Transcript Support Vector Machines

Data Mining
Classification: Alternative Techniques
Lecture Notes for Chapter 5
Introduction to Data Mining
by
Tan, Steinbach, Kumar
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
1
Instance-Based Classifiers
Set of Stored Cases
Atr1
……...
AtrN
Class
A
• Store the training records
• Use training records to
predict the class label of
unseen cases
B
B
C
A
Unseen Case
Atr1
……...
AtrN
C
B
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
‹#›
Instance Based Classifiers

Examples:
– Rote-learner
Memorizes entire training data and performs
classification only if attributes of record match one of
the training examples exactly

– Nearest neighbor
Uses k “closest” points (nearest neighbors) for
performing classification

© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
‹#›
Nearest Neighbor Classifiers

Basic idea:
– If it walks like a duck, quacks like a duck, then
it’s probably a duck
Compute
Distance
Training
Records
© Tan,Steinbach, Kumar
Test
Record
Choose k of the
“nearest” records
Introduction to Data Mining
4/18/2004
‹#›
Nearest-Neighbor Classifiers
Unknown record

Requires three things
– The set of stored records
– Distance Metric to compute
distance between records
– The value of k, the number of
nearest neighbors to retrieve

To classify an unknown record:
– Compute distance to other
training records
– Identify k nearest neighbors
– Use class labels of nearest
neighbors to determine the
class label of unknown record
(e.g., by taking majority vote)
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
‹#›
Definition of Nearest Neighbor
X
(a) 1-nearest neighbor
X
X
(b) 2-nearest neighbor
(c) 3-nearest neighbor
K-nearest neighbors of a record x are data points
that have the k smallest distance to x
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
‹#›
1 nearest-neighbor
Voronoi Diagram
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
‹#›
Nearest Neighbor Classification

Compute distance between two points:
– Euclidean distance
d (r , s) 
 (r  s )
i
2
i
i

Determine the class from nearest neighbor list
– take the majority vote of class labels among
the k-nearest neighbors
– Weigh the vote according to distance

weight factor, w = 1/d2
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
‹#›
Nearest Neighbor Classification…

Choosing the value of k:
– If k is too small, sensitive to noise points
– If k is too large, neighborhood may include points from
other classes
X
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
‹#›
Nearest Neighbor Classification…

Scaling issues
– Attributes may have to be scaled to prevent
distance measures from being dominated by
one of the attributes
– Example:
height of a person may vary from 1.5m to 1.8m
 weight of a person may vary from 90lb to 300lb
 income of a person may vary from $10K to $1M

© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
‹#›
Nearest Neighbor Classification…

Problem with Euclidean measure:
– High dimensional data

curse of dimensionality

Solution: Normalize the vectors to unit length
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
‹#›
Nearest neighbor Classification…

k-NN classifiers are lazy learners
– It does not build models explicitly
– Unlike eager learners such as decision tree
induction and rule-based systems
– Classifying unknown records are relatively
expensive
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
‹#›
Example: PEBLS

PEBLS: Parallel Examplar-Based Learning
System (Cost & Salzberg)
– Works with both continuous and nominal
features
For
nominal features, distance between two
nominal values is computed using modified value
difference metric (MVDM)
– Each record is assigned a weight factor
– Number of nearest neighbor, k = 1
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
‹#›
Example: PEBLS
Tid Refund Marital
Status
Taxable
Income Cheat
1
Yes
Single
125K
No
d(Single,Married)
2
No
Married
100K
No
= | 2/4 – 0/4 | + | 2/4 – 4/4 | = 1
3
No
Single
70K
No
d(Single,Divorced)
4
Yes
Married
120K
No
5
No
Divorced 95K
Yes
6
No
Married
No
d(Married,Divorced)
7
Yes
Divorced 220K
No
= | 0/4 – 1/2 | + | 4/4 – 1/2 | = 1
8
No
Single
85K
Yes
d(Refund=Yes,Refund=No)
9
No
Married
75K
No
= | 0/3 – 3/7 | + | 3/3 – 4/7 | = 6/7
10
No
Single
90K
Yes
60K
Distance between nominal attribute values:
= | 2/4 – 1/2 | + | 2/4 – 1/2 | = 0
10
Marital Status
Class
Refund
Single
Married
Divorced
Yes
2
0
1
No
2
4
1
© Tan,Steinbach, Kumar
Class
Yes
No
Yes
0
3
No
3
4
Introduction to Data Mining
d (V1 ,V2 )  
i
4/18/2004
n1i n2i

