Transcript i290_280I_Lecture_4

Classification Methods:
k-Nearest Neighbor
Naïve Bayes
Ram Akella
Lecture 4
February 9, 2011
UC Berkeley
Silicon Valley Center/SC
1
Overview
 Example
 The Naïve rule
 Two data-driven methods (no model)
 K-nearest neighbors
 Naïve Bayes
2
Example: Personal Loan Offer
As part of customer acquisition efforts,
Universal bank wants to run a campaign for
current customers to purchase a loan.
In order to improve target marketing, they
want to find customers that are most likely to
accept the personal loan offer.
They use data from a previous campaign on
5000 customers, 480 of them accepted.
3
Personal Loan Data Description
ID
Age
Experience
Income
ZIPCode
Family
CCAvg
Education
Mortgage
Personal Loan
Securities Account
CD Account
Online
CreditCard
Customer ID
Customer's age in completed years
#years of professional experience
Annual income of the customer ($000)
Home Address ZIP code.
Family size of the customer
Avg. spending on credit cards per month ($000)
Education Level. 1: Undergrad; 2: Graduate; 3: Advanced/Professional
Value of house mortgage if any. ($000)
Did this customer accept the personal loan offered in the last campaign?
Does the customer have a securities account with the bank?
Does the customer have a certificate of deposit (CD) account with the bank?
Does the customer use internet banking facilities?
Does the customer use a credit card issued by UniversalBank?
File: “UniversalBank KNN NBayes.xls”
4
The Naïve Rule
 Classify a new observation as a
member of the majority class
 In the personal loan example, the
majority of customers did not accept the
loan
5
K-Nearest Neighbor: Idea
Find the k closest records to the one to
be classified, and let them “vote”.
100
90
80
70
Age
60
Regular beer
50
Light beer
40
30
20
10
0
$0
$20,000
$40,000
$60,000
$80,000
Income
6
What does the algorithm do?
 Computes the distance between the
record to be classified and each of
records in the training set
 Finds the k shortest distances
 Computes the vote of these k neighbors
 This is repeated for every record in the
validation set
7
Experiment
We have 100 training points : 60 pink and 40
blue. Then we have 50 test points,
 For each point, we voted, using 5-nearest
neighbor
How do we measure how well the classifier
did?
 We compare the predicted with actual value
in each of the 50 point validation/test set
8
Distance between 2 observations
 Single variable case: each item has 1 value.
 Customer 1 has income = 49K
 Multivariate case: Each observation is a vector of
values.
 Customer1 = (Age=25,Exp=1,Income=49,…,CC=0)
 Customer2 = (Age=49,Exp=19,Income=34,…,CC=0)
 The distance between obs i and j is denoted dij.
 Distance Requirements:
 Non-negative ( dij > 0 )
 dii = 0
 Symmetry (dij = dji )
 Triangle inequality ( dij + djk  dik )
9
Types of Distances
Notation:
Example:
xi  ( xi1 , xi 2 ,, xip )
x j  ( x j1 , x j 2 ,, x jp )
 Customer1=(Age=25,Exp=1, Inc=49,
fam=4,CCAvg=1.6)
 Customer2=(Age=49,Exp=19,Inc=34,
fam=3,CCAvg=1.5)
10
Euclidean Distance
dij 
x
i1  x j1   xi 2  x j 2     xip  x jp 
2
2
2
 The Euclidean distance between the age of
customer1 (25) and customer2 (49):
 [ (25-49)2 ] = 24
 The Euclidean distance between the two on the 5dimensions (Age, Exper, Income, Family, CCAvg):
 [ (25-49)2 + (1-19)2 + (49-34)2 + (4-3)2 + (1.6-1.5)2]= =30.82
11
which pair is closest ?
Carry
Sam
Miranda
Income
$31,779
$32,739
$33,880
Age
36
40
38
55%
27%
1. Carry & Sam
2. Sam & Miranda
18%
3. Carry & Miranda
Carry & Sam:  (31.779-32.739)2 + (36-40)2 = 960.00
Now, income is in
$000. Which pair is
closest?
12%
84%
4%
Carry
Sam
Miranda
Income
$31.779
$32.739
$33.880
1. Carry & Sam
2. Sam & Miranda
3. Carry & Miranda
Sam & Miranda: √(32.739-33.88)2 + (40-38)2 = 5.30
Age
36
40
38
Why do we need to standardize the variables?
The distance measure is influenced by the
units of the different variables, especially if
there is a wide variation in units. Variables
with “larger” units will influence the distances
more than others.
The solution:
standardize each variable before measuring
distances!
14
Other distances
 Squared Euclidean distance
 Correlation-based distance: the correlation between
two vectors of (standardized) items/observations,
rij, measures their similarity. We can define a
distance measure as dij = 1- rij2
 Statistical distance (no need to standardize)
dij 
x  x S x  x 
1
i
j
T
i
j
The only measure that
accounts for covariance!
 Manhattan distance (“city-block”)
d ij  xi1  x j1  xi 2  x j 2    xip  x jp
Note: some software use “similarities” instead of “distances”.
15
Distances for Binary Data
 Are obtained from the 2x2 table of
counts.
Carrie
0 1
0 a b
0
Miranda
1 c d
Married?
Carrie
Sam
Miranda
Smoker?
1
0
0
Manager?
1
1
1
0
0
1
1
0
0
2
1
0
1
16
Choosing the number or neighbors (K)




