math 5364 notes chapter 5: alternative classification techniques

Math 5364 NotesChapter 5: Alternative Classification Techniques

Jesse Crawford

Department of MathematicsTarleton State University

Today's Topics

• The k-Nearest Neighbors Algorithm

• Methods for Standardizing Data in R

• The class package, knn, and knn.cv

k-Nearest Neighbors

• Divide data into training and test data.• For each record in the test data

• Find the k closest training records• Find the most frequently occurring

class label among them• The test record is classified into that

category• Ties are broken at random

• Example• If k = 1, classify green point as p• If k = 3, classify green point as n• If k = 2, classify green point as p

or n (chosen randomly)

k-Nearest Neighbors Algorithm

Algorithm depends on a distance metric d

Euclidean Distance Metric

1 2 1 2

1 2 1 2

2 211 21 1 2

( , )

( ) ( )

( () )p p

d x x x x

x x x x

x xx x

‖ ‖

=

1 11 1

2 21 2

( , , )

( , , )

p

p

x

x

x x

x x

Example 1• x = (percentile rank, SAT)• x1 = (90, 1300)• x2 = (85, 1200)• d(x1, x2) = 100.12

Example 2• x1 = (70, 950)• x2 = (40, 880)• d(x1, x2) = 76.16

• Euclidean distance is sensitive to measurement scales.• Need to standardize variables!

Standardizing Variables

1

1

Data set , ,

sample mean

sample standard deviation

Then , , has mean 0

and standard deviation 1

n

ii

n

x x

x

s

x xz

s

z z

mean percentile rank = 67.04st dev percentile rank = 18.61

mean SAT = 978.21st dev SAT = 132.35

Example 1• x = (percentile rank, SAT)• x1 = (90, 1300)• x2 = (85, 1200)• z1 = (1.23, 2.43)• z2 = (0.97, 1.68)• d(z1, z2) = 0.80

Example 2• x1 = (70, 950)• x2 = (40, 880)• z1 = (0.16, -0.21)• z2 = (-1.45, -0.74)• d(z1, z2) = 1.70

Standardizing iris Data

x=iris[,1:4]xbar=apply(x,2,mean)xbarMatrix=cbind(rep(1,150))%*%xbars=apply(x,2,sd)sMatrix=cbind(rep(1,150))%*%s

z=(x-xbarMatrix)/sMatrixapply(z,2,mean)apply(z,2,sd)

plot(z[,3:4],col=iris$Species)

Another Way to Split Data

#Split iris into 70% training and 30% test data.set.seed=5364train=sample(nrow(z),nrow(z)*.7)

z[train,] #This is the training dataz[-train,] #This is the test data

The class Package and knn Function

library(class)

Species=iris$SpeciespredSpecies=knn(train=z[train,],test=z[-train,],cl=Species[train],k=3)

confmatrix(Species[-train],predSpecies)

Accuracy = 93.33%

Leave-one-out CV with knn

predSpecies=knn.cv(train=z,cl=Species,k=3)confmatrix(Species,predSpecies)

CV estimate for accuracy is 94.67%

Optimizing k with knn.cv

accvect=1:10

for(k in 1:10){ predSpecies=knn.cv(train=z,cl=Species,k=k) accvect[k]=confmatrix(Species,predSpecies)$accuracy}

which.max(accvect)

For binary classification problems, odd values of k avoid ties.

General Comments about k

• Smaller values of k result in greater model complexity.• If k is too small, model is sensitive to noise.• If k is too large, many records will start to be classified

simply into the most frequent class.

Today's Topics

• Weighted k-Nearest Neighbors Algorithm

• Kernels

• The kknn package

• Minkowski Distance Metric

Indicator Functions

True) 1

False) 0

Example:

Assume 5

( 10) 1

(x 20) 0

(

(

I

x

I x

I

I

max and argmax

2Let ( ) 10 ( 4)

max

arg m

(

ax

) 10

( ) 4

x

x

f x

f

f x x

x

k-Nearest Neighbors Algorithm

Kernel Functions

23(1( ) ) (| | 1)

4K Id d d

2 215(1 ) (| | 1)