n1 n2
‹#›
Example: PEBLS
Tid Refund Marital
Status
Taxable
Income Cheat
X
Yes
Single
125K
No
Y
No
Married
100K
No
10
Distance between record X and record Y:
d
( X , Y )  wX wY  d ( X i , Yi )
2
i 1
where:
Number of times X is used for prediction
wX 
Number of times X predicts correctly
wX  1 if X makes accurate prediction most of the time
wX > 1 if X is not reliable for making predictions
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
‹#›
Bayes Classifier
A probabilistic framework for solving classification
problems
 Conditional Probability:
P ( A, C )
P (C | A) 
P ( A)

P ( A, C )
P( A | C ) 
P (C )

Bayes theorem:
P( A | C ) P(C )
P(C | A) 
P( A)
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
‹#›
Example of Bayes Theorem

Given:
– A doctor knows that meningitis causes stiff neck 50% of the
time
– Prior probability of any patient having meningitis is 1/50,000
– Prior probability of any patient having stiff neck is 1/20

If a patient has stiff neck, what’s the probability
he/she has meningitis?
P( S | M ) P( M ) 0.5 1 / 50000
P( M | S ) 

 0.0002
P( S )
1 / 20
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
‹#›
Bayesian Classifiers
Consider each attribute and class label as random
variables
 Given a record with attributes (A1, A2,…,An)=A
– Goal is to predict class C
– Specifically, we want to find the value of C that
maximizes P(C= cj | A=a )
Maximum posterior classifier:
optimal=minimizes error probability
 Can we estimate P(C= cj | A=a ) directly from data?

© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
‹#›
Bayesian Classifiers

Approach:
– compute the posterior probability P(C= cj | A=a ) for all
values cj of C using the Bayes theorem
P(C  c j | A  a) 
P( A  a | C  c j ) P(C  c j )
P( A  a )
– Choose value of C that maximizes
P(C= cj | A=a )
– Equivalent to choosing value of C that maximizes
P(A=a|C= cj) P(C= cj)

How to estimate likelihood P(A=a|C= cj)?
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
‹#›
Naïve Bayes Classifier

Assume independence among attributes Ai when class is
given:
P(A=a|C=cj)=P(A1=a1|C=cj)P(A2=a2|C=cj)…P(An=an|C=cj)
Can estimate P(Ai=ai|C=cj) for all Ai and cj.
New point is classified to cj
if P(C=cj)  P(Ai=ai|C=cj) is maximal.
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
‹#›
How to Estimate
Probabilities
from
Data?
l
l
c
Tid
at
Refund
o
eg
a
c
i
r
c
at
o
eg
a
c
i
r
c
on
u
it n
s
u
o
s
s
a 
cl
Marital
Status
Taxable
Income
Evade
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
Class: P(C) = Nc/N
– e.g., P(No) = 7/10,
P(Yes) = 3/10

For discrete attributes:
P(Ai | Ck) = |Aik|/ Nc k
– where |Aik| is number of
instances having attribute
Ai and belongs to class Ck
– Examples:
10
© Tan,Steinbach, Kumar
Introduction to Data Mining
P(Status=Married|No) = 4/7
P(Refund=Yes|Yes)=0
4/18/2004
‹#›
Naïve Bayes Classifier
If one of the conditional probability is zero, then
the entire expression becomes zero
 Probability estimation:

N ij
ˆ
Original : P( Ai  ai | C  c j ) 
Nj
si : number of values of Ai
N ij  1
ˆ
Laplace : P( Ai  ai | C  c j ) 
N j  si
m: parameter
p(ai): prior probability
N ij  mp(ai )
ˆ
m - estimate : P( Ai  ai | C  c j ) 
Nj m
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
‹#›
How to Estimate Probabilities from Data?