Too small: under-smoothing
Too large: over-smoothing
Typically k<20
K should be odd (to avoid ties)
Solution:
Use validation set to find “best” k
17
Training Data scoring - Summary Report (for k=4)
Cut off Prob.Val. for Success (Updatable)
Output
0.5
Classification Confusion Matrix
Predicted Class
Actual Class
1
0
1
243
20
0
43
2694
Error Report
We’re using the validation data
here to choose the best k
Validation error log for different k
% Error
Training
Value of k
1
2
3
4
5
4.15
4.45
4.10
3.80 <--- Best k
4.50
Cumulative
150off Prob.Val. for Success
Cut
100
50
Classification
Confusion Matrix
0
Predicted Class
0
1000
2000
Actual Class# cases
1
3000
1
243
Cumulative
Personal Loan
using average
0
43
% Error
15.03
0.74
2.10
0.5
Classification Confusion Matrix
Predicted Class
Actual Class
1
0
1
134
16
Class
1
0
Overall
# Cases
194
1806
2000
0
60
1790
Error Report
250
Training Data scoring
- Summary Report
(for k=4)
Cumulative
Personal Loan
when sorted
(Updatable)
using predicted
values
# Errors
43
20
63
Cut off Prob.Val. for Success (Updatable)
Lift chart (validation dataset)
200
# Cases
286
2714
3000
Validation Data scoring - Summary Report (for k=4)
% Error
Validation
0.00
1.30
2.47
2.10
3.40
Class
1
0
Overall
0.5
# Errors
60
16
76
% Error
30.93
0.89
3.80
18
Advantages and Disadvantages of K nearest neighbors
 The Good
 Very flexible, data-driven
 Simple
 With large amount of data, where predictor levels are
well represented, has good performance
 Can also be used for continuous y: instead of voting,
take average of neighbors (XLMiner: Prediction > KNN)
 The bad




No insight about importance/role of each predictor
Beware of over-fitting! Need a test set
Can be computationally intensive for large k
Need LOTS of data (exponential in #predictors)
19
Conditional Probability - reminder
 A = the event “customer accepts loan”
 B = the event “customer has credit card”
 P( A | B)
denotes the probability of A
given B (the conditional probability that A
occurs given that B occurred)
P( A  B)
P( A | B) 
P( B)
If P(B)>0
20
Naïve Bayes
 Naive Bayes is one of the
most efficient and effective
inductive learning algorithms
for machine learning and data
mining.
 It calculates the probability of
a point E to belong to a
certain class Ci based on its
attributes (x1, x2, …, xn)
 It assumes that the
attributes are conditional
independent on the class Ci
C
x1
x2
xn
….
21
Illustrative Example
 The example E is represented by a set of attribute
values (x1, x2, · · · , xn), where xi is the value of
attribute Xi. Let C represents the classification
variable, and let c be the value of C.
 In this example we assume that there are only two
classes: + (the positive class) or − (the negative
class).
 A classifier is a function that assigns a class label to
an example. From the probability perspective,
according to Bayes Rule, the probability of an
example E = (x1, x2, · · · , xn) being class c is
22
Naïve Bayes Classifier
E is classified as the class C = +if and only if:
where fb(E) is called a Bayesian classifier.
Assume that all attributes are independent given the value of
the class variable, that is:
The function fb(E) is called a naive Bayesian classifier,
or simply naive Bayes (NB).
23
Augmented Naïve Bayes
 Naive Bayes is the simplest form of Bayesian network,
in which all attributes are independent given the value
of the class variable.
 This conditional independence assumption is rarely
true in most real-world applications.
 A straightforward approach to overcome the
limitation of naive Bayes is to extend its structure to
represent explicitly the dependencies among
attributes.
24
Augmented Naïve Bayes
An augmented naive Bayes (ANB), is an extended classifier,
in which the class node directly points to all attribute nodes,
and there exist links among attribute nodes. An ANB
represents a joint probability distribution represented by:
where pa(xi) denotes an assignment to values of the parents
of Xi.
C
x1
x2
…. Xn-1
xn
25
Why does this classifier work?
 The basic idea comes from


