Transcript Classification: Basic Concepts and Decision Trees

Classification: Basic Concepts and Decision
Trees
A programming task
Classification: Definition
• Given a collection of records (training set )
– Each record contains a set of attributes, one of the
attributes is the class.
• Find a model for class attribute as a function
of the values of other attributes.
• Goal: previously unseen records should be
assigned a class as accurately as possible.
– A test set is used to determine the accuracy of the
model. Usually, the given data set is divided into
training and test sets, with training set used to build the
model and test set used to validate it.
Illustrating Classification Task
Tid
Attrib1
Attrib2
Attrib3
Class
1
Yes
Large
125K
No
2
No
Medium
100K
No
3
No
Small
70K
No
4
Yes
Medium
120K
No
5
No
Large
95K
Yes
6
No
Medium
60K
No
7
Yes
Large
220K
No
8
No
Small
85K
Yes
9
No
Medium
75K
No
10
No
Small
90K
Yes
Learning
algorithm
Induction
Learn
Model
Model
10
Training Set
Tid
Attrib1
Attrib2
11
No
Small
55K
?
12
Yes
Medium
80K
?
13
Yes
Large
110K
?
14
No
Small
95K
?
15
No
Large
67K
?
10
Test Set
Attrib3
Apply
Model
Class
Deduction
Examples of Classification Task
• Predicting tumor cells as benign or malignant
• Classifying credit card transactions
as legitimate or fraudulent
• Classifying secondary structures of protein
as alpha-helix, beta-sheet, or random
coil
• Categorizing news stories as finance,
weather, entertainment, sports, etc
Classification Using Distance
• Place items in class to which they are
“closest”.
• Must determine distance between an item
and a class.
• Classes represented by
– Centroid: Central value.
– Medoid: Representative point.
– Individual points
• Algorithm: KNN
Classification Techniques
•
•
•
•
•
•
Decision Tree based Methods
Rule-based Methods
Memory based reasoning
Neural Networks
Naïve Bayes and Bayesian Belief Networks
Support Vector Machines
A first example
Database of 20,000 images of handwritten digits, each
labeled by a human
[28 x 28 greyscale; pixel values 0-255; labels 0-9]
Use these to learn a classifier which will label digit-images automatically…
The learning problem
Input space X = {0,1,…,255}784
Output space Y = {0,1,…,9}
Training set
(x1, y1), …, (xm, ym)
m = 20000
Learning
Algorithm
To measure how good f is: use a test set
[Our test set: 100 instances of each digit.]
Classifier
f: X ! Y
A possible strategy
Input space X = {0,1,…,255}784
Output space Y = {0,1,…,9}
Treat each image as a point in 784-dimensional Euclidean space
To classify a new image: find its nearest neighbor in the database (training set) and
return that label
f = entire training set + search engine
K Nearest Neighbor (KNN):
• Training set includes classes.
• Examine K items near item to be classified.
• New item placed in class with the most
number of close items.
• O(q) for each tuple to be classified. (Here q
is the size of the training set.)
KNN
Nearest neighbor
Image to label
Nearest neighbor
Overall:
error rate = 6%
(on test set)
Question: what is
the error rate
for random
guessing?
What does it get wrong?
Who knows… but here’s a hypothesis:
Each digit corresponds to some connected region of R784. Some of the regions come
close to each other; problems occur at these boundaries.
R1
R2
e.g. a random point in
this ball has only a
70% chance of
being in R2
Nearest neighbor: pros and cons
Pros
Simple
Flexible
Excellent performance on a wide range of tasks
Cons
Algorithmic:
Statistical:
time consuming – with n training
points in Rd, time to label a new
point is O(nd)
memorization, not learning!
no insight into the domain
would prefer a compact classifier
Prototype selection
A possible fix: instead of the entire training set, just keep a “representative
sample”
Voronoi
cells
“Decision boundary”
Prototype selection
A possible fix: instead of the entire training set, just keep a “representative
sample”
Voronoi
cells
“Decision boundary”
How to pick prototypes?
They needn’t be actual data points
Idea 2: one prototype per class: mean of training points
Examples:
Error = 23%
Postscript: learning models
Batch learning
Training data
Learning
Algorithm
On-line learning
See a new point x
predict label
Classifier f
See y
test
Update
classifer
Example of a Decision Tree
T id
