Predictive analytics using 'R' Programming

Classification: Decision Tree Induction

This presentation slides includes…
 Concept of Decision Tree
 Use of Decision Tree to classify data
 Basic algorithm to build Decision Tree
 Some illustrations
 Concept of Entropy
 Basic concept of entropy in information theory
 Mathematical formulation of entropy
 Calculation of entropy of a training set
 Decision Tree induction algorithms
 ID3
 CART
 C4.5
2

Basic Concept
 A Decision Tree is an important data structure known to solve
many computational problems
Example 9.1: Binary Decision Tree
A B C f
0 0 0 m0
0 0 1 m1
0 1 0 m2
0 1 1 m3
1 0 0 m4
1 0 1 m5
1 1 0 m6
1 1 1 m7
3

Basic Concept
 In Example 9.1, we have considered a decision tree where values
of any attribute if binary only. Decision tree is also possible where
attributes are of continuous data type
Example 9.2: Decision Tree with numeric data
4

Some Characteristics
 Decision tree may be n-ary, n 2.
≥
 There is a special node called root node.
 All nodes drawn with circle (ellipse) are called internal nodes.
 All nodes drawn with rectangle boxes are called terminal nodes
or leaf nodes.
 Edges of a node represent the outcome for a value of the node.
 In a path, a node with same label is never repeated.
 Decision tree is not unique, as different ordering of internal nodes
can give different decision tree. 5

Decision Tree and Classification Task
 Decision tree helps us to classify data.
 Internal nodes are some attribute
 Edges are the values of attributes
 External nodes are the outcome of classification
 Such a classification is, in fact, made by posing questions starting
from the root node to each terminal node.
6

Example 9.3 : Vertebrate Classification
What are the class label of Dragon and Shark?
Name Body
Temperatur
e
Skin
Cover
Gives
Birth
Aquatic
Creature
Aerial
Creatur
e
Has
Legs
Hibernate
s
Class
Human Warm hair yes no no yes no Mammal
Python Cold scales no no no no yes Reptile
Salmon Cold scales no yes no no no Fish
Whale Warm hair yes yes no no no Mammal
Frog Cold none no semi no yes yes Amphibian
Komodo Cold scales no no no yes no Reptile
Bat Warm hair yes no yes yes yes Mammal
Pigeon Warm feathers no no yes yes no Bird
Cat Warm fur yes no no yes no Mammal
Leopard Cold scales yes yes no no no Fish
Turtle Cold scales no semi no yes no Reptile
Penguin Warm feathers no semi no yes no Bird
Porcupine Warm quills yes no no yes yes Mammal
Eel Cold scales no yes no no no Fish
Salamande
r
Cold none no semi no yes yes Amphibian
7

Example 9.3 : Vertebrate Classification
 Suppose, a new species is discovered as follows.
 Decision Tree that can be inducted based on the data (in Example
9.3) is as follows.
Name Body
Temperatur
e
Skin
Cover
Gives
Birth
Aquatic
Creatur
e
Aerial
Creatur
e
Has
Legs
Hibernate
s
Class
Gila Monster cold scale no no no yes yes
?
8

 Example 9.3 illustrates how we can solve a classification problem
by asking a series of question about the attributes.
 Each time we receive an answer, a follow-up question is asked until
we reach a conclusion about the class-label of the test.
 The series of questions and their answers can be organized in the
form of a decision tree
 As a hierarchical structure consisting of nodes and edges
 Once a decision tree is built, it is applied to any test to classify it.
9

Definition of Decision Tree
Given a database D = here denotes a tuple, which is defined by a set
of attribute set of classes C = .
A decision tree T is a tree associated with D that has the following
properties:
• Each internal node is labeled with an attribute Ai
• Each edges is labeled with predicate that can be applied to the
attribute associated with the parent node of it
• Each leaf node is labeled with class cj
Definition 9.1: Decision Tree
10

Building Decision Tree
 In principle, there are exponentially many decision tree that can
be constructed from a given database (also called training data).
 Some of the tree may not be optimum
 Some of them may give inaccurate result
 Two approaches are known
 Greedy strategy
 A top-down recursive divide-and-conquer
 Modification of greedy strategy
 ID3
 C4.5
 CART, etc.
11

Built Decision Tree Algorithm
 Algorithm BuiltDT
 Input: D : Training data set
 Output: T : Decision tree
Steps
1. If all tuples in D belongs to the same class Cj
Add a leaf node labeled as Cj
Return // Termination condition
2. Select an attribute Ai (so that it is not selected twice in the same branch)
3. Partition D = { D1, D2, …, Dp} based on p different values of Ai in D
4. For each Dk ϵ D
Create a node and add an edge between D and Dk with label as the Ai’s attribute value in Dk
5. For each Dk ϵ D
BuildTD(Dk) // Recursive call
6. Stop 12

Node Splitting in BuildDT Algorithm
 BuildDT algorithm must provides a method for expressing an attribute test
condition and corresponding outcome for different attribute type
 Case: Binary attribute
 This is the simplest case of node splitting
 The test condition for a binary attribute generates only two outcomes
13