16( ) d IK d d

Weighted k-Nearest Neighbors

kknn Package

• train.kknn uses leave-one-out cross-validation to optimize k and the kernel

• kknn gives predictions for a specific choice of k and kernel (see R script)

• R Documentation

http://cran.r-project.org/web/packages/kknn/kknn.pdf

• Hechenbichler, K. and Schliep, K.P. (2004) "Weighted k-Nearest-Neighbor Techniques and Ordinal Classification".

http://epub.ub.uni-muenchen.de/1769/1/paper_399.pdf



Minkowski Distance Metric

12

1

2

1

1

Euclidian Distance

( , ) ( )

Minkowski Distance

( , ) ( )q

p

i j is jss

pq

i j is jss

d x x x x

d x x x x

Euclidean distance is Minkowski distance with q = 2

Today's Topics

• Naïve Bayes Classification

HouseVotes84 Data

• Want to calculate

P(Y = Republican | X1 = no, X2 = yes, …, X16 = yes)

• Possible Method• Look at all records where X1 = no, X2 = yes, …., X16 = yes• Calculate the proportion of those records with Y = Republican

• Problem: There are 216 = 65,536 combinations of Xj's, but only 435 records

• Possible solution: Use Bayes' Theorem

Setting for Naïve Bayes

1 1 1

1 1 1

( ) ( ),

| | )

(

|

for 1,2, ,

( , , ) ( , ,

| ) ( | )

( , , ( | ) ( | ))

p p p

j j j j

p p p

y c

P X x X x Y y f x x

x Y y f x y

f x x f

P Y y f y

y

P X

y x y f x y

p.m.f. for YPrior distribution for Y

Joint conditional distributionof Xj's given Y

Conditional distribution of Xj given Y

Assumption: Xj's are conditionally independent given Y

1 11 1

1 1

( , , ,, , )

( ,

)|

),( p p

p pp p

P Y y x X xx X

Xx

P x X xP Y y X

X

1 1

1 11

( ) ( | )

( ) (

,

, |

,

, )

p p

c

p py

P Y y P X Y y

P Y P

x

X Y

X x

y x X x y

1

11

, ,( ) (

( ) (

| )

, , | )

p

c

py

xf y f x

f fy xx

y

y

1 11

1 11

( ) ( | )( |

( | ), , )

(( ) ( | ) | )

p pp c

p py

y f x yf y f xf y x

f f xx

y y f x y

Bayes' Theorem

Prior Probabilities

ConditionalProbabilities

Posterior Probability

How can we estimate prior probabilities?

), for Democrat, Republican

(Democrat) 0.614

(Republican) 0.38

(

6

y y

f

f

f

1 11

1 11

( ) ( | )( |

( | ), , )

(( ) ( | ) | )

p pp c

p py

y f x yf y f xf y x

f f xx

y y f x y

Prior Probabilities



How can we calculate posterior probabilities?

1

7

| , , ), for Democrat, Republican

(Democrat | n, y,n, y, y, y,n,n,n, y, NA, y, y, y,n, y) 1.03 10

(Republican | n, y,n, y, y, y,n,n,n, y, NA, y, y, y,n, y) 0.99999 9

(

9

px x y

f

f

f y

1 11

1 11

( ) ( | )( |

( | ), , )

(( ) ( | ) | )

p pp c

p py

y f x yf y f xf y x

f f xx

y y f x y

Naïve Bayes Classification

1 11

1 11

( ) ( | )( |

( | ), , )

(( ) ( | ) | )

p pp c

p py

y f x yf y f xf y x

f f xx

y y f x y

1 1

1

If , ,

ˆ arg max ( | , , )

p p

y p

X x X x

Y f y x x

7(Democrat | n, y, , y) 1.03 10

(Republican | n, y, , y) 0.9999999

ˆ Republican

f

f

Y

Naïve Bayes with Quantitative Predictors

2|

2||

3|setosa

3|setosa

Option 1: Assume has a certain type of

conditional distribution, e.g., normal

(1( | ) exp

2

Example: Iris data

)

2

1.462

0.174

j

j j yj j

j yj y

X

xf x y

Testing Normality

0 A

0

Shaprio-Wilk Test

H : Distribution is normal vs. H : Distribution is not normal

If -value , reject H (statistically significant evidence distribution is not normal)p

qq Plots• Straight line: evidence of normality• Deviates from straight line: evidence against normality

Naïve Bayes with Quantitative Predictors

Option 2: Discretize predictor variables using cut function.

