Data Mining Classification: Alternative Techniques 第 5 章 分類技術

Data Mining Classification: Alternative

Techniques

第 5 章 分類技術

2

Bayes Classifier

1. 貝氏理論2. 利用貝氏理論分類3. 單純貝氏分類法（ Naïve Bayes ）4. 貝氏信念網路

（ Bayesian belief network ， BBN）

3

某一事件發生的機率常受其他相關事件是否發生影響，某一事件發生的機率常受其他相關事件是否發生影響，稱為條件機率稱為條件機率 (conditional probability)(conditional probability) 記作記作 P P ((A A | | BB)) 。。某一事件發生的機率常受其他相關事件是否發生影響，某一事件發生的機率常受其他相關事件是否發生影響，稱為條件機率稱為條件機率 (conditional probability)(conditional probability) 記作記作 P P ((A A | | BB)) 。。

條件機率：條件機率：

或或

條件機率：條件機率：

或或

( )( | )

( )P A B

P A BP B

( )

( | )( )

P A BP A B

P B

條件機率 & 乘法律

AP

BAPABP

|

AP

BAPABP

|

乘法律乘法律 (multiplication law)(multiplication law) 用來計算兩事件交集的機率用來計算兩事件交集的機率

及及

乘法律乘法律 (multiplication law)(multiplication law) 用來計算兩事件交集的機率用來計算兩事件交集的機率

及及PP((AA ∩∩BB) = ) = PP((BB))PP((A A | | BB)) PP((AA ∩∩BB) = ) = PP((AA))PP((B B | | AA))

4一般形式的貝氏定理

若事件若事件 BB11,B,B22,…,B,…,Bkk 構成樣本空間構成樣本空間 SS 的一組分割，的一組分割，且且 PP((BBii)≠0,)≠0,ii=1,2,…,=1,2,…,kk ；則對；則對 SS 中的任一事件中的任一事件 PP((AA)≠0)≠0 而言，而言，若事件若事件 BB11,B,B22,…,B,…,Bkk 構成樣本空間構成樣本空間 SS 的一組分割，的一組分割，且且 PP((BBii)≠0,)≠0,ii=1,2,…,=1,2,…,kk ；則對；則對 SS 中的任一事件中的任一事件 PP((AA)≠0)≠0 而言，而言，

1 1 2 2

( ) ( )( )

( ) ( ) ( ) ( ) ( ) ( )i i

ik k

P B P A BP B A

P B P A B P B P A B P B P A B

5

(1) (2) (3) (4) (5)

事件 事前機率 條件機率 聯合機率 事後機率Bi P(Bi) P(A|Bi) P(Bi∩A) P(Bi|A)

B1 0.2 0.05 0.010 0.010/0.032=0.3125

B2 0.3 0.04 0.012 0.012/0.032=0.3750

B3 0.5 0.02 0.010 0.010/0.032=0.3125

1.00 P(A)=0.032 　　　　　 1.000

貝氏定理的列表分析貝氏定理的列表分析

一般形式的貝氏定理例題

6

如果事件 如果事件 AA 發生的機率不受事件發生的機率不受事件 BB 的影響，稱事件的影響，稱事件 AA 和和BB 為獨立事件為獨立事件 (independent events)(independent events) 。。如果事件 如果事件 AA 發生的機率不受事件發生的機率不受事件 BB 的影響，稱事件的影響，稱事件 AA 和和BB 為獨立事件為獨立事件 (independent events)(independent events) 。。

兩事件 兩事件 AA 和 和 BB 是獨立事件，則是獨立事件，則

且且

兩事件 兩事件 AA 和 和 BB 是獨立事件，則是獨立事件，則

且且PP((AA||BB) = ) = PP((AA))

獨立事件 / 條件獨立事件條件獨立事件

條件獨立事件條件獨立事件 (Conditional independence)(Conditional independence) ：：在在 MM 事件發生的狀況下，事件發生的狀況下，A,B,CA,B,C 事件發生的機率不受彼此的影響事件發生的機率不受彼此的影響

條件獨立事件條件獨立事件 (Conditional independence)(Conditional independence) ：：在在 MM 事件發生的狀況下，事件發生的狀況下，A,B,CA,B,C 事件發生的機率不受彼此的影響事件發生的機率不受彼此的影響

PP((AA ∩∩BB) = ) = PP((AA))PP((BB))