In a given dataset, two attributes may depend on each
other, but the dependence may distribute evenly in each
class.
Clearly, in this case, the conditional independence
assumption is violated, but naive Bayes is still the
optimal classifier.
 What eventually affects the classification is the
combination of dependencies among all attributes.


If we just look at two attributes, there may exist strong
dependence between them that affects the
classification.
When the dependencies among all attributes work
together, however, they may cancel each other out and
no longer affect the classification.
26
Why does this classifier work?
Definition 1:
Given an example E, two classifiers f1
and f2 are said to be equal under zeroone loss on E, if f1(E) ≥ 0 if and only if
f2(E) ≥ 0, denoted by f1(E) = f2(E) for
every example E in the example space.
27
Local Dependence Distribution
Definition 2:
For a node X on ANB, the local dependence derivative
of X in classes + and − are defined as:
where dd+G(x|pa(x)) reflects the strength of the local
dependence of node X in class +,
 This measures the influence of X’s local dependence
on the classification in class +.
dd−G (x|pa(x)) is similar for the negative class.
28
Local Dependence Distribution
1. When X has no parent, then:
dd+ G(x|pa(x)) = dd−G(x|pa(x)) = 1.
2. When dd+G(x|pa(x)) ≥ 1,
 X’s local dependence in class + supports
the classification of C = +. Otherwise, it
supports the classification of C = −
3. When dd−G(x|pa(x)) ≥ 1,
 X’s local dependence in class − supports
the classification of C = −. Otherwise, it
supports the classification of C = +.
29
Local Dependence Distribution
When the local dependence derivatives in both
classes support the different classifications,
the local dependencies in the two classes
cancel partially each other out,
 The final classification that the local
dependence supports, is the class with the
greater local dependence derivative.
 Another case is that the local dependence
derivatives in the two classes support the same
classification. Then, the local dependencies in
the two classes work together to support the
classification.
30
Local Dependence Derivative Ratio
Definition 3
For a node X on ANB G, the local
dependence derivative ratio at node X,
denoted by ddrG(x) is defined by:
ddrG(x) quantifies the influence of X’s
local dependence on the classification.
31
Local Dependence Derivative Ratio
We have:
1. If X has no parents,
ddrG(x) = 1.
2. If dd+G(x|pa(x)) = dd−G (x|pa(x)),
ddrG(x) = 1.
This means that x’s local dependence distributes evenly
in class + and class −.
Thus, the dependence does not affect the classification,
no matter how strong the dependence is.
3. If ddrG(x) > 1,
X’s local dependence in class + is stronger than that in
class −.
ddrG(x) < 1 means the opposite.
32
Global Dependence Distribution
Let us explore under what condition an ANB works exactly
the same as its correspondent naive Bayes.
Theorem 1 Given an ANB G and its correspondent naïve
Bayes Gnb (i.e., remove all the arcs among attribute nodes
from G) on attributes X1, X2, ..., Xn, assume that fb and fnb
are the classifiers corresponding to G and Gnb, respectively.
For a given example E = (x1, x2, · · ·, xn), the equation
below is true.
where the product of ddrG(xi) for i=1..N is called the
dependence distribution factor at example E, denoted by
DFG(E).
33
Global Dependence Distribution
 Proof:
34
Global Dependence Distribution
Theorem 2
Given an example E = (x1, x2, ..., xn),
an ANB G is equal to its correspondent
naive Bayes Gnb under zero-one loss if
and only if when fb(E) ≥ 1, DFG(E) ≤
fb(E); or when fb(E) < 1, DFG(E) > fb(E).
35
Global Dependence Distribution
Applying the theorem 2 we have the following
results:
1. When DFG(E) = 1, the dependencies in ANB
G has no influence on the classification.
The classification of G is exactly the same as that
of its correspondent naïve Bayes Gnb.
 There exist three cases for DFG(E) = 1.
 no dependence exists among attributes.
 for each attribute X on G, ddrG(x) = 1; that is, the
local distribution of each node distributes evenly
in both classes.
 the influence that some local dependencies
support classifying E into C = +is canceled out by
the influence that other local dependences
support classifying E into C = −.