(convert variable into a categorical variables by breaking into bins)

Today's Topics

• The Class Imbalance Problem

• Sensitivity, Specificity, Precision, and Recall

• Tuning probability thresholds

Class Imbalance Problem

Confusion MatrixPredicted Class

+ -ActualClass

+ f++ f+-

- f-+ f--

• Class Imbalance: One class is much less frequent than the other

• Rare class: Presence of an anomaly (fraud, disease, loan default, flight delay, defective product).• + Anomaly is present• - Anomaly is absent


+ -ActualClass

+ f++ (TP) f+- (FN)- f-+ (FP) f-- (TN)

• TP = True Positive

• FP = False Positive

• TN = True Negative

• FN = False Negative

ˆAccura

C

c

lassific

y (

ation Accurac

)

y

TP TNP Y Y

TP TN FP FN


+ -ActualClass

+ f++ (TP) f+- (FN)- f-+ (FP) f-- (TN)





True Positive Rate (Sensitivity)

True Negative Rate (Specificity)

ˆ( | )

ˆ( | )

TPTPR P Y Y

TP FN

TNTNR P Y Y

TN FP


+ -ActualClass

+ f++ (TP) f+- (FN)- f-+ (FP) f-- (TN)





False Positive Rate

False Negative Rate

ˆ( | )

ˆ( | )

FPFPR P Y Y

FP TN

FNFNR P Y Y

TP FN


+ -ActualClass

+ f++ (TP) f+- (FN)- f-+ (FP) f-- (TN)





Precision

Recall (Same as sensitivity)

ˆ( | )

ˆ( | )

TPp P Y Y

TP FP

TPr TPR P Y Y

TP FN

1

1 1 1

ˆ( | )

ˆ

Precision

Recal

( | )

2 2 2

2

l

measure

r p

TPp P Y Y

TP FP

TPr TPR P Y Y

TP FN

rp TPF

r p TP FP

F

FN

• F1 is the harmonic mean of p and r• Large values of F1 ensure reasonably large values of p and r

2

2

0

1 1

1 4

1 2 3 4

( 1)

= , where each 0

Accuracy, sensitivity

measure

Weighted Ac

, specificity, prec

cur

ision, recall, and are all

special cases of weight

ac

ed a

y

c

i

rpF

r p

F p

F F

wTP w TNw

wTP w FP w FN w

F

r

T

F

N

F

curacy

Probability Threshold

1 1 1ˆˆ ( , , ) ( | , , )

ˆˆ ( )

Typical Classification Scheme

ˆˆIf 0.5 then

ˆˆIf 0.5 then

p p pp x x P Y X x X x

p P Y

p Y

p Y


1 1 1ˆˆ ( , , ) ( | , , )

ˆˆ ( )

Typical classification scheme

ˆÎf 0.5 then

ˆÎf 0.5 then


p P Y

p Y

p Y

0

0

0

More general scheme

ˆÎf then

ˆÎf then

Probability threshold

p p Y

p p Y

p


1 1 1ˆˆ ( , , ) ( | , , )

ˆˆ ( )

Typical classification scheme

ˆÎf 0.5 then

ˆÎf 0.5 then


p P Y

p Y

p Y

0

0

0

More general scheme

ˆÎf then

ˆÎf then

Probability threshold

p p Y

p p Y

p

We can modify the probability threshold p0

to optimize performance metrics

Today's Topics

• Receiver Operating Curves (ROC)

• Cost Sensitive Learning

• Oversampling and Undersampling

Receiver Operating Curves (ROC)

• Plot of True Positive Rate vs False Positive Rate

• Plot of Sensitivity vs 1 – Specificity

• AUC = Area under curve

AUC = Area under curve

ˆ ˆAUC Percentage of the time that

when and

AUC 1

AUC 0.5 Worse than Random Guessing

AUC 0.5 Random Guessing

AUC 0.7 Acceptable Discrimination

0

i j

i j

p p

Y Y

AUC 0.8 Good Discrimination

AUC 0.9 Excellent Discrimination

• AUC is a measure of model discrimination• How good is the model at discriminating between +'s and –'s