PP((AA ∩ ∩ BB ∩ ∩ C | MC | M)) = P = P((A|MA|M)) P P((B|MB|M)) P P((C|MC|M))

7Bayes theorem

貝氏理論（ Bayes theorem ），它是一個從資料當中結合類別知識的方法。

A probabilistic framework for solving classification problems

Conditional Probability:

Bayes theorem:

)()()|(

)|(AP

CPCAPACP

)(),(

)|(

)(),(

)|(

CPCAP

CAP

APCAP

ACP

8Example of Bayes Theorem

Given: – 醫生知道，腦膜炎有 50 ％的機率會導致頸部僵硬 – 先驗機率：任何病人患有腦膜炎的機率是 P(M)=1 /

50000

– 先驗機率：有任何病人頸部僵硬的機率是 P(S)=1 / 20

後驗機率：如果病人頸部僵硬，該病人患腦膜炎的機率為何 P(M|S) ?

0002.020/150000/15.0

)()()|(

)|( SP

MPMSPSMP

9Bayesian Classifiers

Consider each attribute and class label as random variables

Given a record with attributes (A1, A2,…,An)

– Goal is to predict class C

– Specifically, we want to find the value of C that maximizes P(C| A1, A2,…,An )

Can we estimate P(C| A1, A2,…,An ) directly from data?

10Bayesian Classifiers

Approach:

– compute the posterior probability P(C | A1, A2, …, An) for all values of C using the Bayes theorem

– Choose value of C that maximizes P(C | A1, A2, …, An)

– Equivalent to choosing value of C that maximizes P(A1, A2, …, An|C) P(C)

How to estimate P(A1, A2, …, An | C )?

)()()|(

)|(21

21

21

n

n

n AAAPCPCAAAP

AAACP

11Naïve Bayes Classifier

Assume independence among attributes Ai when

class is given:

– P(A1, A2, …, An |Cj) = P(A1| Cj) P(A2| Cj)… P(An| Cj)

– Can estimate P(Ai| Cj) for all Ai and Cj.

– New point is classified to Cj if P(Cj) P(Ai| Cj) is maximal.

12How to Estimate Probabilities from Data?

Class: P(C) = Nc/N– e.g., P(No) = 7/10,

P(Yes) = 3/10

For discrete attributes:

P(Ai | Ck) = |Aik|/ Nck

– where |Aik| is number of instances having attribute Ai and belongs to class Ck

– Examples:

P(Status=Married|No) = 4/7P(Refund=Yes|Yes)=0

13Example

P(O)=3/10 P(No|O)=1 P(Div|O)=1/3 P(Low|O)=1/3 P(X)=7/10 P(No|X)=4/7 P(Div|X)=1/7 P(Low|X)=3/7

W={No,Div,Low} Class(W)=O or X ?

Tid RefundMaritalStatus

TaxableIncome

Class

1 Yes Single Mid X

2 No Married Mid X

3 No Single Low X

4 Yes Married Mid X

5 No Divorced Mid O

6 No Married Low X

7 Yes Divorced High X

8 No Single Low O

9 No Married Low X

10 No Single Mid O

14Example

P(O)=3/10 P(No|O)=1 P(Div|O)=1/3 P(Low|O)=1/3

Tid RefundMaritalStatus

TaxableIncome

Class

1 Yes Single Mid X

2 No Married Mid X

3 No Single Low X

4 Yes Married Mid X

5 No Divorced Mid O

6 No Married Low X

7 Yes Divorced High X

8 No Single Low O

9 No Married Low X

10 No Single Mid O

P(X)=7/10 P(No|X)=4/7 P(Div|X)=1/7 P(Low|X)=3/7

W={No,Div,Low} P(O|W)= P(No|O)P(Div|O)P(Low|O)P(O)/P(W)= 1/30/P(W) P(X|W)= P(No|X)P(Div|X)P(Low|X)P(X)/P(W)= 2/245/P(W) ∵P(O|W) > P(X|W) Class(W)=O∴

)(

)()|()|(

21

2121

n

nn AAAP

CPCAAAPAAACP

P(A1, A2, …, An |Cj) = P(A1| Cj) P(A2| Cj)… P(An| Cj)