R e fu n d
Splitting Attributes
M a rita l
S ta tu s
T a x a b le
In c o m e
C heat
1
Yes
S in g le
125K
No
2
No
M a rrie d
100K
No
3
No
S in g le
70K
No
4
Yes
M a rrie d
120K
No
5
No
D iv o rc e d
95K
Y es
6
No
M a rrie d
60K
No
7
Yes
D iv o rc e d
220K
No
8
No
S in g le
85K
Y es
9
No
M a rrie d
75K
No
10
No
S in g le
90K
Y es
Refund
Yes
No
NO
MarSt
Single, Divorced
TaxInc
< 80K
NO
NO
> 80K
YES
10
Training Data
Married
Model: Decision Tree
Another Example of Decision Tree
MarSt
T id
10
R e fu n d
Married
M a rita l
S ta tu s
T a x a b le
In c o m e
C heat
1
Yes
S in g le
125K
No
2
No
M a rrie d
100K
No
3
No
S in g le
70K
No
4
Yes
M a rrie d
120K
No
5
No
D iv o rc e d
95K
Y es
6
No
M a rrie d
60K
No
7
Yes
D iv o rc e d
220K
No
8
No
S in g le
85K
Y es
9
No
M a rrie d
75K
No
10
No
S in g le
90K
Y es
NO
Single,
Divorced
Refund
No
Yes
NO
TaxInc
< 80K
NO
> 80K
YES
There could be more than one tree that
fits the same data!
Decision Tree Classification Task
Tid
Attrib1
Attrib2
Attrib3
Class
1
Yes
Large
125K
No
2
No
Medium
100K
No
3
No
Small
70K
No
4
Yes
Medium
120K
No
5
No
Large
95K
Yes
6
No
Medium
60K
No
7
Yes
Large
220K
No
8
No
Small
85K
Yes
9
No
Medium
75K
No
10
No
Small
90K
Yes
Tree
Induction
algorithm
Induction
Learn
Model
Model
10
Training Set
Tid
Attrib1
Attrib2
11
No
Small
55K
?
12
Yes
Medium
80K
?
13
Yes
Large
110K
?
14
No
Small
95K
?
15
No
Large
67K
?
10
Test Set
Attrib3
Apply
Model
Class
Deduction
Decision
Tree
Apply Model to Test Data
Test Data
Start from the root of tree.
R e fu n d
No
Refund
Yes
10
No
NO
MarSt
Single, Divorced
TaxInc
< 80K
NO
Married
NO
> 80K
YES
M a rita l
S ta tu s
T a x a b le
In c o m e
C heat
M a rrie d
80K
?
Apply Model to Test Data
Test Data
R e fu n d
No
Refund
Yes
10
No
NO
MarSt
Single, Divorced
TaxInc
< 80K
NO
Married
NO
> 80K
YES
M a rita l
S ta tu s
T a x a b le
In c o m e
C heat
M a rrie d
80K
?
Apply Model to Test Data
Test Data
R e fu n d
No
Refund
Yes
10
No
NO
MarSt
Single, Divorced
TaxInc
< 80K
NO
Married
NO
> 80K
YES
M a rita l
S ta tu s
T a x a b le
In c o m e
C heat
M a rrie d
80K
?
Apply Model to Test Data
Test Data
R e fu n d
No
Refund
Yes
10
No
NO
MarSt
Single, Divorced
TaxInc
< 80K
NO
Married
NO
> 80K
YES
M a rita l
S ta tu s
T a x a b le
In c o m e
C heat
M a rrie d
80K
?
Apply Model to Test Data
Test Data
R e fu n d
No
Refund
Yes
10
No
NO
MarSt
Single, Divorced
TaxInc
< 80K
NO
Married
NO
> 80K
YES
M a rita l
S ta tu s
T a x a b le
In c o m e
C heat
M a rrie d
80K
?
Apply Model to Test Data
Test Data
R e fu n d
No
Refund
Yes
M a rita l
S ta tu s
T a x a b le
In c o m e
C heat
M a rrie d
80K
?
10
No
NO
MarSt
Single, Divorced
TaxInc
< 80K
NO
Married
NO
> 80K
YES
Assign Cheat to “No”
Decision Tree Classification Task
Tid
Attrib1
Attrib2
Attrib3
Class
1
Yes
Large
125K
No
2
No
Medium
100K
No
3
No
Small
70K
No
4
Yes
Medium
120K
No
5
No
Large
95K
Yes
6
No
Medium
60K
No
7
Yes
Large
220K
No
8
No
Small
85K
Yes
9
No
Medium
75K
No
10
No
Small
90K
Yes
Tree
Induction
algorithm
Induction
Learn
Model
Model
10
Training Set
Tid
Attrib1
Attrib2
11
No
Small
55K
?
12
Yes
Medium
80K
?
13
Yes
Large
110K
?
14
No
Small
95K
?
15
No
Large
67K
?
10
Test Set
Attrib3
Apply
Model
Class
Deduction
Decision
Tree
Decision Tree Induction
• Many Algorithms:
– Hunt’s Algorithm (one of the earliest)
– CART
– ID3, C4.5
– SLIQ,SPRINT
General Structure of Hunt’s Algorithm
• Let Dt be the set of training
records that reach a node t
• General Procedure:
– If Dt contains records that
belong the same class yt, then t
is a leaf node labeled as yt
– If Dt is an empty set, then t is a
leaf node labeled by the
default class, yd
– If Dt contains records that
belong to more than one class,
use an attribute test to split
the data into smaller subsets.
Recursively apply the
procedure to each subset.
Tid Refund Marital
Status
Taxable
Income Cheat
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
No
7
Yes
Divorced 220K
No
8
No
Single
85K
Yes
9
No
Married
75K
No
10
No
Single
90K
Yes
10
Dt
?
60K
Hunt’s Algorithm
T id
Refund
Don’t
Cheat
Yes
No
Don’t
Cheat
Don’t
Cheat
Refund
Refund
Yes
Yes
No
Don’t
Cheat
Don’t
Cheat
Marital
Status
Single,
Divorced
Cheat
Married
No
Marital
Status
Single,
Divorced
>= 80K
Don’t
Cheat
Cheat
T a x a b le
In c o m e
C heat
Yes
S in g le
125K
No
2
No
M a rrie d