ˆPlotting vs. p p

Cost Sensitive Learning


+ -ActualClass

+ f++ (TP) f+- (FN)- f-+ (FP) f-- (TN)

Cost ( , ) ( , ) ( , ) ( , )

Usually, we have

( , ) 0

( , ) 0

C TP C TN C FN C FP

C

C

( , ) 0

( , ) 0

C

C

Example: Flight Delays


Delay + Ontime -ActualClass

Delay + f++ (TP) f+- (FN) Ontime - f-+ (FP) f-- (TN)

( , ) 0

( , ) 0

( , ) 1

( , ) 5

C

C

C

C

0

Optimal Probability Thresho

( , )

( ,

ld

1 1.

1 5 6) ( , )

Cp

C C

0

0

0

0 0

0

Assume ( , ) ( , ) 0

and ( , ), ( , ) 0

ˆˆIf then

ˆ(Cost) ( , )

ˆˆIf then

ˆ(Cost) ( , )(1 )

ˆWhen we are indifferent to classifying as or

( , ) ( , )(1 )

( , )

C C

C C

p p Y

E C p

p p Y

E C p

p p Y

C p C p

Cp

( , ) ( , )C C

Key Assumption:

ˆModel probabilities are accuratep

0

Assume ( , ), ( , ) 0

and ( , ), ( , ) 0

( , ) C( , )

( , ) ( , ) C( , ) ( , )

C C

C C

Cp

C C C

Undersampling and Oversampling

• Split training data into cases with Y = + and Y = -• Take a random sample with replacement from each group• Combine samples together to create new training set

• Undersampling: decreasing frequency of one of the groups• Oversampling: increasing frequency of one of the groups

Today's Topics

• Support Vector Machines

( 1)-dimensional subspace of

an -dimensional space.

E

Hyperplane

line in 2-dimen

xamples:

A space

A

sional

plane in 3-dimension

: A

al s

n

pace

n

n

Hyperplanes

2 1

1 2

1

2

4

2 4 0

24 0

1

2 and

0

1

2

4

w

x x

x

x

w

x

x

b

x

b

Equation of a Hyperplane

Rank-nullity Theorem

If is linear and onto then

dim(ker )

( )

1 dim(ker )

1 dim(ker ) dim({ | 0})

{ | 0} is a hyper lane

:

p

:

n m

n

n m T

T x w x

n T

n T x w x

x w x

T

T

{ | 0}

{ | 0}

x w x

x w x b

Support Vector Machines

Goal: Separate different classeswith a hyperplane


Goal: Separate different classeswith a hyperplane

Here, it's possible

This is a linearly separable problem


Another hyperplane that works


Many possible hyperplanes


Which one is better?


Want the hyperplane with the

maximal margin


Want the hyperplane with the

maximal margin

How can we find this hyperplane?


1

2

1 2

0

0

0

( ) 0

x b

w x b

w x

w

x

b

w x

2x

1xw


support vect

0

an orsd

, where 0

, where

are

' 0

c s

c

s

x b

x x

x b k k

w x b k k

w

w

cx

sx

w


support vect

0

an ors ared

1

1

c s

c

s

w

w

x b

x x

x b

w x b

cx

sx

w

Modify and w b


0

1

1

(

2

2

) 2

c

s

c s

x b

x b

w x b

w

w

w d

dw

w

xx

‖ ‖

‖ ‖

cx

sx

w

d

margind

Want to maximize

Want to minimize

d

w‖ ‖


0

1

1

1, if 1

1,

)

1

1(

if

c

s

i i

i i

i i

x b

x b

w x b

w x b y

w x b y

y w x

w

w

b

cx

sx

w

d


Linear SVM Problem (Separable Case)

minimiFind and that

subject to the constraints

( ) 1, for 1,2 ,

ze

i i

w b

w

y w x b i N

‖ ‖


2


minimizFind and that

2


( ) 1, for 1 ,

e

, 2i i

w b

w

y w x b i N

‖ ‖


2


minimizFind and that

2


( ) 1, for 1 ,

e

, 2i i

w b

w

y w x b i N

‖ ‖2

1

1) 1)]

