Data mining classification-2009-v0

Data Mining

Classification

Prithwis Mukerjee, Ph.D.

Prithwis Mukerjee 2

Classification

Definition The separation or ordering of objects ( or things ) in

classes

A Priori Classification When the classification is done before you have

looked at the data

Post Priori Classification When the classification is done after you have

looked at the data

Prithwis Mukerjee 3

General approach

You decide on the classes without looking at the data For example : High risk, medium risk, low risk classes

You “train” system Take a small set of objects – the training set

Each object has a set of attributes Classify the objects in this small (“training”) set into

the three classes, without looking at the attributes You will need human expertise here, to classify objects

Now find a set of rules based on the attributes such that the system classifies the objects just as you have done without looking at the attributes

Use these rules to classify the full set of attributes

Prithwis Mukerjee 4

If we have this data ...

Name Eggs Pouch Flies Feathers ClassCockatoo Yes No Yes Yes Bird

No No No No MammalYes Yes No No Marsupial

Emu Yes No No Yes BirdKangaroo No Yes No No Marsupial

Koala No Yes No No MarsupialYes No Yes Yes Bird

Owl Yes No Yes Yes BirdPenguin Yes No No Yes BirdPlatypus Yes No No No MammalPossum No Yes No No MarsupialWombat No Yes No No Marsupial

DugongEchidna

Kokkabura

Prithwis Mukerjee 5

We need to build a decision tree like ....

Pouch ?Pouch ?

Feathers ?Feathers ?

Bird Mammal

Marsupial

YES

YES

NO

NO

Prithwis Mukerjee 6

Question is ...

Why did we ignore two attributes ? Flies ? Feathers ?

Why did we use the attribute called POUCH first ? And then we used the

attribute called FEATHERS

A rigorous classification process should tell us If there are lots of

attributes to be looked at then which are the important ones ?

In which order should we look at the attributes

So that the classification arrived at is very similar to the classification done with the training set

Prithwis Mukerjee 7

Decision Tree : Tree Induction Algorithm

Step 1 : Place all members into one node If all members belong to the same class

Stop : there is nothing to be done

Step 2 : Else Choose one attribute and based on its value split

the node into two nodes For each of the two nodes

If all members belong to the same class Stop

Else : Recursively go to Step 1

Big question : How do you choose which attribute to split a node on ? Information Theory GINI Index

Prithwis Mukerjee 8

Information Theory : Recapitulate

Information Content I Of an event E That has n possible outcomes Where outcome i happens with probability pi

Is defined as I = Σi ( - pi log2 pi )

Example : Event EA has two possible outcomes

P1 = 0, P2 = 0 : Outcome 1 is a certainty

I = 0 because there is NO information in the outcome Event EB has two possible outcomes

P1 = 0.5, P2 = 0.5 : Both outcomes are equally likely

I = -0.5 log2(0.5) – 0.5 log2(0.5) = 1

Maximum possible information that is possible for an event with two outcomes

Prithwis Mukerjee 9

Information in the roll of a dice

Fair dice All numbers 1 – 6 equally probable ( pi = 1/6)

I = 6 x (- 1/6) log2(1/6) = 2.585

Loaded Dice Case 1 P6 = 0.5; P1 = P2 = P3 = P4 = P5 = 0.1

I = 5 x (-0.1) log2(0.1) – 0.5 x log2(0.5) = 2.16

Loaded Dice Case 2 P6 = 0.75; P1 = P2 = P3 = P4 = P5 = 0.05

I = 5 x (-0.05) log2(0.1) – 0.75 x log2(0.75) = 1.39

Point to note ... We can change the information in the roll of the

dice by changing the probabilities of the various outcomes !

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How do we change the information ?

In a dice We make mechanical

modifications so that the probabilities of each outcome changes This is higly illegal

In a set of individuals We regroup the

individuals into the classes so that the probability of each class changes This is highly permitted

in our algorithm

H


Consider the following scenario ..

Probability of each outcome ( or class ) P(A) = 3/10 , P(B) = 3/10 , P(C) = 4/10

Total Information Content of Set S -(3/10) log2(3/10) – (3/10) log2(3/10) – (4/10) log2(4/10) =

1.57

ID Home Married Gender Employed Credit Class1 Yes Yes Male Yes A B2 No No Female Yes A A3 Yes Yes Female Yes B C4 Yes No Male No B B5 No Yes Female Yes B C6 No No Female Yes B A7 No No Male No B B8 Yes No Female Yes A A9 No Yes Female Yes A C

10 Yes Yes Female Yes A C


Suppose we split this set on HOME

I1 : Information in set S1

-(2/5)log2(2/5) – (1/5) log2(1/5) – (2/5) log2(2/5) = 1.52


-(1/5)log2(1/5) – (2/5) log2(2/5) – (2/5) log2(2/5) = 1.52

Total Information in S1 and S2 0.5 I1 + 0.5I2 = 0.5 x 1.52 + 0.5 x 1.52 = 1.52

ID Home Married Gender Employed Credit Class2 No No Female Yes A A5 No Yes Female Yes B C6 No No Female Yes B A7 No No Male No B B9 No Yes Female Yes A C

ID Home Married Gender Employed Credit Class1 Yes Yes Male Yes A B3 Yes Yes Female Yes B C4 Yes No Male No B B8 Yes No Female Yes A A


P1(A) = 2/5

P1(B) = 1/5

P1(C) = 2/5

P2(A) = 1/5

P2(B) = 2/5

P2(C) = 2/5


Impact of HOME attribute

In sets S1 and S2, the attribute HOME was the same

But in set S the attribute HOME is not the same and so is of some significance

What is the significance of the HOME attribute ?