For continuous attributes:
– Discretize the range into bins
one ordinal attribute per bin
 violates independence assumption

k
– Two-way split: (A < v) or (A > v)

choose only one of the two splits as new attribute
– Probability density estimation:
Assume attribute follows a normal distribution
 Use data to estimate parameters of distribution
(e.g., mean and standard deviation)

© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
‹#›
l
s
al
uProbabilities
ca
c
How tooriEstimate
from
Data?
i
o
r
u
o
c
Tid
e
at
Refund
g
c
e
at
g
c
t
n
o
in
as
l
c
Marital
Status
Taxable
Income
Evade
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
s

Normal distribution:
f (ai | c j ) 

1
2
2
ij
( ai   ij ) 2
e
2 ij2
– One for each (Ai,cj) pair

For (Income, Class=No):
– If Class=No

sample mean = 110

sample variance = 2975
10
1
f ( Income  120 | No) 
e
2 (54.54)
© Tan,Steinbach, Kumar
Introduction to Data Mining
(120110) 2

2 ( 2975)
 0.0072
4/18/2004
‹#›
Example of Naïve Bayes Classifier
Given a Test Record:
X  (Refund  No, Married, Income  120K)
with original estimate:
naive Bayes Classifier:
P(Refund=Yes|No) = 3/7
P(Refund=No|No) = 4/7
P(Refund=Yes|Yes) = 0
P(Refund=No|Yes) = 1
P(Marital Status=Single|No) = 2/7
P(Marital Status=Divorced|No)=1/7
P(Marital Status=Married|No) = 4/7
P(Marital Status=Single|Yes) = 2/7
P(Marital Status=Divorced|Yes)=1/7
P(Marital Status=Married|Yes) = 0
For taxable income:
If class=No:
sample mean=110
sample variance=2975
If class=Yes: sample mean=90
sample variance=25
© Tan,Steinbach, Kumar

P(X|Class=No) = P(Refund=No|Class=No)
 P(Married| Class=No)
 f(Income=120K| Class=No)
= 4/7  4/7  0.0072 = 0.0024

P(X|Class=Yes) = P(Refund=No| Class=Yes)
 P(Married| Class=Yes)
 f(Income=120K| Class=Yes)
= 1  0  1.2  10-9 = 0
Since P(X|No)P(No) > P(X|Yes)P(Yes)
Therefore P(No|X) > P(Yes|X)
=> Class = No
Introduction to Data Mining
4/18/2004
‹#›
Example of Naïve Bayes Classifier
with Laplace estimate:
Name
human
python
salmon
whale
frog
komodo
bat
pigeon
cat
leopard shark
turtle
penguin
porcupine
eel
salamander
gila monster
platypus
owl
dolphin
eagle
Give Birth
yes
Give Birth
yes
no
no
yes
no
no
yes
no
yes
yes
no
no
yes
no
no
no
no
no
yes
no
Can Fly
no
no
no
no
no
no
yes
yes
no
no
no
no
no
no
no
no
no
yes
no
yes
Can Fly
no
© Tan,Steinbach, Kumar
Live in Water Have Legs
no
no
yes
yes
sometimes
no
no
no
no
yes
sometimes
sometimes
no
yes
sometimes
no
no
no
yes
no
Class
yes
no
no
no
yes
yes
yes
yes
yes
no
yes
yes
yes
no
yes
yes
yes
yes
no
yes
mammals
non-mammals
non-mammals
mammals
non-mammals
non-mammals
mammals
non-mammals
mammals
non-mammals
non-mammals
non-mammals
mammals
non-mammals
non-mammals
non-mammals
mammals
non-mammals
mammals
non-mammals
Live in Water Have Legs
yes
no
Class
?
Introduction to Data Mining
A: attributes
M: mammals
N: non-mammals
7 7 3 3
P( A | M )      0.0604
9 9 10 9
2 11 4 5
P( A | N )      0.0081
15 15 16 15
7
P( A | M ) P( M )  0.0602 
 0.02114
20
13
P( A | N ) P( N )  0.0081
 0.005265
20
P(A|M)P(M) > P(A|N)P(N)
=> Mammals
4/18/2004
‹#›
Naïve Bayes (Summary)