36
Global Dependence Distribution
2. fb(E) = fnb(E) does not require that DFG(E)
= 1. The precise condition is given by
Theorem 2. That explains why naive Bayes
still produces accurate classification even
in the datasets with strong dependencies
among attributes (Domingos & Pazzani
1997).
3. The dependencies in an ANB flip (change)
the classification of its correspondent naive
Bayes, only if the condition given by
Theorem 2 is no longer true.
37
Conditions of the optimality of the
Naïve Bayes
Naive Bayes classifier is optimal if the dependencies
among attributes cancel each other out.
 The classifier is still optimal even though the
dependencies do exist
38
Optimality of the Naïve Bayes
Example:
We have two attributes X1 and X2, and assume that the class density is a
multivariate Gaussian in both the positive and negative classes. That is:
where







x = (x1, x2)
∑+ and ∑ − are the covariance matrices in the positive and negative
classes respectively,
| ∑ − | and | ∑ + | are the determinants of ∑ − and ∑ +,
∑ −1 + and ∑−1 − are the inverses of ∑ − and ∑ +
μ+ = (μ+1 , μ+2 ) and μ− = (μ−1 , μ−2 ),
μ+ i and μ−i are the means of attribute Xi in the positive and negative
classes respectively,
(x−μ+)T and (x−μ−)T are the transposes of (x−μ+) and (x−μ−).
39
Optimality of the Naïve Bayes
We assume:
The two classes have a common covariance matrix
∑+ = ∑− = ∑ ,
X1 and X2 have the same variance σ in both classes.
Then, when applying a logarithm to the Bayesian
classifier, defined previously, we obtain the following fb
classifier
40
Optimality of the Naïve Bayes
 Then, because of the conditional independence
assumption, we have the correspondent naive
Bayesian classifier fnb
 Assume that
 X1 and X2 are independent if σ12 = 0. If σ ≠ σ12, we
have:
41
Optimality of the Naïve Bayes
An example E is classified into the positive
class by fb, if and only if fb ≥ 0. fnb is similar.
When fb or fnb is divided by a non-zero positive
constant, the resulting classifier is the same
as fb or fnb. Then
42
Optimality of the Naïve Bayes
where a = − (1/σ2)(μ+ + μ−)Σ−1(μ+ − μ−), is
a constant independent of x.
For any x1 and x2, Naive Bayes has the
same classification as that of the
underlying classifier if:
43
Optimality of the Naïve Bayes
 This is:
1
44
Optimality of the Naïve Bayes
Assuming that:
We can simplify the equation
to:
1
where
45
Optimality of the Naïve Bayes
The shaded area of the figure shows the
region in which the Naïve Bayes Classifier is
optimal
46
Example with 2 predictors: CC, Online
P(accept =1 | CC=1, online=1) =
50/286
286/3000
P(CC  1, Online  1 | accept  1) P(accept  1)
P(CC  1, Online  1 | accept  1) P(accept  1)  P(CC  1, Online  1 | accept  0) P(accept  0)
Count of Personal Loan
CreditCard
Personal Loan
0
Online
0
1
0 Total
1
1 Total
Grand Total
0
1
0
769
71
840
321
36
357
1197
1 Grand Total
1163
1932
129
200
1292
2132
461
782
50
86
511
868
1803
3000
47
P(CC=1, Online=1 | accept=0) is
approx
20%
20%
20%
20%
20%
1.
2.
3.
4.
5.
50/286
1-50/286
461/3000
461/(3000-286)
129/(3000-286)
Example with 2 predictors: CC, Online
P(accept =1 | CC=1, online=1) =
P(CC  1, Online  1 | accept  1) P(accept  1)
P(CC  1, Online  1 | accept  1) P(accept  1)  P(CC  1, Online  1 | accept  0) P(accept  0)
50 286

286 3000

 0.0978
50 286 461 2714



286 3000 2714 3000
49
The practical difficulty
 We need to have ALL the
combinations of predictor categories