15Example of Naïve Bayes Classifier

　 Class 　 　

Give Birthmammal

s

non-mammal

s總計

no 1 12 13

yes 6 1 7

總計 7 13 20

Lay Eggsmammal

s

non-mammal

s總計

no 6 1 7

yes 1 12 13

總計 7 13 20

Can Flymammal

s

non-mammal

s總計

no 6 10 16

yes 1 3 4

總計 7 13 20

Live in Water

mammals

non-mammal

s總計

no 5 6 11

sometimes 　 4 4

yes 2 3 5

總計 7 13 20

Have Legsmammal

s

non-mammal

s總計

no 2 4 6

yes 5 9 14

總計 7 13 20

Name Give Birth Can Fly Live in Water Have Legs Class

human yes no no yes mammalspython no no no no non-mammalssalmon no no yes no non-mammalswhale yes no yes no mammalsfrog no no sometimes yes non-mammalskomodo no no no yes non-mammalsbat yes yes no yes mammalspigeon no yes no yes non-mammalscat yes no no yes mammalsleopard shark yes no yes no non-mammalsturtle no no sometimes yes non-mammalspenguin no no sometimes yes non-mammalsporcupine yes no no yes mammalseel no no yes no non-mammalssalamander no no sometimes yes non-mammalsgila monster no no no yes non-mammalsplatypus no no no yes mammalsowl no yes no yes non-mammalsdolphin yes no yes no mammalseagle no yes no yes non-mammals

Give Birth Can Fly Live in Water Have Legs Class

yes no yes no ?

16Example of Naïve Bayes Classifier

Name Give Birth Can Fly Live in Water Have Legs Class

human yes no no yes mammalspython no no no no non-mammalssalmon no no yes no non-mammalswhale yes no yes no mammalsfrog no no sometimes yes non-mammalskomodo no no no yes non-mammalsbat yes yes no yes mammalspigeon no yes no yes non-mammalscat yes no no yes mammalsleopard shark yes no yes no non-mammalsturtle no no sometimes yes non-mammalspenguin no no sometimes yes non-mammalsporcupine yes no no yes mammalseel no no yes no non-mammalssalamander no no sometimes yes non-mammalsgila monster no no no yes non-mammalsplatypus no no no yes mammalsowl no yes no yes non-mammalsdolphin yes no yes no mammalseagle no yes no yes non-mammals

Give Birth Can Fly Live in Water Have Legs Class

yes no yes no ?

0027.02013

004.0)()|(

021.0207

06.0)()|(

0042.0134

133

1310

131

)|(

06.072

72

76

76

)|(

NPNAP

MPMAP

NAP

MAP

A: attributes

M: mammals

N: non-mammals

P(A|M)P(M) > P(A|N)P(N)

Mammals

18How to Estimate Probabilities from Data?

Normal distribution:

– One for each (Ai,ci) pair

For (Income, Class=No):

– If Class=No μ= 110

σ2= 2975

2

2

2

)(

22

1)|( ij

ijiA

ij

ji ecAP

0072.0)2975(2

1)|120( )2975(2

)110120( 2

eNoIncomeP

20Example

P(O)=4/10 P(Yes|O)=1/4 P(Single|O)=1/4 P(Mid|O)=1/4

Tid RefundMaritalStatus

TaxableIncome

Class

1 Yes Single High X

2 Yes Single High X

3 Yes Single High X

4 Yes Single High X

5 Yes Single High X

6 No Married High X

7 Yes Married Low O

8 No Single Low O

9 No Married Low O

10 No Married Mid O

P(X)=6/10 P(Yes|X)=5/6 P(Single|X)=5/6 P(Mid|X)=0

W={Yes,Single,Mid} P(O|W)= (4/10)(1/4)3/P(W)= (1/160)/P(W) P(X|W)= (6/10)(5/6)2*0/P(W)= 0/P(W) = 0 ∵P(O|W) > P(X|W) Class(W)=O∴

)(

)()|()|(

21

2121

n

nn AAAP

CPCAAAPAAACP

P(A1, A2, …, An |Cj) = P(A1| Cj) P(A2| Cj)… P(An| Cj)

21Naïve Bayes Classifier

If one of the conditional probability is zero, then the entire expression becomes zero