100K
No
3
No
S in g le
70K
No
4
Yes
M a rrie d
120K
No
5
No
D ivo rc e d
95K
Yes
6
No
M a rrie d
60K
No
7
Yes
D ivo rc e d
220K
No
8
No
S in g le
85K
Yes
9
No
M a rrie d
75K
No
10
No
S in g le
90K
Yes
10
Don’t
Cheat
< 80K
M a rita l
S ta tu s
1
Married
Taxable
Income
Don’t
Cheat
R e fu n d
Tree Induction
• Greedy strategy.
– Split the records based on an attribute test that
optimizes certain criterion.
• Issues
– Determine how to split the records
• How to specify the attribute test condition?
• How to determine the best split?
– Determine when to stop splitting
Tree Induction
• Greedy strategy.
– Split the records based on an attribute test that
optimizes certain criterion.
• Issues
– Determine how to split the records
• How to specify the attribute test condition?
• How to determine the best split?
– Determine when to stop splitting
How to Specify Test Condition?
• Depends on attribute types
– Nominal
– Ordinal
– Continuous
• Depends on number of ways to split
– 2-way split
– Multi-way split
Splitting Based on Nominal Attributes
• Multi-way split: Use as many partitions as distinct values.
CarType
Family
Luxury
Sports
• Binary split: Divides values into two subsets.
Need to find optimal partitioning.
{Sports,
Luxury}
CarType
{Family}
OR
{Family,
Luxury}
CarType
{Sports}
Splitting Based on Ordinal Attributes
• Multi-way split: Use as many partitions as distinct values.
Size
Small
Large
Medium
• Binary split: Divides values into two subsets.
Need to find optimal partitioning.
{Small,
Medium}
Size
{Large}
OR
{Medium,
Large}
Size
• What about this split?
{Small,
Large}
Size
{Medium}
{Small}
Splitting Based on Continuous Attributes
• Different ways of handling
– Discretization to form an ordinal categorical
attribute
• Static – discretize once at the beginning
• Dynamic – ranges can be found by equal interval
bucketing, equal frequency bucketing
(percentiles), or clustering.
– Binary Decision: (A < v) or (A  v)
• consider all possible splits and finds the best cut
• can be more compute intensive
Splitting Based on Continuous Attributes
Taxable
Income
> 80K?
Taxable
Income?
< 10K
Yes
> 80K
No
[10K,25K)
(i) Binary split
[25K,50K)
[50K,80K)
(ii) Multi-way split
Tree Induction
• Greedy strategy.
– Split the records based on an attribute test that
optimizes certain criterion.
• Issues
– Determine how to split the records
• How to specify the attribute test condition?
• How to determine the best split?
– Determine when to stop splitting
How to determine the Best Split
Before Splitting: 10 records of class 0,
10 records of class 1
Own
Car?
Yes
Car
Type?
No
Family
Student
ID?
Luxury
c1
Sports
C0: 6
C1: 4
C0: 4
C1: 6
C0: 1
C1: 3
C0: 8
C1: 0
C0: 1
C1: 7
Which test condition is the best?
C0: 1
C1: 0
...
c10
C0: 1
C1: 0
c11
C0: 0
C1: 1
c20
...
C0: 0
C1: 1
How to determine the Best Split
• Greedy approach:
– Nodes with homogeneous class distribution are
preferred
• Need a measure of node impurity:
C0: 5
C1: 5
C0: 9
C1: 1
Non-homogeneous,
Homogeneous,
High degree of impurity
Low degree of impurity
Measures of Node Impurity
• Gini Index
• Entropy
• Misclassification error
How to Find the Best Split
Before Splitting:
C0
C1
N00
N01
M0
A?
B?
Yes
No
Node N1
C0
C1
Node N2
N10
N11
C0
C1
N20
N21
M2
M1
Yes
No
Node N3
C0
C1
Node N4
N30
N31
C0
C1
M3
M12
M4
M34
Gain = M0 – M12 vs M0 – M34
N40
N41
Measure of Impurity: GINI
• Gini Index for a given node t :
GINI ( t )  1   [ p ( j | t )]
2
j
(NOTE: p( j | t) is the relative frequency of class j at node t).
– Maximum (1 - 1/nc) when records are equally distributed among all
classes, implying least interesting information
– Minimum (0.0) when all records belong to one class, implying most
interesting information
C1
0
C1
1
C1
2
C1
3
C2
6
C2
5
C2
4
C2
3
G in i= 0 .0 0 0
G in i= 0 .2 7 8
G in i= 0 .4 4 4
G in i= 0 .5 0 0
Examples for computing GINI
GINI ( t )  1   [ p ( j | t )]
2
j
C1
0
P(C1) = 0/6 = 0
C2
6
Gini = 1 – P(C1)2 – P(C2)2 = 1 – 0 – 1 = 0
C1
1
P(C1) = 1/6
C2
5
Gini = 1 – (1/6)2 – (5/6)2 = 0.278
C1
2
P(C1) = 2/6
C2
4
Gini = 1 – (2/6)2 – (4/6)2 = 0.444
P(C2) = 6/6 = 1
P(C2) = 5/6
P(C2) = 4/6
Splitting Based on GINI
• Used in CART, SLIQ, SPRINT.
• When a node p is split into k partitions (children), the quality of
split is computed as,
k
GINI
split