 Case: Nominal attribute
 Since a nominal attribute can have many values, its test condition can be expressed
in two ways:
 A multi-way split
 A binary split
 Muti-way split: Outcome depends on the number of distinct values for the
corresponding attribute
 Binary splitting by grouping attribute values
14

 Case: Ordinal attribute
 It also can be expressed in two ways:
 A multi-way split
 A binary split
 Muti-way split: It is same as in the case of nominal attribute
 Binary splitting attribute values should be grouped maintaining the order property
of the attribute values
15

 Case: Numerical attribute
 For numeric attribute (with discrete or continuous values), a test condition can be
expressed as a comparison set
 Binary outcome: A > v or A ≤ v
 In this case, decision tree induction must consider all possible split positions
 Range query : vi A < v
≤ i+1 for i = 1, 2, …, q (if q number of ranges are chosen)
 Here, q should be decided a priori
16

Illustration : BuildDT Algorithm
Example 9.4: Illustration of BuildDT Algorithm
 Consider a training data set as shown.
Person Gender Height Class
1 F 1.6 S
2 M 2.0 M
3 F 1.9 M
4 F 1.88 M
5 F 1.7 S
6 M 1.85 M
7 F 1.6 S
8 M 1.7 S
9 M 2.2 T
10 M 2.1 T
11 F 1.8 M
12 M 1.95 M
13 F 1.9 M
14 F 1.8 M
15 F 1.75 S
Attributes:
Gender = {Male(M), Female (F)} // Binary attribute
Height = {1.5, …, 2.5} // Continuous
attribute
Class = {Short (S), Medium (M), Tall (T)}
Given a person, we are to test in which class s/he
belongs
17

 To built a decision tree, we can select an attribute in two different orderings:
<Gender, Height> or <Height, Gender>
 Further, for each ordering, we can choose different ways of splitting
 Different instances are shown in the following.
 Approach 1 : <Gender, Height>
18

19

 Approach 2 : <Height, Gender>
20

Example 9.5: Illustration of BuildDT Algorithm
 Consider an anonymous database as shown.
A1 A2 A3 A4 Class
a11 a21 a31 a41 C1
a12 a21 a31 a42 C1
a11 a21 a31 a41 C1
a11 a22 a32 a41 C2
a11 a22 a32 a41 C2
a12 a22 a31 a41 C1
a11 a22 a32 a41 C2
a11 a22 a31 a42 C1
a11 a21 a32 a42 C2
a11 a22 a32 a41 C2
a12 a22 a31 a41 C1
a12 a22 a31 a42 C1
• Is there any “clue” that enables to
select the “best” attribute first?
• Suppose, following are two
attempts:
• A1A2A3A4 [Naïve]
• A3A2A4A1 [Random]
• Draw the decision trees in the above-
mentioned two cases.
• Are the trees different to classify any test
data?
• If any other sample data is added into the
database, is that likely to alter the decision
tree already obtained?
21

Concept of Entropy
If a point represents a gas molecule,
then which system has the more
entropy?
How to measure? ?
More organized or Less organized or
ordered (less probable) disordered (more probable)
More ordered Less ordered
less entropy higher entropy
23

Concept of Entropy
Universe!
What was its entropy value at its starting point?
24

An Open Challenge!
Two sheets showing the tabulation of marks obtained in a course
are shown.
Which tabulation of marks shows the “good” performance of the
class?
How you can measure the same?
Roll No. Assignment Project Mid-Sem End-Sem
12BT3FP06 89 99 56 91
10IM30013 95 98 55 93
12CE31005 98 96 58 97
12EC35015 93 95 54 99
12GG2005 90 91 53 98
12MI33006 91 93 57 97
13AG36001 96 94 58 95
13EE10009 92 96 56 96
13MA20012 88 98 59 96
14CS30017 94 90 60 94
14ME10067 90 92 58 95
14MT10038 99 89 55 93
Roll No. Assignment Project Mid-Sem End-Sem
12BT3FP06 19 59 16 71
10IM30013 37 38 25 83
12CE31005 38 16 48 97
12EC35015 23 95 54 19
12GG2005 40 71 43 28
12MI33006 61 93 47 97
13AG36001 26 64 48 75
13EE10009 92 46 56 56
13MA20012 88 58 59 66
14CS30017 74 20 60 44
14ME10067 50 42 38 35
14MT10038 29 69 25 33
25

Entropy and its Meaning
 Entropy is an important concept used in Physics in the context of heat and
thereby uncertainty of the states of a matter.
 At a later stage, with the growth of Information Technology, entropy becomes an
important concept in Information Theory.
 To deal with the classification job, entropy is an important concept, which is
considered as
 an information-theoretic measure of the “uncertainty” contained in a training data
 due to the presence of more than one classes.
26

Entropy in Information Theory
 The entropy concept in information theory first time coined by Claude
Shannon (1850).
 The first time it was used to measure the “information content” in messages.
 According to his concept of entropy, presently entropy is widely being used
as a way of representing messages for efficient transmission by
Telecommunication Systems.
27