Probability estimation:

mN

mpNCAP

cN

NCAP

N

NCAP

c

ici

c

ici

c

ici

)|(:estimate-m

1)|(:Laplace

)|( :Originalc: number of classes

p: prior probability

m: parameter

22Example

P(O)=4/10 P(Yes|O)=1/4 P(Single|O)=1/4 P(Mid|O)=1/4

Tid RefundMaritalStatus

TaxableIncome

Class

1 Yes Single High X

2 Yes Single High X

3 Yes Single High X

4 Yes Single High X

5 Yes Single High X

6 No Married High X

7 Yes Married Low O

8 No Single Low O

9 No Married Low O

10 No Married Mid O

P(X)=6/10 P(Yes|X)=5/6 P(Single|X)=5/6 P(Mid|X)=0

m

mXMidP

mN

mpNCAP

XMidPcN

NCAP

XMidPN

NCAP

c

ici

c

ici

c

ici

9

1.00)|(,)|(:estimate-m

9

1

36

10)|(,

1)|(:Laplace

6

0)|(,)|( :Original

23Laplace probability estimate

where k is the number of classes.

Problems with Laplace:Assumes all classes a priori equally likelyDegree of pruning depends on number of classes

kNnp C

C 1

24m-estimate of probability

pC = ( nC + pCa m ) / ( N + m)

where:

pCa = a prior probability of class C

m is a non-negative parameter tuned

by expert

25m-estimate

Important points: Takes into account prior probabilities Pruning not sensitive to number of classes Varying m: series of differently pruned

trees Choice of m depends on confidence in

data

26m-estimate in pruning

Choice of m:

Low noise low m little pruning

High noise high m much pruning

Note: Using m-estimate is as if examples at

node were a random sample, which they are

not. Suitably adjusting m compensates for this.

27貝氏信念網路 (Bayesian belief network ， BBN)

28貝氏信念網路 範例 2- 已知高血壓

1967.08033.01)|(

8033.05185./49.*85.)(

)()|()|(

5185.051.2.49..85)()|()(

HighBPNoHDP

HighBPP

YesHDPYesHDHighBPPHighBPYesHDP

HDPHDHighBPPHighBPP

29貝氏信念網路 範例 3- 高血壓 ,健康飲食 ,規律運動

4138.05862.01),,|(

5862.075.*2.25.*85.

25.*85.

),|()|(

),|()|(

),|(),,(

),,|(

),,|(

YesEHealthyDHighBPNoHDP

YesEHealthyDHDPHDHighBPP

YesEHealthyDYesHDPYesHDHighBPP

YesEHealthyDYesHDPYesEHealthyDHighBPP

YesEHealthyDYesHDHighBPP

YesEHealthyDHighBPYesHDP

30

Artificial Neural Networks (ANN)

31Artificial Neural Networks (ANN)

X1 X2 X3 Y1 0 0 01 0 1 11 1 0 11 1 1 10 0 1 00 1 0 00 1 1 10 0 0 0

X1

X2

X3

Y

Black box

Output

Input

Output Y is 1 if at least two of the three inputs are equal to 1.

32Artificial Neural Networks (ANN)

X1 X2 X3 Y1 0 0 01 0 1 11 1 0 11 1 1 10 0 1 00 1 0 00 1 1 10 0 0 0

X1

X2

X3

Y

Black box

0.3

0.3

0.3 t=0.4

Outputnode

Inputnodes

otherwise0

trueis if1)( where

)04.03.03.03.0( 321

zzI

XXXIY

33Artificial Neural Networks (ANN)

Model is an assembly of inter-connected nodes and weighted links

Output node sums up each of its input value according to the weights of its links

Compare output node against some threshold t

X1

X2

X3

Y

Black box

w1

t

Outputnode

Inputnodes

w2

w3

)( tXwIYi

ii Perceptron Model

)( tXwsignYi

ii

or

34General Structure of ANN

Activationfunction

g(Si )Si Oi

I1

I2

I3

wi1

wi2

wi3

Oi

Neuron iInput Output

threshold, t

InputLayer

HiddenLayer

OutputLayer

x1 x2 x3 x4 x5

y

Training ANN means learning the weights of the neurons

35Algorithm for learning ANN

Initialize the weights (w0, w1, …, wk)

Adjust the weights in such a way that the output of ANN is consistent with class labels of training examples

– Objective function:

– Find the weights wi’s that minimize the above objective function e.g., backpropagation algorithm (see lecture notes)

2),( i

iii XwfYE

36Support Vector Machines

Find a linear hyperplane (decision boundary) that will separate the data

37Support Vector Machines

One Possible Solution

B1

38Support Vector Machines

Another possible solution

B2

39Support Vector Machines

Other possible solutions

B2

40Support Vector Machines

Which one is better? B1 or B2? How do you define better?