i 1
where,
ni
GINI ( i )
n
ni = number of records at child i,
n = number of records at node p.
Binary Attributes: Computing GINI Index


Splits into two partitions
Effect of Weighing partitions:

Larger and Purer Partitions are sought for.
P a re n t
B?
Yes
No
C1
6
C2
6
G in i = 0 .5 0 0
Gini(N1)
= 1 – (5/6)2 – (2/6)2
= 0.194
Gini(N2)
= 1 – (1/6)2 – (4/6)2
= 0.528
Node N1
Node N2
N1
N2
C1
5
1
C2
2
4
G in i= 0 .3 3 3
Gini(Children)
= 7/12 * 0.194 +
5/12 * 0.528
= 0.333
Categorical Attributes: Computing Gini Index
• For each distinct value, gather counts for each class in the
dataset
• Use the count matrix to make decisions
Multi-way split
Two-way split
(find best partition of values)
C a rT y p e
C a rT y p e
F a m ily S p o rts L u x u ry
{S p o rts ,
{F a m ily }
L u x u ry }
C1
1
2
1
C2
4
1
1
G in i
0 .3 9 3
C a rT y p e
{S p o rts }
{F a m ily ,
L u x u ry }
C1
3
1
C1
2
2
C2
2
4
C2
1
5
G in i
0 .4 0 0
G in i
0 .4 1 9
Continuous Attributes: Computing Gini
Index
• Use Binary Decisions based on one value
• Several Choices for the splitting value
– Number of possible splitting values
= Number of distinct values
• Each splitting value has a count matrix
associated with it
– Class counts in each of the partitions, A < v
and A  v
• Simple method to choose best v
– For each v, scan the database to gather
count matrix and compute its Gini index
– Computationally Inefficient! Repetition of
work.
Tid Refund Marital
Status
Taxable
Income Cheat
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
No
7
Yes
Divorced 220K
No
8
No
Single
85K
Yes
9
No
Married
75K
No
10
No
Single
90K
Yes
60K
10
Taxable
Income
> 80K?
Yes
No
Continuous Attributes: Computing Gini
Index...
• For efficient computation: for each attribute,
– Sort the attribute on values
– Linearly scan these values, each time updating the count matrix and
computing gini index
– Choose the split position that has the least gini index
Cheat
No
No
No
Yes
Yes
Yes
No
No
No
No
100
120
125
220
T a x a b le In c o m e
60
Sorted Values
70
55
Split Positions
75
65
85
72
90
80
95
87
92
97
110
122
172
230
<=
>
<=
>
<=
>
<=
>
<=
>
<=
>
<=
>
<=
>
<=
>
<=
>
<=
>
Yes
0
3
0
3
0
3
0
3
1
2
2
1
3
0
3
0
3
0
3
0
3
0
No
0
7
1
6
2
5
3
4
3
4
3
4
3
4
4
3
5
2
6
1
7
0
G in i
0 .4 2 0
0 .4 0 0
0 .3 7 5
0 .3 4 3
0 .4 1 7
0 .4 0 0
0 .3 0 0
0 .3 4 3
0 .3 7 5
0 .4 0 0
0 .4 2 0
Alternative Splitting Criteria based on INFO
• Entropy at a given node t:
Entropy ( t )    p ( j | t ) log p ( j | t )
j
(NOTE: p( j | t) is the relative frequency of class j at node t).
– Measures homogeneity of a node.
• Maximum (log nc) when records are equally distributed
among all classes implying least information
• Minimum (0.0) when all records belong to one class,
implying most information
– Entropy based computations are similar to the GINI
index computations
Examples for computing Entropy
Entropy ( t )    p ( j | t ) log p ( j | t )
j
2
C1
0
P(C1) = 0/6 = 0
C2
6
Entropy = – 0 log 0 – 1 log 1 = – 0 – 0 = 0
C1
1
P(C1) = 1/6
C2
5
Entropy = – (1/6) log2 (1/6) – (5/6) log2 (1/6) = 0.65
C1
2
P(C1) = 2/6
C2
4
Entropy = – (2/6) log2 (2/6) – (4/6) log2 (4/6) = 0.92
P(C2) = 6/6 = 1
P(C2) = 5/6
P(C2) = 4/6
Splitting Based on INFO...
• Information Gain:
 n

 Entropy ( p )    Entropy ( i ) 
 n

k
GAIN
i
split
i 1
Parent Node, p is split into k partitions;
ni is number of records in partition i
– Measures Reduction in Entropy achieved because of the split. Choose
the split that achieves most reduction (maximizes GAIN)
– Used in ID3 and C4.5
– Disadvantage: Tends to prefer splits that result in large number of
partitions, each being small but pure.
Splitting Based on INFO...
• Gain Ratio:
GainRATIO
split