 People, in general, are information hungry!
 Everybody wants to acquire information (from newspaper, library, nature,
fellows, etc.)
 Think how a crime detector do it to know about the crime from crime spot and
criminal(s).
 Kids annoyed their parents asking questions.
 In fact, fundamental thing is that we gather information asking questions (and
decision tree induction is no exception).
 We may note that information gathering may be with certainty or uncertainty.
Measure of Information Content
28

Example 9.6
a) Guessing a birthday of your classmate
It is with certainty ~
Whereas guessing the day of his/her birthday is .
This uncertainty, we may say varies between 0 to 1, both inclusive.
b) As another example, a question related to event with eventuality (or
impossibility) will be answered with 0 or 1 uncertainty.
 Does sun rises in the East? (answer is with 0 uncertainty)
 Will mother give birth to male baby? (answer is with ½ uncertainty)
 Is there a planet like earth in the galaxy? (answer is with an extreme uncertainty)
Measure of Information Content
29

Definition of Entropy
Suppose there are m distinct objects, which we want to identify by asking a
series of Yes/No questions. Further, we assume that m is an exact power of 2,
say , where .
The entropy of a set of m distinct values is the minimum number of yes/no
questions needed to determine an unknown values from these m possibilities.
Definition 9.2: Entropy
30

 How can we calculate the minimum number of questions, that is,
entropy?
 There are two approaches:
 Brute –force approach
 Clever approach.
Example 9.7: City quiz
Suppose, Thee is a quiz relating to guess a city out of 8 cities, which are as follows:
Bangalore, Bhopal, Bhubaneshwar, Delhi, Hyderabad, Kolkata, Madras, Mumbai
The question is, “Which city is called city of joy”?
Entropy Calculation
31

 Brute force approach
 We can ask “Is it city X?”,
 if yes stop, else ask next …
In this approach, we can ask such questions randomly choosing one city at a time. As a matter
of randomness, let us ask the questions, not necessarily in the order, as they are in the list.
Q.1: Is the city Bangalore? No
Q.2: Is the city Bhubaneswar? No
Q.3: Is the city Bhopal? No
Q.4: Is the city Delhi? No
Q.5: Is the city Hyderabad? No
Q.6: Is the city Madras? No
Q.7: Is the city Mumbai? No
No need to ask further question! Answer is already out by the Q.7. If asked randomly, each of
these possibilities is equally likely with probability . Hence on the average, we need
questions.
Approach 1: Brute-force search
32

 Clever approach (binary search)
 In this approach, we divide the list into two halves, pose a question for a half
 Repeat the same recursively until we get yes answer for the unknown.
Q.1: Is it Bangalore, Bhopal, Bhubaneswar or Delhi? No
Q.2: Is it Madras or Mumbai? No
Q.3: Is it Hyderabad? No
So after fixing 3 questions, we are able to crack the answer.
Note:
Approach 2 is considered to be the best strategy because it will invariably find the answer and
will do so with a minimum number of questions on the average than any other strategy.
Approach 1 occasionally do better (when you are lucky enough!)
 It is no coincidence that , and the minimum number of yes/no questions needed is 3.
 If m = 16, then , and we can argue that we need 4 questions to solve the problem. If m = 32,
then 5 questions, m = 256, then 8 questions and so on.
Approach 2: Clever approach
33

The minimum number of yes/no questions needed to identify an unknown
object from equally likely possible object is n.
Thus, if m is a power of 2, then the entropy of a set of m distinct objects
that are equally likely is
Lemma 9.1: Entropy calculation
Entropy Calculation
34

Entropy in Messages
 We know that the most conventional way to code information is using binary bits, that
is, using 0s and 1s.
 The answer to a question that can only be answered yes/no (with equal probability)
can be considered as containing one unit of information, that is, one bit.
 In other words, the unit of information can also be looked at as the amount of
information that can be coded using only 0s and 1s.
35

Entropy in Messages
Example 9.7: Information coding
 If we have two possible objects say male and female, then we use the coding
0 = female
1 = male
 We can encode four possible objects say East, West, North, South using two bits, for example
00 : North
01 : East
10 : West
11 : South
 We can encode eight values say eight different colours, we need to use three bits, such as
000 : Violet
001 : Indigo
010 : Blue
011 : Green
100 : Yellow
101 : Orange
110 : Red
111 : White
Thus, in general, to code m values, each in a distinct manner, we need n bits such that.
𝑚=2(¿2𝑛
,𝑛=1)
𝑚=4(¿ 2𝑛
,𝑛=2)
𝑚=8 (¿2𝑛
,𝑛=3)
36

Entropy in Messages
 In this point, we can note that to identify an object, if it is encoded with bits, then we
have to ask questions in an alternative way. For example
 Is the first bit 0?
 Is the second bit 0?
 Is the third bit 0? and so on
 Thus, we need n questions, if m objects are there such that .
 The above leads to (an alternative) and equivalent definition of entropy
The entropy of a set of m distinct values is the number of bits needed to encode
all the values in the most efficient way.
Definition 9.3: Entropy
37

