Test Phase: Given an unknown instance ,
Look up tables to assign the label c* to X’ if
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Example
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• Example: Play Tennis
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Example
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• Learning Phase
Outlook Play=Yes
Play=No
Sunny 2/9 3/5Overcast 4/9 0/5
Rain 3/9 2/5
Temperature
Play=Yes Play=No
Hot 2/9 2/5Mild 4/9 2/5Cool 3/9 1/5
Humidity Play=Yes
Play=No
High 3/9 4/5Normal 6/9 1/5
Wind Play=Yes
Play=No
Strong 3/9 3/5Weak 6/9 2/5
P(Play=Yes) = 9/14P(Play=No) = 5/14
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Example
120
• Test Phase– Given a new instance, x’=(Outlook=Sunny, Temperature=Cool, Humidity=High,
Wind=Strong)– Look up tables
– MAP rule
P(Outlook=Sunny|Play=No) = 3/5
P(Temperature=Cool|Play==No) = 1/5
P(Huminity=High|Play=No) = 4/5
P(Wind=Strong|Play=No) = 3/5
P(Play=No) = 5/14
P(Outlook=Sunny|Play=Yes) = 2/9
P(Temperature=Cool|Play=Yes) = 3/9
P(Huminity=High|Play=Yes) = 3/9
P(Wind=Strong|Play=Yes) = 3/9
P(Play=Yes) = 9/14
P(Yes|x’): [P(Sunny|Yes)P(Cool|Yes)P(High|Yes)P(Strong|
Yes)]P(Play=Yes) = 0.0053 P(No|x’): [P(Sunny|No) P(Cool|No)P(High|No)P(Strong|No)]P(Play=No) = 0.0206
Given the fact P(Yes|x’) < P(No|x’), we label x’ to be
“No”.
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Example
121
• Test Phase– Given a new instance, x’=(Outlook=Sunny, Temperature=Cool, Humidity=High,
Wind=Strong)– Look up tables
– MAP rule
P(Outlook=Sunny|Play=No) = 3/5
P(Temperature=Cool|Play==No) = 1/5
P(Huminity=High|Play=No) = 4/5
P(Wind=Strong|Play=No) = 3/5
P(Play=No) = 5/14
P(Outlook=Sunny|Play=Yes) = 2/9
P(Temperature=Cool|Play=Yes) = 3/9
P(Huminity=High|Play=Yes) = 3/9
P(Wind=Strong|Play=Yes) = 3/9
P(Play=Yes) = 9/14
P(Yes|x’): [P(Sunny|Yes)P(Cool|Yes)P(High|Yes)P(Strong|
Yes)]P(Play=Yes) = 0.0053 P(No|x’): [P(Sunny|No) P(Cool|No)P(High|No)P(Strong|No)]P(Play=No) = 0.0206
Given the fact P(Yes|x’) < P(No|x’), we label x’ to be
“No”.
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Relevant Issues
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• Violation of Independence Assumption– For many real world tasks,
– Nevertheless, naïve Bayes works surprisingly well anyway!
• Zero conditional probability Problem– If no example contains the attribute value
– In this circumstance, during test
– For a remedy, conditional probabilities estimated with
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Relevant Issues
123
• Continuous-valued Input Attributes– Numberless values for an attribute
– Conditional probability modeled with the normal distribution
– Learning Phase: Output: normal distributions and – Test Phase:
• Calculate conditional probabilities with all the normal distributions
• Apply the MAP rule to make a decision
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Conclusions• Naïve Bayes based on the independence assumption
– Training is very easy and fast; just requiring considering each attribute in each class separately
– Test is straightforward; just looking up tables or calculating conditional probabilities with normal distributions
• A popular generative model– Performance competitive to most of state-of-the-art classifiers even
in presence of violating independence assumption– Many successful applications, e.g., spam mail filtering– A good candidate of a base learner in ensemble learning– Apart from classification, naïve Bayes can do more…
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