2

0

Lagrangi n

[ (

a

N

i i ii

i

L w x by w

‖ ‖

2

1

1

1

1) 1

Lagran

[ (

0

i n

2

a

0

]

g

N

i i ii

N

i i ii

N

i ii

L w x by w

y x

y

Lw

w

L

b

‖ ‖

1

1

0

N

i i ii

N

i ii

w y x

y

2

1 1 ,

1

1

1) 1)

Lagrang

[ (

0

i

]

0

2

an

N N

i i i i i j i j i ji i i j

N

i i ii

N

i ii

y w yL w x b x

Lw

w

L

y x

y

yb

x

‖ ‖

1

1

0

0

N

i i ii

N

i i

i

i

w y x

y

Karush-Kuhn-Tucker Theorem

Want to maximize this

subject to these constraints

Karush-Kuhn-Tucker TheoremKuhn, H.W. and Tucker, A.W. (1951). "Nonlinear Programming". Proceedings of 2nd Berkeley Symposium. pp. 481–492.

Derivations of SVM'sCortes, C. and Vapnik, V. (1995). "Support Vector Networks". Machine Learning, 20, p. 273—297.

) 1] 0

0

If 0 then ) 1

If 0 then is

Important Result of K

[ (

(

KT Theorem

support vectoa r

i i i

i

i i i

i i

y w

y

x

w

x b

x b

Key Results

1

support If 0 t vectorhen is a

N

i i i

i i

i

w x

x

y

Today's Topics

• Soft Margin Support Vector Machines

• Nonlinear Support Vector Machines

• Kernel Methods

Soft Margin SVM

• Allows points to be on the wrong side of hyperplane

• Uses slack variables

1 , if 1

1 ,

( )

if 1

1

i i i

i i i

i i i

w x b y

w x b y

y xw b

Soft Margin SVM

2

1

Constraints

Objective Function

( 1

2

)i i i

N

ii

y x

wC

w b

‖ ‖

Want to minimize this

2

1 1 1

1

1

Lagrangian

Resul

[1

) 1 ]2

0

ts

0

(

0

0

0

N N N

i i i i i i ii i i

i i i

N

i i ii

N

i ii

i i

L w x bC y w

y

w x

C

y

‖ ‖

) 1

KKT Conditions

support vector

] 0

0

If 0 then is

[

a

(i i i i

i i

i i

b

x

y w x

Soft Margin SVM

) 1

KKT Conditions

support vector

] 0

0

If 0 then is

[

a

(i i i i

i i

i i

b

x

y w x

Relationship Between Soft and Hard Margins

2

1

0

Soft Margin

Objective Functio

0

2

n

0i i

i i

i i

i

N

ii

C

C

C

wC

‖ ‖


2

1

0

Soft Margin

Objective Functio

0

2

n

0i i

i i

i i

i

N

ii

C

C

C

wC

‖ ‖ 2

Hard Margin

Objective Functi

0

on

2

i

w

‖ ‖


2

1

0

Soft Margin

Objective Functio

0

2

n

0i i

i i

i i

i

N

ii

C

C

C

wC

‖ ‖ 2

Hard Margin

Objective Functi

0

on

2

i

w

‖ ‖

lim Soft Margin) Hard Ma( rginC

Nonlinear SVM

Nonlinear SVM

2 21 2 1 2 1 2 1 2F , ,eature Mappin ,g: ( ) ( , )2 , ,12 2x xx x x x x x

Nonlinear SVM

2

1

Minimize

2

subject to the constrai

(

nts

( 1) )

N

ii

i i i

wC

y xw b

‖ ‖

,

1( ) ( ) other te

Lag

rms2

rangian

i j i j i ji j

y yL x x

Can be computationallyexpensive

Kernel Trick

2 21 2 1 2 1 2 1 2

2 2 2 21 2 1 2 1 2 1 2 1 2 1 2

2 2 2 21 1 2 2 1 1 2 2 1 2 1 2

2 21 1 2 2

, , , , ,1)