 In the previous discussion, we have assumed that m, the number of distinct objects is exactly a
power of 2, that is for some and all m objects are equally likely.
 This is mere an assumption to make the discussion simplistic.
 In the following, we try to redefine the entropy calculation in more general case, that is, when
and not necessarily m objects are equally probable. Let us consider a different instance of yes/no
question game, which is as follows.
Example 9.8: Name game
 There are seven days: Sun, Mon, Tue, Wed, Thu, Fri, Sat.
•
 We are to identify a sequence of such values (each one chosen independently of the others, that is,
repetitions are allowed). Note that if k = 1, it is the type of game, we have already dealt with.
 We denote the minimum number of questions needed to identify a sequence of unknown values drawn
independently from possibilities as , the entropy in this case.
 In other words, is the number of questions required to discriminate amongst distinct possibilities.
Messages when ()
38

 Here, (as stated in the game of sequence of days) and k = 6 (say).
 An arbitrary sequence may be {Tue, Thu, Tue, Mon, Sun, Tue}, etc. There are possible sequences of six
days.
 From our previous understanding, we can say that the minimum number of questions that is required to
identify such a sequence is.
 Since, this is a non integer number, and the number of question should be an integer, we can say 17
questions are required. Thus,
 In general,
 Alternatively, the above can be written as,
Or
Messages when ()
39

Note that here is the average number of questions needed to determine each of the values
in a sequence of k values. By choosing a large enough value of k, that is, a long enough
sequence, the value of can be made as small as we wish. Thus, the average number of
questions required to determine each value can be made arbitrarily close to . This is
evident from our earlier workout, for example, tabulated below, for m = 7.
Entropy of Messages when ()
k No. Q
6 117649 16.84413 17 2.8333
21 58.95445 59 2.8095
1000 2807.3549 2808 2.8080
….. ….. ….. ….. …..
No. Q = Number of questions
Note that and . Further, i.e. (is independent of k and is a constant!)
40

Entropy of Messages when ()
The entropy of a set of m distinct objects is even when m is not exactly a power
of 2.
• We have arrived at a conclusion that for any value of m, irrespective of weather it
is a power of 2 or not.
Note: E is not necessarily be an integer always.
• Next, we are to have our observation, if all m objects are not equally probable.
• Suppose, denotes the frequency with which the of the m objects occurs, where
for all such that
Lemma 9.4: Entropy Calculation
41

Example 9.8: Discriminating among objects
• Suppose four objects which occur with frequencies , , and , respectively.
• Thus, in this example, and , , and .
• Using standard 2-bit encoding, we can represent them as .
• Also, we can follow variable length coding (also called Huffman coding) as an
improved way of representing them.
• The Huffman coding of with their frequencies , , and are shown below.
A = 1
B = 01
C = 001
D = 000
Discriminating amongst m values ()
42

• With the above representation, say if A is to be identified, then we
need to examine only one question, for B it is 2 and for C and D both, it
is 3.
• Thus, on the average, we need
• This is the number of yes/no questions to identify any one of the four
objects, whose frequency of occurrences are not uniform.
• This is simply in contrast to 2-bit encoding, where we need 2-bits
(questions) on the average.
43

• It may be interesting to note that even with variable length encoding, there are
several ways of encoding. Few of them are given below.
• The calculation of entropy in the observed cases can be obtained as:
• Anyway, key to finding the most efficient way of encoding is to assign
a smallest number of bits to the object with highest frequency and
so on.
• The above observation is also significant in the sense that it provides a
systematic way of finding a sequence of well-chosen question in
order to identify an object at a faster rate.
1) 1.75 2) 2 3) 3.875
44

Information Content
If an object occurs with frequency p, then the most efficient way to represent it
with bits.
Lemma 9.3: Information content
Example 9.9: Information content
• A which occurs with frequency is represented by 1-bit, B which occurs with
frequency represented by 2-bits and both C and D which occurs with frequency
are represented by 3 bits each.
Based on the previous discussion, we can easily prove the following lemma.
45

Entropy Calculation
If pi denotes the frequencies of occurrences of m distinct objects, then the entropy
E is
Theorem 9.4: Entropy calculation
We can generalize the above understanding as follows.
• If there are m objects with frequencies , ……., , then the average number of bits (i.e.
questions) that need to be examined a value, that is, entropy is the frequency of
occurrence of the value multiplied by the number of bits that need to be determined,
summed up values of from 1 to m.
Note:
• If all are equally likely, then and ; it is the special case.
46

Entropy of a Training Set
• If there are k classes , ……., and for denotes the number of occurrences of classes
divided by the total number of instances (i.e., the frequency of occurrence of ) in the
training set, then entropy of the training set is denoted by
Here, E is measured in “bits” of information.
Note:
• The above formula should be summed over the non-empty classes only, that is, classes
for which
• E is always a positive quantity
• E takes it’s minimum value (zero) if and only if all the instances have the same class
(i.e., the training set with only one non-empty class, for which the probability 1).
• Entropy takes its maximum value when the instances are equally distributed among k
possible classes. In this case, the maximum value of E is .
47