Robust to isolated noise points

Handle missing values by ignoring the instance
during probability estimate calculations

Robust to irrelevant attributes

Independence assumption may not hold for some
attributes
– Use other techniques such as Bayesian Belief
Networks (BBN)
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
‹#›
Artificial Neural Networks (ANN)
X1
X2
X3
Y
Input
1
1
1
1
0
0
0
0
0
0
1
1
0
1
1
0
0
1
0
1
1
0
1
0
0
1
1
1
0
0
1
0
X1
Black box
Output
X2
Y
X3
Output Y is 1 if at least two of the three inputs are equal to 1.
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
‹#›
Artificial Neural Networks (ANN)
X1
X2
X3
Y
1
1
1
1
0
0
0
0
0
0
1
1
0
1
1
0
0
1
0
1
1
0
1
0
0
1
1
1
0
0
1
0
Input
nodes
Black box
X1
Output
node
0.3
0.3
X2
X3
0.3

Y
t=0.4
Y  I (0.3 X 1  0.3 X 2  0.3 X 3  0.4  0)
1
where I ( z )  
0
© Tan,Steinbach, Kumar
if z is true
otherwise
Introduction to Data Mining
4/18/2004
‹#›
Artificial Neural Networks (ANN)

Model is an assembly of
inter-connected nodes
and weighted links
Input
nodes
Black box
X1
w1
w2
X2


Output node sums up
each of its input value
according to the weights
of its links
Compare output node
against some threshold t
Output
node

Y
w3
X3
t
Perceptron Model
yˆ  I ( wi xi  t )
or
i
yˆ  sgn(  wi xi  t )  sgn( w  x)
i
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
‹#›
Perceptron algorithm
Minden x attributumvektort kiegészítjük egy d+1-edik értékkel (mindig 1)
Legyen w=(0,0,...0)
while van helytelenül klasszifikált eleme a tanító adathalmaznak
for all x if x rosszul klasszifikált then
if x az első osztályba tartozik then w=w+x
else w=w-x
Lineárisan szeparálható osztályok esetén a perceptron tanulás véges
iteráció után megáll.
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
‹#›
Linearly separable
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
‹#›
Not linearly separable
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
‹#›
General Structure of ANN
x1
x2
x3
Input
Layer
x4
x5
Input
I1
I2
Hidden
Layer
I3
Neuron i
Output
wi1
wi2
wi3
Si
Activation
function
g(Si )
Oi
Oi
threshold, t
Output
Layer
Training ANN means learning
the weights of the neurons
y
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
‹#›
Activation function
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
‹#›
Algorithm 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
2
– Objective function: E   Yi  f ( wi , X i )
i
– Find the weights wi’s that minimize the above
objective function

e.g., backpropagation algorithm (see lecture notes)
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
‹#›
Support Vector Machines

Find a linear hyperplane (decision boundary) that will separate the data
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
‹#›
Support Vector Machines
B1

One Possible Solution
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
‹#›
Support Vector Machines
B2

Another possible solution
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
‹#›
Support Vector Machines
B2

Other possible solutions
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
‹#›
Support Vector Machines
B1
B2


Which one is better? B1 or B2?
How do you define better?
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
‹#›
Support Vector Machines
B1
B2
b21
b22
margin
b11
b12

Find hyperplane maximizes the margin => B1 is better than B2
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
‹#›
Support Vector Machines
B1
 
w x  b  0
 
w  x  b  1
 
w  x  b  1
b11
 
if w  x  b  1
1

f ( x)  
 
 1 if w  x  b  1
© Tan,Steinbach, Kumar
Introduction to Data Mining
b12
2
Margin   2
|| w ||
4/18/2004
‹#›
Support Vector Machines

We want to maximize:
2
Margin   2
|| w ||
 2
|| w ||
– Which is equivalent to minimizing: L( w) 
2
– But subjected to the following constraints:
 
if w  x i  b  1
1

f ( xi )  
 