GAIN
Split
SplitINFO
 
k
SplitINFO
i 1
n
i
log
n
Parent Node, p is split into k partitions
ni is the number of records in partition i
– Adjusts Information Gain by the entropy of the partitioning (SplitINFO).
Higher entropy partitioning (large number of small partitions) is
penalized!
– Used in C4.5
– Designed to overcome the disadvantage of Information Gain
n
i
n
Splitting Criteria based on Classification Error
• Classification error at a node t :
Error ( t )  1  max P ( i | t )
i
• Measures misclassification error made by a node.
• Maximum (1 - 1/nc) when records are equally distributed among all
classes, implying least interesting information
• Minimum (0.0) when all records belong to one class, implying most
interesting information
Examples for Computing Error
Error ( t )  1  max P ( i | t )
i
C1
0
P(C1) = 0/6 = 0
C2
6
Error = 1 – max (0, 1) = 1 – 1 = 0
C1
1
P(C1) = 1/6
C2
5
Error = 1 – max (1/6, 5/6) = 1 – 5/6 = 1/6
C1
2
P(C1) = 2/6
C2
4
Error = 1 – max (2/6, 4/6) = 1 – 4/6 = 1/3
P(C2) = 6/6 = 1
P(C2) = 5/6
P(C2) = 4/6
Comparison among Splitting Criteria
For a 2-class problem:
Misclassification Error vs Gini
P a re n t
A?
Yes
No
Node N1
Gini(N1)
= 1 – (3/3)2 – (0/3)2
=0
Gini(N2)
= 1 – (4/7)2 – (3/7)2
= 0.489
Node N2
C1
C2
N1
3
0
N2
4
3
C1
7
C2
3
G in i = 0 .4 2
Gini(Children)
= 3/10 * 0
+ 7/10 * 0.489
= 0.342
Tree Induction
• Greedy strategy.
– Split the records based on an attribute test that
optimizes certain criterion.
• Issues
– Determine how to split the records
• How to specify the attribute test condition?
• How to determine the best split?
– Determine when to stop splitting
Stopping Criteria for Tree Induction
• Stop expanding a node when all the records
belong to the same class
• Stop expanding a node when all the records
have similar attribute values
• Early termination (to be discussed later)
Decision Tree Based Classification
• Advantages:
– Inexpensive to construct
– Extremely fast at classifying unknown records
– Easy to interpret for small-sized trees
– Accuracy is comparable to other classification
techniques for many simple data sets
Example: C4.5
•
•
•
•
•
Simple depth-first construction.
Uses Information Gain
Sorts Continuous Attributes at each node.
Needs entire data to fit in memory.
Unsuitable for Large Datasets.
– Needs out-of-core sorting.
• You can download the software from:
http://www.cse.unsw.edu.au/~quinlan/c4.5r8.tar.gz
Practical Issues of Classification
• Underfitting and Overfitting
• Missing Values
• Costs of Classification
Underfitting and Overfitting (Example)
500 circular and 500
triangular data points.
Circular points:
0.5  sqrt(x12+x22)  1
Triangular points:
sqrt(x12+x22) > 0.5 or
sqrt(x12+x22) < 1
Underfitting and Overfitting
Overfitting
Underfitting: when model is too simple, both training and test errors are large
Overfitting due to Noise
Decision boundary is distorted by noise point
Overfitting due to Insufficient Examples
Lack of data points in the lower half of the diagram makes it difficult
to predict correctly the class labels of that region
- Insufficient number of training records in the region causes the
decision tree to predict the test examples using other training
records that are irrelevant to the classification task
Notes on Overfitting
• Overfitting results in decision trees that are
more complex than necessary
• Training error no longer provides a good
estimate of how well the tree will perform on
previously unseen records
• Need new ways for estimating errors
Estimating Generalization Errors
• Re-substitution errors: error on training ( e(t) )
• Generalization errors: error on testing ( e’(t))
• Methods for estimating generalization errors:
– Optimistic approach: e’(t) = e(t)
– Pessimistic approach:
•
•
•
For each leaf node: e’(t) = (e(t)+0.5)
Total errors: e’(T) = e(T) + N  0.5 (N: number of leaf nodes)
For a tree with 30 leaf nodes and 10 errors on training
(out of 1000 instances):
Training error = 10/1000 = 1%
Generalization error = (10 + 300.5)/1000 = 2.5%
– Reduced error pruning (REP):
• uses validation data set to estimate generalization
error
Occam’s Razor
• Given two models of similar generalization
errors, one should prefer the simpler model
over the more complex model
• For complex models, there is a greater chance
that it was fitted accidentally by errors in data
• Therefore, one should include model
complexity when evaluating a model
Minimum Description Length
(MDL)
X
X1
X2
X3
X4
y
1
0
0
1
…
…
Xn
1
A?
Yes
No
0
B?
B1
A
B2
C?
1
C1
C2
0
1
B
X
X1
X2
X3
X4
y
?
?
?
?
…
…
Xn
?
• Cost(Model,Data) = Cost(Data|Model) + Cost(Model)
– Cost is the number of bits needed for encoding.
– Search for the least costly model.
• Cost(Data|Model) encodes the misclassification errors.
• Cost(Model) uses node encoding (number of children) plus splitting
condition encoding.
How to Address Overfitting
• Pre-Pruning (Early Stopping Rule)
– Stop the algorithm before it becomes a fully-grown tree
– Typical stopping conditions for a node:
• Stop if all instances belong to the same class
• Stop if all the attribute values are the same
– More restrictive conditions:
• Stop if number of instances is less than some user-specified threshold
• Stop if class distribution of instances are independent of the available features
(e.g., using  2 test)
• Stop if expanding the current node does not improve impurity
measures (e.g., Gini or information gain).
How to Address Overfitting…
• Post-pruning
– Grow decision tree to its entirety
– Trim the nodes of the decision tree in a bottom-up