Example 9.10: OPTH dataset
Consider the OTPH data shown in the following table with total 24 instances in it.
Entropy of a Training Set
Age Eye sight Astigmatic Use Type Class
1
1
1
1
1
1
1
1
1
1
2
2
1
1
2
2
1
1
1
2
1
2
1
2
3
2
3
1
3
2
1
1
2
2
2
2
2
2
1
1
1
1
2
2
1
1
2
2
1
2
1
2
1
2
3
1
3
2
3
1
2
2
2
2
3
3
2
2
2
2
1
1
1
1
2
2
1
1
1
2
1
2
1
2
3
2
3
3
3
3
3
3
3
3
3
3
1
1
2
2
2
2
2
2
1
1
2
2
1
2
1
2
1
2
3
1
3
2
3
3
A coded
forms for
all values
of
attributes
are used to
avoid the
cluttering
in the table.
48

Specification of the attributes are as follows.
Entropy of a training set
Age Eye Sight Astigmatic Use Type
1: Young 1: Myopia 1: No 1: Frequent
2: Middle-aged 2: Hypermetropia 2: Yes 2: Less
3: Old
Class: 1: Contact Lens 2:Normal glass 3: Nothing
In the OPTH database, there are 3 classes and 4 instances with class 1, 5 instances
with class 2 and 15 instances with class 3. Hence, entropy E of the database is:
49

Note:
 The entropy of a training set implies the number of yes/no questions, on the average,
needed to determine an unknown test to be classified.
 It is very crucial to decide the series of questions about the value of a set of attribute,
which collectively determine the classification. Sometimes it may take one question,
sometimes many more.
 Decision tree induction helps us to ask such a series of questions. In other words, we
can utilize entropy concept to build a better decision tree.
How entropy can be used to build a decision tree is our next topic of discussion.
50

Decision Tree Induction Techniques
 Decision tree induction is a top-down, recursive and divide-and-conquer
approach.
 The procedure is to choose an attribute and split it into from a larger training
set into smaller training sets.
 Different algorithms have been proposed to take a good control over
1. Choosing the best attribute to be splitted, and
2. Splitting criteria
 Several algorithms have been proposed for the above tasks. In this lecture, we
shall limit our discussions into three important of them
• ID3
• C 4.5 51

ID3: Decision Tree Induction Algorithms
 Quinlan [1986] introduced the ID3, a popular short form of Iterative
Dichotomizer 3 for decision trees from a set of training data.
 In ID3, each node corresponds to a splitting attribute and each arc is a
possible value of that attribute.
 At each node, the splitting attribute is selected to be the most informative
among the attributes not yet considered in the path starting from the root.
53

Algorithm ID3
 In ID3, entropy is used to measure how informative a node is.
 It is observed that splitting on any attribute has the property that average entropy
of the resulting training subsets will be less than or equal to that of the previous
training set.
 ID3 algorithm defines a measurement of a splitting called Information Gain
to determine the goodness of a split.
 The attribute with the largest value of information gain is chosen as the splitting
attribute and
 it partitions into a number of smaller training sets based on the distinct values of
attribute under split.
54

Defining Information Gain
 We consider the following symbols and terminologies to define information gain,
which is denoted as α.
 D denotes the training set at any instant
 |D| denotes the size of the training set D
 E(D) denotes the entropy of the training set D
 The entropy of the training set D
E(D) = -)
 where the training set D has , , … , , the k number of distinct classes and
 , 0< is the probability that an arbitrary tuple in D belongs to class (i = 1, 2, … , k).
55

 pi can be calculated as
=
 where is the set of tuples of class in D.
 Suppose, we want to partition D on some attribute A having m distinct values
{
 to split D into m partitions {}, where (j = 1,2, … ,m) contains those tuples in
D that have outcome of A.
56

The weighted entropy denoted as 𝐸𝐴( ) for all partitions of
𝐷 D with respect to A is
given by:
= E()
Here, the term denotes the weight of the j-th training set.
More meaningfully, 𝐸𝐴( ) is the expected information required to classify a tuple
𝐷
from D based on the splitting of A.
Definition 9.4: Weighted Entropy
57

 Our objective is to take A on splitting to produce an exact classification (also
called pure), that is, all tuples belong to one class.
 However, it is quite likely that the partitions is impure, that is, they contain
tuples from two or more classes.
 In that sense, is a measure of impurities (or purity). A lesser value of
implying more power the partitions are.
Information gain, of the training set D splitting on the attribute A is given by
= E(D) -
In other words, gives us an estimation how much would be gained by splitting on
A. The attribute A with the highest value of should be chosen as the splitting
attribute for D.
Definition 9.5: Information Gain
58

Example 9.11 : Information gain on splitting OPTH
 Let us refer to the OPTH database discussed in Slide #49.
 Splitting on Age at the root level, it would give three subsets and as shown in
the tables in the following three slides.
 The entropy and of training sets and and corresponding weighted entropy
and are also shown alongside.
 The Information gain is then can be calculated as 0.0394.
 Recall that entropy of OPTH data set, we have calculated as E(OPTH) = 1.3261
(see Slide #50)
Information Gain Calculation
59

Example 9.11 : Information gain on splitting OPTH
Training set: (Age = 1)
Age Eye-sight Astigmatism Use type Class
1 1 1 1 3
1 1 1 2 2
1 1 2 1 3
1 1 2 2 1
1 2 1 1 3
1 2 1 2 2
1 2 2 1 3
1 2 2 2 1
Information Gain Calculation
() ()
() = 1.5
× 1.5 = 0.5000
60