, , , , ,1) , , ,

Feature Mapping

( ) ( , ) ( 2 2

, ,1)

2 2

2

( ) (

2 1

( 1)

) ( 2 2 2 ( 2 2

( ) (1 , )

2

x

x y

x y x y x y x y

x x x x x x x x

x y x x

x x y y

x x y

x x x y y y y

x Ky y

y

xy

Kernels

2

degree

Polynomial Kernel

Radial Basi

( ) ( )

(

s Ker

, ) ( coef0)

( , )

( , ) tanh( coef0)

(

nel

Sigmoid Kernel

, )

x y

x y

K x y x y

K x y e

K x y

K x

x y

y

‖ ‖

Kernels

2

degree

Polynomial Kernel

Radial Basi

( ) ( )

(

s Ker

, ) ( coef0)

( , )

( , ) tanh( coef0)

(

nel

Sigmoid Kernel

, )

x y

x y

K x y x y

K x y e

K x y

K x

x y

y

‖ ‖

2( ( , )

Polynomial kern

( ) ( ) 1)

1

coef0 1

degre

el with

e 2

Example

x Kx y yy x

Today's Topics

• Neural Networks

The Logistic Function

:

(

(0,1)

1)

1

1

x

x x

f

f xe

e e

Neural Networks

0.92

0.13

0.02

8.94

13.29

10.17

0.66

Neural Networks

0.92

0.13

0.02

8.94

13.29

10.17

0.66100

0.92 0.13(100) 13.95

0.66 0.02(100) 1.66

(13.95) 1.00logistic

logist (1.66 4i ) 8c 0.

13.95

1.66

1.00

0.84

Neural Networks

0.92

0.13

0.02

8.94

13.29

10.17

0.66100

13.95

1.66

1.00

0.84

10.17 8.94(1.00) 13.29(0.84) 9.94 (9.94)linout 9.94

9.94 9.94

1.5

0.2

1

Softmax Function (Generalized Logistic Function)

( )j

i

x

j nx

i

ef x

e

16.00

6.49

22.50

0.99993

57.37 10

171.89 10

16.00

16.00 6.49 22.500.99993

e

e e e

Probabilities

This flower would beclassified as setosa

Let be the parameter / weight vector of a learning algorithm

( ) Error Function

At the Global Minimum

0

points in the direction of steepest descent

Idea of Gradient Descent: Take steps in dir

w

E w

EE

w

E

( )

(0)

( 1) ( )

ection of steepest descent

Initial guess for parameter/weight vector

k

k k

w

w

Ew w

w

Gradient Descent

Learning Rate

2

1

( 1) ( ) ( )

1 1( ' 2 ' ' ' )

2 2

' ' 0

( ' )

Gradient Descen

( )

)

t

( ' 'k k k

Y Xw Y Y Y Xw w X Xw

EX Y X Xw

w

w X X X Y

w w X Y X w

E w

X

‖ ‖

Gradient Descent for Multiple Regression Models

Neural Network (Perceptron)

1ix

ˆiy

2ix

i px

1w

2w

pw

2

's are binary (0/1)

Assume activation functio

ˆ ( )

Assume

is sigmoid/logistic

1( )

n

1

( )(1 ( ))(1 )

i i

i

u

u

u

y w x

y

ue

eu u

u e

Input

Layer

Output

Layer

Single Layer Neural Network

(No hidden layers)

Gradient for Neural Network

1ix

ˆiy

2ix

i px

1w

2w

pw

2 2

1 1

1

1

ˆ ( )

1 1ˆ( ) ) ( ))

2 2

( )) ( )(1 (

( (

( ))

( ) ) ( )(1 ( ))

(Notation different from Gale s ha dout

(

n )

i i

n n

i i i ii i

n

i i i i ii

n

i i i i ii

y w x

E w y w x

Ew x w x w x x

w

w x y w x w x

y

x

y y

Input

Layer

Output

Layer

Single Layer Neural Network

(No hidden layers)

• Neural network with one hidden layer• 30 neurons in hidden layer• Classification accuracy = 98.7%

Two-layer Neural Networks

Input

Layer

Output

Layer

Hidden

Layer

• Two-layer Neural Network (One hidden layer)

• A two-layer neural network with sigmoid (logistic) activation functions can model any decision boundary

Multi-layer Perceptron

91% Accuracy

Gradient Descent for Multi-layer Perceptron

Error Back Propagation Algorithm

At each iteration• Feed inputs forward through the neural network

using current weights.• Use a recursion formula (back propagation) to

obtain the gradient with respect to all weights in the neural network.