fashion
– If generalization error improves after trimming,
replace sub-tree by a leaf node.
– Class label of leaf node is determined from
majority class of instances in the sub-tree
– Can use MDL for post-pruning
Example of Post-Pruning
Training Error (Before splitting) = 10/30
Class = Yes
20
Pessimistic error = (10 + 0.5)/30 = 10.5/30
Class = No
10
Training Error (After splitting) = 9/30
Error = 10/30
Pessimistic error (After splitting)
= (9 + 4  0.5)/30 = 11/30
A?
A1
PRUNE!
A4
A3
A2
Class =
Yes
8
Class =
Yes
3
Class =
Yes
4
Class =
Yes
5
Class =
No
4
Class =
No
4
Class =
No
1
Class =
No
1
Examples of Post-pruning
Case 1:
– Optimistic error?
Don’t prune for both cases
– Pessimistic error?
C0: 11
C1: 3
C0: 2
C1: 4
C0: 14
C1: 3
C0: 2
C1: 2
Don’t prune case 1, prune case 2
– Reduced error pruning?
Case 2:
Depends on validation set
Handling Missing Attribute Values
• Missing values affect decision tree
construction in three different ways:
– Affects how impurity measures are computed
– Affects how to distribute instance with missing
value to child nodes
– Affects how a test instance with missing value is
classified
Computing Impurity Measure
Before Splitting:
Entropy(Parent)
= -0.3 log(0.3)-(0.7)log(0.7) = 0.8813
Tid Refund Marital
Status
Taxable
Income Class
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
No
7
Yes
Divorced 220K
No
8
No
Single
85K
Yes
Entropy(Refund=Yes) = 0
9
No
Married
75K
No
10
?
Single
90K
Yes
Entropy(Refund=No)
= -(2/6)log(2/6) – (4/6)log(4/6) = 0.9183
60K
C la ss
= Yes
C la ss
= No
R e fu n d = Y e s
0
3
R e fu n d = N o
2
4
R e fu n d = ?
1
0
Split on Refund:
10
Missing
value
Entropy(Children)
= 0.3 (0) + 0.6 (0.9183) = 0.551
Gain = 0.9  (0.8813 – 0.551) = 0.3303
Distribute Instances
Tid Refund Marital
Status
Taxable
Income Class
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
No
7
Yes
Divorced 220K
No
8
No
Single
85K
Yes
9
No
Married
75K
No
60K
Tid Refund Marital
Status
Taxable
Income Class
10
90K
Single
?
Yes
10
Refund
Yes
No
Class=Yes
0 + 3/9
Class=Yes
2 + 6/9
Class=No
3
Class=No
4
Probability that Refund=Yes is 3/9
10
Refund
Yes
Probability that Refund=No is 6/9
No
Class=Yes
0
Cheat=Yes
2
Class=No
3
Cheat=No
4
Assign record to the left child with
weight = 3/9 and to the right child
with weight = 6/9
Classify Instances
Married
New record:
Tid Refund Marital
Status
Taxable
Income Class
11
85K
No
?
?
10
Single
Divorce
d
Total
Class=No
3
1
0
4
Class=Yes
6/9
1
1
2.67
Total
3.67
2
1
6.67
Refund
Yes
NO
No
Single,
Divorced
MarSt
Married
TaxInc
< 80K
NO
NO
> 80K
YES
Probability that Marital Status
= Married is 3.67/6.67
Probability that Marital Status
={Single,Divorced} is 3/6.67
Scalable Decision Tree Induction Methods
• SLIQ (EDBT’96 — Mehta et al.)
– Builds an index for each attribute and only class list and the current
attribute list reside in memory
• SPRINT (VLDB’96 — J. Shafer et al.)
– Constructs an attribute list data structure
• PUBLIC (VLDB’98 — Rastogi & Shim)
– Integrates tree splitting and tree pruning: stop growing the tree earlier
• RainForest (VLDB’98 — Gehrke, Ramakrishnan & Ganti)
– Builds an AVC-list (attribute, value, class label)
• BOAT (PODS’99 — Gehrke, Ganti, Ramakrishnan & Loh)
– Uses bootstrapping to create several small samples
Postscript: learning models
Batch learning
Training data
Learning
Algorithm
On-line learning
See a new point x
predict label
Classifier f
See y
test
Update
classifer
Preprocessing step
Generative and Discriminative
Classifiers
Generative vs. Discriminative Classifiers
Training classifiers involves estimating f: X  Y, or P(Y|X)
Discriminative classifiers (also called ‘informative’ by
Rubinstein&Hastie):
1. Assume some functional form for P(Y|X)
2. Estimate parameters of P(Y|X) directly from training data
Generative classifiers
1. Assume some functional form for P(X|Y), P(X)
2. Estimate parameters of P(X|Y), P(X) directly from training data
3. Use Bayes rule to calculate P(Y|X= xi)
Bayes Formula
Generative Model
•
•
•
•
•
Color
Size
Texture
Weight
…
Discriminative Model
• Logistic Regression
•
•
•
•
•
Color
Size
Texture
Weight
…
Comparison
• Generative models
– Assume some functional form for P(X|Y), P(Y)
– Estimate parameters of P(X|Y), P(Y) directly from
training data
– Use Bayes rule to calculate P(Y|X= x)
• Discriminative models
– Directly assume some functional form for P(Y|X)
– Estimate parameters of P(Y|X) directly from
training data
Probability Basics
•
Prior, conditional and joint probability for random
variables
– Prior probability:
P(X )
– Conditional probability:
– Joint probability:
– Relationship:
•
X  ( X 1 , X 2 ), P ( X )  P(X
P(X
– Independence:
P ( X 1 | X 2 ) , P(X 2 | X 1 )
1
1
,X 2 )
,X 2 )  P ( X 2 | X 1 ) P ( X 1 )  P ( X 1 | X 2 ) P ( X 2 )
P ( X 2 | X 1 )  P ( X 2 ), P ( X 1 | X 2 )  P ( X 1 ), P(X
1
,X 2 )  P ( X 1 ) P ( X 2 )
Bayesian Rule
P (C | X ) 
P ( X |C ) P (C )
P(X )
Posterior 
Likelihood  Prior
Evidence
111
Probability Basics
•
Quiz:
We have two six-sided dice. When they are tolled, it could end up
with the following occurance: (A) dice 1 lands on side “3”, (B) dice 2 lands
on side “1”, and (C) Two dice sum to eight. Answer the following
questions:
1) P ( A )  ?
2) P(B)  ?
3) P(C)
?
4) P ( A | B )  ?
5) P ( C | A )  ?
6) P ( A , B )  ?
7) P ( A , C )  ?
8) Is P ( A , C ) equals
P(A)  P(C) ?
112
Probabilistic Classification
•
Establishing a probabilistic model for classification
– Discriminative model
P ( C | X ) C  c 1 ,   , c L , X  (X 1 ,   , X n )
P(c1 | x )
P(c 2 | x )