2 1 1 1 3
2 1 1 2 2
2 1 2 1 3
2 1 2 2 1
2 2 1 1 3
2 2 1 2 2
2 2 2 1 3
2 2 2 2 3
Calculating Information Gain
() () ()
= 1.2988
× 1.2988 = 0.4329
61

3 1 1 1 3
3 1 1 2 3
3 1 2 1 3
3 1 2 2 1
3 2 1 1 3
3 2 1 2 2
3 2 2 1 3
3 2 2 2 3
= 1.3261 – (0.5000 + 0.4329 + 0.3504) = 0.0394
Calculating Information Gain
() ()
() = 1.0613
× 1.0613 = 0.3504
62

Information Gains for Different Attributes
• In the same way, we can calculate the information gains, when splitting the
OPTH database on Eye-sight, Astigmatic and Use Type. The results are
summarized below.
• Splitting attribute: Age
• Splitting attribute: Eye-sight
• Splitting attribute: Astigmatic
770
• Splitting attribute: Use Type
5488
63

Decision Tree Induction : ID3 Way
• The ID3 strategy of attribute selection is to choose to split on the attribute that
gives the greatest reduction in the weighted average entropy
• The one that maximizes the value of information gain
• In the example with OPTH database, the larger values of information gain is
5488
• Hence, the attribute should be chosen for splitting is “Use Type”.
• The process of splitting on nodes is repeated for each branch of the evolving
decision tree, and the final tree, which would look like is shown in the following
slide and calculation is left for practice.
64

Decision Tree Induction : ID3 Way
Age
Eye-
sight
Astigmati
c
𝑂𝑃𝑇𝐻
Use
Type
Age Eye Ast Use Clas
s
1 1 1 1 3
1 1 2 1 3
1 2 1 1 3
1 2 2 1 3
2 1 1 1 3
2 2 1 1 3
2 2 2 1 3
3 1 1 1 3
3 1 2 1 3
3 2 1 1 3
3 2 2 1 3
Age Eye Ast Use Clas
s
1 1 1 2 2
1 1 2 2 1
1 2 1 2 2
1 2 2 2 1
2 1 1 2 2
2 1 2 2 1
2 2 1 2 2
3 1 1 2 3
3 1 2 2 3
3 2 1 2 2
3 2 2 2 3
× × ×

𝐹𝑟𝑒𝑞𝑢𝑒𝑛𝑡 (1) 𝐿𝑒𝑠𝑠(2)
𝐷 1 𝐷 2
𝛼=0.0394 𝛼=0.395 𝛼=0.0394
𝐸(𝐷¿¿1)=?¿ 𝐸(𝐷¿¿2)=?¿
𝑬 (𝑶𝑷𝑻𝑯)=𝟏.𝟑𝟐𝟔𝟏
Age Eye-
sight
Astigmati
c
Age Eye-
sight
Astigmati
c
? ? ? ? ? ?
𝛼=? 𝛼=? 𝛼=? 𝛼=? 𝛼=? 𝛼=?
𝛼=0.3770
65

Frequency Table : Calculating α
• Calculation of entropy for each table and hence information gain for a
particular split appears tedious (at least manually)!
• As an alternative, we discuss a short-cut method of doing the same using
a special data structure called Frequency Table.
• Frequency Table: Suppose, denotes an attribute with attribute values in
it. For a given database , there are a set of say . Given this, a frequency
table will look like as follows.
66

Frequency Table : Calculating α
Class
𝑋
• Number of rows = Number
of classes
• Number of columns =
Number of attribute values
• = Frequency of for class
Assume that , the number of total instances of .
67

Calculation of α using Frequency Table
Example 9.12 : OTPH Dataset
With reference to OPTH dataset, and for the attribute Age, the frequency table
would look like
Column Sums
Age=1 Age=2 Age=3 Row
Sum
Class 1 2 1 1 4
Class 2 2 2 1 5
Class 3 4 5 6 15
Column
Sum
8 8 8 24
N=24
68

Calculation of α using Frequency Table
• The weighted average entropy then can be calculated from the frequency
table following the
• Calculate for all
(Entry Sum) and
• Calculate for all
(Column Sum) in the row of column sum
• Calculate
Example 9.13: OTPH Dataset
For the frequency table in Example 9.12, we have
69

Proof of Equivalence
• In the following, we prove the equivalence of the short-cut of entropy
calculation using Frequency Table.
• Splitting on an attribute with produces of the training dataset (of size ).
The subset contains all the instances for which the attribute takes its
value. Let denotes the number of instances in the subset. Then
• Let denotes the number of instances for which the classification is and
attribute takes its value. Then
70

Denoting as the entropy of the subset, we have
Therefore, the weighted average entropy of the splitting attribute is given by
71

Hence, the equivalence is
proved.
72

Limiting Values of Information Gain
• The Information gain metric used in ID3 always should be positive or
zero.
• It is always positive value because information is always gained (i.e.,
purity is improved) by splitting on an attribute.
• On the other hand, when a training set is such that if there are classes,
and the entropy of training set takes the largest value i.e., (this occurs
when the classes are balanced), then the information gain will be zero.
73