 1 if w  x i  b  1

This is a constrained optimization problem
– Numerical approaches to solve it (e.g., quadratic programming)
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
‹#›
Support Vector Machines

What if the problem is not linearly separable?
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
‹#›
Support Vector Machines

What if the problem is not linearly separable?
– Introduce slack variables
 2
 Need to minimize:
|| w ||
 N k
L( w) 
 C   i 
2
 i 1 

Subject to:
 
if w  x i  b  1 - i
1

f ( xi )  
 
 1 if w  x i  b  1  i
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
‹#›
Nonlinear Support Vector Machines

What if decision boundary is not linear?
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
‹#›
Nonlinear Support Vector Machines

Transform data into higher dimensional space
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
‹#›
Ensemble Methods

Construct a set of classifiers from the training
data

Predict class label of previously unseen records
by aggregating predictions made by multiple
classifiers
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
‹#›
General Idea
D
Step 1:
Create Multiple
Data Sets
Step 2:
Build Multiple
Classifiers
D1
D2
C1
C2
Step 3:
Combine
Classifiers
© Tan,Steinbach, Kumar
....
Original
Training data
Dt-1
Dt
Ct -1
Ct
C*
Introduction to Data Mining
4/18/2004
‹#›
Why 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  i
25i

(
1


)
 0.06



 i 
i 13 

25
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
‹#›
Examples of Ensemble Methods

How to generate an ensemble of classifiers?
– Bagging
– Boosting
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
‹#›
Bagging

Sampling with replacement
Original Data
Bagging (Round 1)
Bagging (Round 2)
Bagging (Round 3)
1
7
1
1
2
8
4
8
3
10
9
5
4
8
1
10
5
2
2
5
6
5
3
5
7
10
2
9
8
10
7
6
9
5
3
3
10
9
2
7

Build classifier on each bootstrap sample

Each sample has probability (1 – 1/n)n of being
selected
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
‹#›
Boosting

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
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
‹#›
Boosting
Records that are wrongly classified will have their
weights increased
 Records that are classified correctly will have
their weights decreased

Original Data
Boosting (Round 1)
Boosting (Round 2)
Boosting (Round 3)
1
7
5
4
2
3
4
4
3
2
9
8
4
8
4
10
5
7
2
4
6
9
5
5
7
4
1
4
8
10
7
6
9
6
4
3
10
3
2
4
• Example 4 is hard to classify
• Its weight is increased, therefore it is more
likely to be chosen again in subsequent rounds
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
‹#›
Example: AdaBoost

Base classifiers: C1, C2, …, CT

Error rate:
1
i 
N

 w  C ( x )  y 
N
j 1
j
i
j
j
Importance of a classifier:
1  1  i 

i  ln 
2  i 
© Tan,Steinbach, Kumar
Introduction to Data Mining
4/18/2004
‹#›
Example: AdaBoost

Weight update:
 j

if C j ( xi )  yi
w exp
( j 1)
wi


j
Zj 
exp
if C j ( xi )  yi

where Z j is the normalizat ion factor
( j)
i
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:
T

C * ( x )  arg max  j C j ( x )  y 
y
© Tan,Steinbach, Kumar
Introduction to Data Mining
j 1
4/18/2004
‹#›
Illustrating AdaBoost
Initial weights for each data point
Original
Data
0.1
0.1
0.1
+++
- - - - -
++
Data points
for training
B1
0.0094
Boosting
Round 1
+++
© Tan,Steinbach, Kumar
0.0094
0.4623
- - - - - - -
Introduction to Data Mining
 = 1.9459
4/18/2004
‹#›
Illustrating AdaBoost
B1
0.0094
Boosting
Round 1
0.0094
+++
0.4623
- - - - - - -
 = 1.9459
B2
Boosting
Round 2
0.0009
0.3037
- - -
- - - - -
0.0422
++
 = 2.9323
B3
0.0276
0.1819
0.0038
Boosting
Round 3
+++
++ ++ + ++
Overall
+++
- - - - -
© Tan,Steinbach, Kumar
Introduction to Data Mining
 = 3.8744
++
4/18/2004
‹#›