• Update the weights using gradient descent.

Today's Topics

• Ensemble Methods• Bagging• Random Forests• Boosting

Ensemble Methods

1 25

25*

1

Idea: Construct a single classifier from many classifiers

Example:

Consider 25 binary classifiers

Error rate for each one is 0.35

Create ensemble by majority vot

• , ,

•

ing

( ) g (

•

ar max y ii

C

x I

C

C C

ò

Ensemble

2525

13

( ) )

13 or more classifiers make an(

25(0.35) (0.65) 0.

error

06

)

i i

i

P

i

x y

ò

Bagging (Bootstrap Aggregating)

Consider a data set

Let bootstrap sa

of size

Bagging Algorithm

1. Repeat the following times:

• witmple of size

Train classifier on bootstrap sample

2. The bagging ensem

h replacem

ble

ent

•

m

i

i i

N

D

D

k

D N

C

*

1

akes predictions with majority vot

arg m

in

ax ( ) )

g:

( ) (k

y ii

xC x I yC

Random Forests

• Uses Bagging

• Uses Decision Trees

• Features used to split decision tree are randomized

Boosting

Idea: • Create classifiers sequentially• Later classifiers focus on mistakes of

previous classifiers

Boosting



1

Training Data {( , ) 1, , }

th classifier

weights for training records

Error rate of classifier

Notat

( )

ion

( )

j j

i

j

i

N

i j i j jj

x y j N

C i

w

C

x yw I C

ò

û

Boosting



1

Training Data {( , ) 1, , }

th classifier

weights for training records

Error rate of classifier

Notat

( )

ion

( )

j j

i

j

i

N

i j i j jj

x y j N

C i

w

C

x yw I C

ò

û

( )( 1)

11ln

2

Weight update fo

Importance of classifie

rmula

, if (

r

)

, if ( )

i

i

i

ii

i

ij i j ji

ji j ji

w C x y

C

e

ew

C x yZ

òò

Today's Topics

• The Multiclass Problem

• One-against-one approach

• One-against-rest approach

The Multiclass Problem

• Binary dependent variable y: Only two possible values

• Multiclass dependent variable y: More than two possible values

How can we deal with multiclass variables?

Classification Algorithms

• Decision Trees• k-Nearest Neighbors• Naïve Bayes• Neural Networks

• Support Vector Machines

• How can we extend SVM to multiclass problems?• How can we extend other algorithms to multiclass

problems?

Deals with multiclass output by default

Only deals with binary classification problems

One-against-one Approach

,

, ,

Dependent variable

training data where

Possible values: 1, 2, ,

For each pair , {1,2, , } where

•

binary classifier trained wi

or

•

Final classification is done by

th

voting

i j

i j i j

y

y k

i j k i j

y i yD

C D

j

*,

1, ,

(ties broken randomly)

( ) a ( ( )rg max )i jy k i j

I C x yC x

One-against-one Approach

,

, ,

Dependent variable

training data where


For each pair , {1,2, , } where

•

binary classifier trained wi

or

•

Final classification is done b

th

y voting

i j

i j i j

y

y k

i j k

y i yD

i j

C D

j

*,

1, ,

(ties broken randomly)

( ) a ( ( )rg max )i jy k i j

I C x yC x

Number of m

(

odels

2

1)

2

k k k

One-against-rest Approach


For each {1,2, , }

• Create training set as follows:

replace that value with

•

ot

Dep

her

endent variable

anytime

binary classifi

Final classi

er trained w

f

i

icati

th

i

i iC

y

y k

i k

D

y

D

y i

( ) one vote for

on is done by

( ) other one vote for each other val

voting (ties broken randomly)

• counts as

• counts as u e of i

i

C x i i

C x y

math 5364 notes chapter 5: alternative classification techniques

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