Limiting Values of Information Gain
Example 9.14: Limiting values of Information gain
Consider a training set shown below.
Data set Table A X Table X
Y
X Y
Clas
s
1 1 A
1 2 B
2 1 A
2 2 B
3 2 A
3 1 B
4 2 A
4 1 B
1 2 3 4
A 1 1 1 1
B 1 1 1 1
2 2 2 2
1 2
A 2 2
B 2 2
4 4
Frequency table of X
Frequency table of Y
Y Table Y
74

Limiting values of Information Gain
• Entropy of Table A is
(The maximum entropy).
• In this example, whichever attribute is chosen for splitting, each of the
branches will also be balanced thus each with maximum entropy.
• In other words, information gain in both cases (i.e., splitting on X as well as
Y) will be zero.
Note:
• The absence of information gain does not imply that there is no profit for
splitting on the attribute.
• Even if it is chosen for splitting, ultimately it will lead to a final decision tree
with the branches terminated by a leaf node and thus having an entropy of
zero.
• Information gain can never be a negative value.
75

Splitting of Continuous Attribute Values
• In the foregoing discussion, we assumed that an attribute to be splitted is
with a finite number of discrete values. Now, there is a great deal if the
attribute is not so, rather it is a continuous-valued attribute.
• There are two approaches mainly to deal with such a case.
1. Data Discretization: All values of the attribute can be discretized into a
finite number of group values and then split point can be decided at each
boundary point of the groups.
So, if there are of discrete values, then we have split points.
𝑣 1 𝑣 2 𝑣 3 𝑣 𝑁
76

Splitting of Continuous attribute values
2. Mid-point splitting: Another approach to avoid the data discretization.
• It sorts the values of the attribute and take the distinct values only in it.
• Then, the mid-point between each pair of adjacent values is considered as a
split-point.
• Here, if n-distinct values are there for the attribute , then we choose split
points as shown above.
• For example, there is a split point in between and
• For each split-point, we have two partitions: , and finally the point with
maximum information gain is the desired split point for that attribute.
𝑣1 𝑣2 𝑣3 𝑣𝑖 𝑣𝑖+1 𝑣𝑛
77

 It is observed that information gain measure used in ID3 is biased towards test with
many outcomes, that is, it prefers to select attributes having a large number of values.
 L. Breiman, J. Friedman, R. Olshen and C. Stone in 1984 proposed an algorithm to build
a binary decision tree also called CART decision tree.
 CART stands for Classification and Regression Tree
 In fact, invented independently at the same time as ID3 (1984).
 ID3 and CART are two cornerstone algorithms spawned a flurry of work on decision tree
induction.
 CART is a technique that generates a binary decision tree; That is, unlike ID3, in
CART, for each node only two children is created.
 ID3 uses Information gain as a measure to select the best attribute to be splitted, whereas
CART do the same but using another measurement called Gini index . It is also known
as Gini Index of Diversity and is denote as .
CART Algorithm
79

Gini Index of Diversity
Suppose, D is a training set with size |D| and be the set of k classifications and
be any attribute with m different values of it. Like entropy measure in ID3, CART
proposes Gini Index (denoted by G) as the measure of impurity of D. It can be
defined as follows.
where is the probability that a tuple in D belongs to class and can be estimated
as
where denotes the number of tuples in D with class .
Definition 9.6: Gini Index
80

Note
 measures the “impurity” of data set D.
 The smallest value of is zero
 which it takes when all the classifications are same.
 It takes its largest value
 when the classes are evenly distributed between the tuples, that is the frequency of
each class is .
81

Suppose, a binary partition on A splits D into and , then the weighted average
Gini Index of splitting denoted by is given by
This binary partition of D reduces the impurity and the reduction in impurity is
measured by
Definition 9.7: Gini Index of Diversity
82

 This is called the Gini Index of diversity.
 It is also called as “impurity reduction”.
 The attribute that maximizes the reduction in impurity (or equivalently, has
the minimum value of) is selected for the attribute to be splitted.
Gini Index of Diversity and CART
83

 The CART algorithm considers a binary split for each attribute.
 We shall discuss how the same is possible for attribute with more than two
values.
 Case 1: Discrete valued attributes
 Let us consider the case where A is a discrete-valued attribute having m
discrete values .
 To determine the best binary split on A, we examine all of the possible
subsets say of A that can be formed using the values of A.
 Each subset can be considered as a binary test for attribute A of the form .
n-ary Attribute Values to Binary Splitting
84

 Thus, given a data set D, we have to perform a test for an attribute value A like
 This test is satisfied if the value of A for the tuples is among the values listed in .
 If A has m distinct values in D, then there are possible subsets, out of which the
empty subset and the power set should be excluded (as they really do not
represent a split).
 Thus, there are possible ways to form two partitions of the dataset D, based on
the binary split of A.
𝐴∈𝑆𝐴
Yes No
𝐷1 𝐷2
D
85