CC=1,Online=1
CC=1, Online=0
CC=0, Online=1
CC=0, Online=0
 With many predictors, this is pretty
unlikely
50
Example with (only) 3 predictors:
CC, Online, CD account
Count of Personal Loan
CreditCard
CD Account
Personal Loan
0
0
1
0 Total
1
1 Total
Grand Total
0
1
0
0
769
69
838
318
30
348
1186
Online
0 Total
1
1152
100
1252
363
363
1615
CD account=0, Online=1, CreditCard=1
1921
169
2090
681
30
711
2801
1
0
2
2
3
6
9
11
1 Total Grand Total
1
11
29
40
98
50
148
188
11
31
42
101
56
157
199
1932
200
2132
782
86
868
3000
51
A practical solution:
From Bayes to Naïve Bayes
 Substitute P(CC=1,Online=1 | accept) with
P(CC=1 | accept) x P(Online=1 | accept)
 This means that we are assuming
independence between CC and Online!
 If the dependence is not extreme, it will
work reasonably well
52
Example with 2 predictors: CC,
Online
P(accept =1 | CC=1, online=1) =
P(CC  1 | accept  1) P(Online 1 | accept  1)
P(CC  1, Online  1 | accept  1) P(accept  1)
P( P(CC  1, Online  1 | accept  1) P(accept  1)  P(CC  1, Online  1 | accept  0) P(accept  0)
P(CC  1 | accept  0) P(Online  1 | accept  0)
Count of Personal Loan
CreditCard
Personal Loan
0
Online
0
1
0 Total
1
1 Total
Grand Total
0
1
0
769
71
840
321
36
357
1197
1 Grand Total
1163
1932
129
200
1292
2132
461
782
50
86
511
868
1803
3000
53
Naïve Bayes for CC, Online:
P(accept =1 | CC=1, online=1) =
P(CC  1 | accept  1) P(Online 1 | accept  1)
P(CC  1, Online  1 | accept  1) P(accept  1)
P( P(CC  1, Online  1 | accept  1) P(accept  1)  P(CC  1, Online  1 | accept  0) P(accept  0)
P(CC  1 | accept  0) P(Online  1 | accept  0)
Count of Personal Loan CreditCard
Personal Loan
0
0
1932
1
200
Grand Total
2132
Count of Personal Loan Online
Personal Loan
0
0
1090
1
107
Grand Total
1197
1 Grand Total
782
2714
86
286
868
3000
1 Grand Total
1624
2714
179
286
1803
3000
86 179 286


286 286 3000
 0.102
86 179 286 782 1642 2714





286 286 3000 2714 2714 3000 54
Naïve Bayes in XLMiner
 Classification> Naïve Bayes
Prior class probabilities
According to relative occurrences in training data
Class
1
0
Prob.
0.095333333 <-- Success Class
0.904666667
P(CC=1| accept=1) = 86/286
Conditional probabilities
Classes-->
Input
Variables
Online
CreditCard
1
Value
0
1
0
1
0
Prob
0.374125874
0.625874126
0.699300699
0.300699301
Value
0
1
0
1
Prob
0.401621223
0.598378777
0.711864407
0.288135593
55
Naïve Bayes in XLMiner
 Scoring the validation data
XLMiner : Naive Bayes - Classification of Validation Data
Data range
['UniversalBank KNN
NBayes.xls']'Data_Partition1'!$C$3019:$O$5018
Cut off Prob.Val. for Success (Updatable)
Row Id.
2
3
7
8
11
13
14
15
16
Predicted
Class
0
0
0
0
0
0
0
0
0
0.5
Actual Class
0
0
0
0
0
0
0
0
0
Prob. for 1
(success)
0.08795125
0.08795125
0.097697987
0.092925663
0.08795125
0.08795125
0.097697987
0.08795125
0.10316131
Back to Navig
( Updating the value here will NOT update value in summary re
Online
CreditCard
0
0
1
0
0
0
1
0
1
0
0
0
1
0
0
0
0
1
56
Advantages and Disadvantages
 The good
 Simple
 Can handle large amount of predictors
 High performance accuracy, when the goal is ranking
 Pretty robust to independence assumption!
 The bad
 Requires large amounts of data
 Need to categorize continuous predictors
 Predictors with “rare” categories -> zero prob (if this
category is important, this is a problem)
 Gives biased probability of class membership
 No insight about importance/role of each predictor
57