P(c L | x )
Discriminative
Probabilistic Classifier
x1
x2

xn
x  ( x 1 , x 2 ,   , x n )
113
Probabilistic Classification
•
Establishing a probabilistic model for classification
(cont.)
– Generative model
P ( X | C ) C  c 1 ,   , c L , X  (X 1 ,   , X n )
P(x | c1 )
P(x |c2 )
P(x |cL )
Generative
Probabilistic Model
Generative
Probabilistic Model
Generative
Probabilistic Model
for Class 1
for Class 2
x1
x2

xn x1
x2


for Class L
xn
x1
x2

x  ( x 1 , x 2 ,   , x n )
114
xn
Probabilistic Classification
•
MAP classification rule
– MAP: Maximum A Posterior
– Assign x to c* if
P (C  c | X  x )  P (C  c | X  x )
*
•
c  c , c  c 1 ,   , c L
*
Generative classification with the MAP rule
– Apply Bayesian rule to convert them into posterior probabilities
P (C  c i | X  x ) 
P ( X  x | C  c i ) P (C  c i )
P(X  x)
 P ( X  x | C  c i ) P (C  c i )
for i  1 , 2 ,   , L
– Then apply the MAP rule
115
Naïve Bayes
•
Bayes classification
P ( C | X )  P ( X | C ) P ( C )  P ( X 1 ,   , X n | C ) P ( C )
Difficulty: learning the joint probability
•
P ( X 1 ,   , X n | C )
Naïve Bayes classification
– Assumption that all input attributes are conditionally
independent!
P ( X 1 , X 2 ,   , X n | C )  P ( X 1 | X 2 ,   , X n ; C ) P ( X 2 ,   , X n | C )
 P ( X 1 | C ) P ( X 2 ,   , X n | C )
 P( X 1 |C )P( X 2 |C )    P( X n |C )
– MAP classification rule: for
x  ( x 1 , x 2 ,   , x n )
[ P ( x 1 | c )    P ( x n | c )] P ( c )  [ P ( x 1 | c )    P ( x n | c )] P ( c ), c  c , c  c 1 ,   , c L
*
*
*
*
116
Naïve Bayes
•
Naïve Bayes Algorithm (for discrete input attributes)
– Learning Phase: Given a training set S,
For each target value of c i (c i  c 1 ,   , c L )
Pˆ ( C  c i )  estimate
For every attribute
P ( C  c i ) with examples
value x jk of each attribute
Pˆ ( X j  x jk | C  c i )  estimate
in S ;
X j ( j  1 ,   , n ; k  1 ,   , N j )
P ( X j  x jk | C  c i ) with examples
Output: conditional probability tables; for
in S ;
X j , N j  Lelements
– Test Phase: Given an unknown instance X   ( a  ,   , a  ) ,
1
n
Look up tables to assign the label c* to X’ if
*
*
*
*
[ Pˆ ( a 1 | c )    Pˆ ( a n | c )] Pˆ ( c )  [ Pˆ ( a 1 | c )    Pˆ ( a n | c )] Pˆ ( c ), c  c , c  c 1 ,   , c L
117
Example
•
Example: Play Tennis
118
Example
•
Learning Phase
Outlook
Play=Yes
Play=No
Sunny
2/9
3/5
4/9
0/5
3/9
2/5
Overcast
Rain
Humidity
Play=Yes
Play=No
High
3/9
Normal
6/9
Temperature
Play=Yes
Play=No
Hot
2/9
2/5
Mild
4/9
2/5
Cool
3/9
1/5
Wind
Play=Yes
Play=No
4/5
Strong
3/9
3/5
1/5
Weak
6/9
2/5
P(Play=Yes) = 9/14
P(Play=No) = 5/14
119
Example
•
Test Phase
– Given a new instance,
x’=(Outlook=Sunny, Temperature=Cool, Humidity=High, Wind=Strong)
– Look up tables
P(Outlook=Sunny|Play=Yes) = 2/9
P(Outlook=Sunny|Play=No) = 3/5
P(Temperature=Cool|Play=Yes) = 3/9 P(Temperature=Cool|Play==No) = 1/5
P(Huminity=High|Play=No) = 4/5
P(Huminity=High|Play=Yes) = 3/9
P(Wind=Strong|Play=Yes) = 3/9
P(Wind=Strong|Play=No) = 3/5