By adding the HOME attribute we have increased the information content FROM : 1.52 TO : 1.57

So HOME attribute adds 0.05 to the overall information content Or HOME attribute

reduces uncertainty by 0.05


Let us go back to the original set S ..

Probability of each outcome ( or class ) P(A) = 3/10 , P(B) = 3/10 , P(C) = 4/10

Total Information Content of Set S -(3/10) log2(3/10) – (3/10) log2(3/10) – (4/10) log2(4/10) =

1.57

ID Home Married Gender Employed Credit Class1 Yes Yes Male Yes A B2 No No Female Yes A A3 Yes Yes Female Yes B C4 Yes No Male No B B5 No Yes Female Yes B C6 No No Female Yes B A7 No No Male No B B8 Yes No Female Yes A A9 No Yes Female Yes A C



This time we split on GENDER


-(3/7)log2(3/7) – (4/7) log2(4/7) = 0.985


= 0

Total Information in S1 and S2 (7/10) I1 + (3/10)I2 = 7/10 x 0.985 + 3/10 x 0 = 0.69

ID Home Married Gender Employed Credit Class2 No No Female Yes A A3 Yes Yes Female Yes B C5 No Yes Female Yes B C6 No No Female Yes B A8 Yes No Female Yes A A9 No Yes Female Yes A C


ID Home Married Gender Employed Credit Class1 Yes Yes Male Yes A B4 Yes No Male No B B7 No No Male No B B

P1(A) = 3/7

P1(B) = 0/7

P1(C) = 4/7

P2(A) = 0/3

P2(B) = 3/3

P2(C) = 0/3


Impact of GENDER attribute

In sets S1 and S2, the attribute GENDER was the same

But in set S the attribute GENDER is not the same and so is of some significance

What is the significance of the GENDER attribute ?

By adding the GENDER attribute we have increased the information content FROM : 0.69 TO : 1.57

So GENDER attribute adds 0.88 to the overall information content Or GENDER attribute

reduces uncertainty by 0.88


If we were to do this for all attributes ...

We would observe that GENDER is the best candidate for the split

Attribute

Home 1.57 1.52 0.05Married 1.57 0.85 0.72Gender 1.57 0.69 0.88

Employed 1.57 1.12 0.45Credit 1.57 1.52 0.05

Information before Split

Information after Split

Information Gain


And the first part of our tree would be ...

GenderGender

What Next ?What Next ? Class B

MaleFemale


Remove GENDER and Class B and continue

ID Home Married Employed Credit Class

2 No No Yes A A3 Yes Yes Yes B C

5 No Yes Yes B C6 No No Yes B A

8 Yes No Yes A A9 No Yes Yes A C10 Yes Yes Yes A C

Probability of each outcome ( or class ) P(A) = 3/7 , P(C) = 4/7

Total Information Content of Set S -(3/7) log2(3/7) – (4/7) log2(4/7) = 1.33


We split this set on HOME ...


-(2/4)log2(2/4) – (2/4) log2(2/4) = 1.00


-(1/3)log2(1/3) – (2/3) log2(2/3) = 0.92

Total Information in S1 and S2 (4/7) I1 + (3/7)I2 = 4/7 x 1.00 + 3/7 x 0.92 = 0.96

ID Home Married Employed Credit Class2 No No Yes A A5 No Yes Yes B C6 No No Yes B A9 No Yes Yes A C

ID Home Married Employed Credit Class3 Yes Yes Yes B C8 Yes No Yes A A10 Yes Yes Yes A C

P1(A) = 2/4

P1(C) = 2/4

P1(A) = 1/3

P1(C) = 2/3

Gain = 1.33 – 0.96= 0.37


But if we were to split on MARRIED


= 0.0


= 0.0

Total Information in S1 and S2 = 0.0

ID Home Married Employed Credit Class2 No No Yes A A8 Yes No Yes A A6 No No Yes B A

ID Home Married Employed Credit Class3 Yes Yes Yes B C9 No Yes Yes A C10 Yes Yes Yes A C5 No Yes Yes B C

P1(A) = 4/4

P1(C) = 0/4

P1(A) = 0/3

P1(C) = 3/3

Gain = 1.33 - 0= 1.33


Two things have happened

With MARRIED We have hit the upper limit of information gain No other attribute can do any better than this

In The TWO sub sets All members belong to the same class

Either A or C

Hence we STOP here and observe ...


That our DECISION TREE looks like

GenderGender

MarriedMarried

Class C Class A

Class B

Male

YES

Female

NO

Data mining classification-2009-v0

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