B1

B2

41Support Vector Machines

Find hyperplane maximizes the margin => B1 is better than B2

B1

B2

b11

b12

b21

b22

margin

42Support Vector Machines

B1

b11

b12

0 bxw

1 bxw 1 bxw

1bxw if1

1bxw if1)(

xf 2||||

2 Margin

w

43Support Vector Machines

We want to maximize:

– Which is equivalent to minimizing:

– But subjected to the following constraints:

This is a constrained optimization problem– Numerical approaches to solve it (e.g., quadratic programming)

2||||

2 Margin

w

1bxw if1

1bxw if1)(

i

i

ixf

2

||||)(

2wwL

44Support Vector Machines

What if the problem is not linearly separable?

45Support Vector Machines

What if the problem is not linearly separable?

– Introduce slack variables Need to minimize:

Subject to:

ii

ii

1bxw if1

-1bxw if1)(

ixf

N

i

kiC

wwL

1

2

2

||||)(

46Nonlinear Support Vector Machines

What if decision boundary is not linear?

47Nonlinear Support Vector Machines

Transform data into higher dimensional space

48Ensemble Methods

Construct a set of classifiers from the training data

Predict class label of previously unseen records by aggregating predictions made by multiple classifiers

49General Idea

OriginalTraining data

....D1D2 Dt-1 Dt

D

Step 1:Create Multiple

Data Sets

C1 C2 Ct -1 Ct

Step 2:Build Multiple

Classifiers

C*Step 3:

CombineClassifiers

50Why does it work?

Suppose there are 25 base classifiers

– Each classifier has error rate, = 0.35

– Assume classifiers are independent

– Probability that the ensemble classifier makes a wrong prediction:

25

13

25 06.0)1(25

i

ii

i

51Examples of Ensemble Methods

How to generate an ensemble of classifiers?

– Bagging

– Boosting

52Bagging

Sampling with replacement

Build classifier on each bootstrap sample

Each sample has probability (1 – 1/n)n of being selected

Original Data 1 2 3 4 5 6 7 8 9 10Bagging (Round 1) 7 8 10 8 2 5 10 10 5 9Bagging (Round 2) 1 4 9 1 2 3 2 7 3 2Bagging (Round 3) 1 8 5 10 5 5 9 6 3 7

53Boosting

An iterative procedure to adaptively change distribution of training data by focusing more on previously misclassified records

– Initially, all N records are assigned equal weights

– Unlike bagging, weights may change at the end of boosting round

54Boosting

Records that are wrongly classified will have their weights increased

Records that are classified correctly will have their weights decreased

Original Data 1 2 3 4 5 6 7 8 9 10Boosting (Round 1) 7 3 2 8 7 9 4 10 6 3Boosting (Round 2) 5 4 9 4 2 5 1 7 4 2Boosting (Round 3) 4 4 8 10 4 5 4 6 3 4

• Example 4 is hard to classify

• Its weight is increased, therefore it is more likely to be chosen again in subsequent rounds

55Example: AdaBoost

Base classifiers: C1, C2, …, CT

Error rate:

Importance of a classifier:

N

jjjiji yxCw

N 1

)(1

i

ii

1ln

2

1

56Example: AdaBoost

Weight update:

If any intermediate rounds produce error rate higher than 50%, the weights are reverted back to 1/n and the resampling procedure is repeated

Classification:

factor ionnormalizat theis where

)( ifexp

)( ifexp)()1(

j

iij

iij

j

jij

i

Z

yxC

yxC

Z

ww

j

j

T

jjj

yyxCxC

1

)(maxarg)(*

57

BoostingRound 1 + + + -- - - - - -

0.0094 0.0094 0.4623B1

= 1.9459

Illustrating AdaBoost

Data points for training

Initial weights for each data point

OriginalData + + + -- - - - + +

0.1 0.1 0.1

58Illustrating AdaBoost

BoostingRound 1 + + + -- - - - - -

BoostingRound 2 - - - -- - - - + +

BoostingRound 3 + + + ++ + + + + +

Overall + + + -- - - - + +

0.0094 0.0094 0.4623

0.3037 0.0009 0.0422

0.0276 0.1819 0.0038

B1

B2

B3

= 1.9459

= 2.9323

= 3.8744

59

Rule-Based Classifier