P(Play=Yes) = 9/14
P(Play=No) = 5/14
– MAP rule
P(Yes|x’): [P(Sunny|Yes)P(Cool|Yes)P(High|Yes)P(Strong|Yes)]P(Play=Yes) =
0.0053
P(No|x’): [P(Sunny|No) P(Cool|No)P(High|No)P(Strong|No)]P(Play=No) =
0.0206
Given the fact P(Yes|x’) < P(No|x’), we label x’ to be “No”.
120
Example
•
Test Phase
– Given a new instance,
x’=(Outlook=Sunny, Temperature=Cool, Humidity=High, Wind=Strong)
– Look up tables
P(Outlook=Sunny|Play=Yes) = 2/9
P(Outlook=Sunny|Play=No) = 3/5
P(Temperature=Cool|Play=Yes) = 3/9 P(Temperature=Cool|Play==No) = 1/5
P(Huminity=High|Play=No) = 4/5
P(Huminity=High|Play=Yes) = 3/9
P(Wind=Strong|Play=Yes) = 3/9
P(Wind=Strong|Play=No) = 3/5
P(Play=Yes) = 9/14
P(Play=No) = 5/14
– MAP rule
P(Yes|x’): [P(Sunny|Yes)P(Cool|Yes)P(High|Yes)P(Strong|Yes)]P(Play=Yes) =
0.0053
P(No|x’): [P(Sunny|No) P(Cool|No)P(High|No)P(Strong|No)]P(Play=No) =
0.0206
Given the fact P(Yes|x’) < P(No|x’), we label x’ to be “No”.
121
Relevant Issues
•
Violation of Independence Assumption
– For many real world tasks,
P ( X 1 ,   , X n | C )  P ( X 1 | C )    P ( X n | C )
– Nevertheless, naïve Bayes works surprisingly well anyway!
•
Zero conditional probability Problem
– If no example contains the attribute value
– In this circumstance,
X j  a jk , Pˆ ( X j  a jk | C  c i )  0
Pˆ ( x 1 | c i )    Pˆ ( a jk | c i )    Pˆ ( x n | c i )  0during
test
– For a remedy, conditional probabilities estimated with
n  mp
Pˆ ( X j  a jk | C  c i )  c
nm
n c : number of training examples for w hich X j  a jk and C  c i
n : number of training examples for w hich C  c i
p : prior estimate (usually, p  1 / t for t possible values of X j )
m : w eight to prior (number of " virtual" examples,
m  1)
122
Relevant Issues
•
Continuous-valued Input Attributes
– Numberless values for an attribute
– Conditional probability modeled with the normal distribution
Pˆ ( X j | C  c i ) 
1
2 
ji
 ( X j   ji ) 2
exp  
2

2  ji





 ji : mean (avearage) of attribute values X j of examples
for w hich C  c i
 ji : standard deviation of attribute values X j of examples
for w hich C  c i
– Learning Phase: for X  ( X 1 ,   , X n ), C  c 1 ,   , c L
Output: n  L normal distributions and P ( C  c i ) i  1 ,   , L
– Test Phase: for X   ( X 1 ,   , X n )
•
•
Calculate conditional probabilities with all the normal distributions
Apply the MAP rule to make a decision
123
Conclusions
•
Naïve Bayes based on the independence assumption
– Training is very easy and fast; just requiring considering each
attribute in each class separately
– Test is straightforward; just looking up tables or calculating
conditional probabilities with normal distributions
•
A popular generative model
– Performance competitive to most of state-of-the-art classifiers even
in presence of violating independence assumption
– Many successful applications, e.g., spam mail filtering
– A good candidate of a base learner in ensemble learning
– Apart from classification, naïve Bayes can do more…
124