Case2: Continuous valued attributes
 For a continuous-valued attribute, each possible split point must be taken into
account.
 The strategy is similar to that followed in ID3 to calculate information gain
for the continuous –valued attributes.
 According to that strategy, the mid-point between ai and ai+1 , let it be vi, then
𝐴 ≤ 𝑣𝑖
Yes No
𝐷1 𝐷2
𝑎1 𝑎2 𝑎𝑖 +1 𝑎𝑛
𝑣𝑖=
𝑎𝑖+ 𝑎𝑖+1
2
𝑎𝑖
86

 Each pair of (sorted) adjacent values is taken as a possible split-point say.
 is the set of tuples in D satisfying and in the set of tuples in D satisfying .
 The point giving the minimum Gini Index is taken as the split-point of the
attribute A.
Note
 The attribute A and either its splitting subset (for discrete-valued splitting
attribute) or split-point (for continuous valued splitting attribute) together
form the splitting criteria.
87

Example 9.15 : CART Algorithm
Suppose we want to build decision tree for the data set EMP as given in the table
below.
CART Algorithm : Illustration
Tuple# Age Salary Job Performance Select
1 Y H P A N
2 Y H P E N
3 M H P A Y
4 O M P A Y
5 O L G A Y
6 O L G E N
7 M L G E Y
8 Y M P A N
9 Y L G A Y
10 O M G A Y
11 Y M G E Y
12 M M P E Y
13 M H G A Y
14 O M P E N
Age
Y : young
M : middle-aged
O : old
Salary
L : low
M : medium
H : high
Job
G : government
P : private
Performance
A : Average
E : Excellent
Class : Select
Y : yes
N : no
88

For the EMP data set,
Now let us consider the calculation of for Age, Salary, Job and Performance.
89

Attribute of splitting: Age
The attribute age has three values, namely Y, M and O. So there are 6 subsets, that should
be considered for splitting as:
?
?
The best value of Gini Index while splitting attribute Age is
Age {O}
{O} {Y,M}
Yes No
90

Attribute of Splitting: Salary
The attribute salary has three values namely L, M and H. So, there are 6 subsets,
that should be considered for splitting as:
Salary {H}
{H} {L,M}
Yes No
91

Attribute of Splitting: job
Job being the binary attribute , we have
?
?
92

Attribute of Splitting: Performance
Job being the binary attribute , we have
?
?
Out of these gives the maximum value and hence, the attribute Salary would be
chosen for splitting subset or .
Note:
It can be noted that the procedure following “information gain” calculation
(i.e. ) and that of “impurity reduction” calculation ( i.e. ) are near about.
93

Calculating γ using Frequency Table
 We have learnt that splitting on an attribute gives a reduction in the average
Gini Index of the resulting subsets (as it does for entropy).
 Thus, in the same way the average weighted Gini Index can be calculated
using the same frequency table used to calculate information gain which is as
follows.
The G() for the subset
94

Calculating γ using Frequency Table
The average weighted Gini Index, () (assuming that attribute has m distinct
values is)
The above gives a formula for m-attribute values; however, it an be fine tuned to
subset of attributes also.
95

Illustration: Calculating γ using Frequency Table
Example 9.16 : Calculating γ using frequency table of OPTH
Let us consider the frequency table for OPTH database considered earlier. Also
consider the attribute with three values 1, 2 and 3. The frequency table is shown
below.
1 2 3
Class 1 2 1 1
Class 2 2 2 1
Class 3 4 5 6
Column sum 8 8 8
96

Now we can calculate the value of Gini Index with the following steps:
1. For each non-empty column, form the sum of the squares of the values in the
body of the table and divide by the column sum.
2. Add the values obtained for all columns and divided by , the size of the
database.
3. Subtract the total from 1.
As an example, with reference to the frequency table as mentioned just.
75
So,
97

Thus, the reduction in the value of Gini Index on splitting attribute is
where
The calculation can be extended to other attributes in the OTPH database and is
left as an exercise.
?
98

Example 9.17 : Comparing Decision Trees of EMP Data set
Compare two decision trees obtained using ID3 and CART for the EMP dataset. The
decision tree according to ID3 is given for your ready reference (subject to the verification)
Decision Tree using ID3
?
Decision Tree using CART
N
Age
Job Performance
Y
Y Y N
Y O
P G A E
Decision Trees with ID3 and CART Algorithms
99

Algorithm C 4.5 : Introduction
 J. Ross Quinlan, a researcher in machine learning developed a decision tree
induction algorithm in 1984 known as ID3 (Iterative Dichotometer 3).
 Quinlan later presented C4.5, a successor of ID3, addressing some
limitations in ID3.
 ID3 uses information gain measure, which is, in fact biased towards splitting
attribute having a large number of outcomes.
 For example, if an attribute has distinct values for all tuples, then it would
result in a large number of partitions, each one containing just one tuple.
 In such a case, note that each partition is pure, and hence the purity measure of the
partition, that is
101

Algorithm C4.5 : Introduction
Example 9.18 : Limitation of ID3
In the following, each tuple belongs to a unique class. The splitting on A is shown.
Thus, is maximum in such a situation.
102

Algorithm: C 4.5 : Introduction
 Although, the previous situation is an extreme case, intuitively, we can infer
that ID3 favours splitting attributes having a large number of values
 compared to other attributes, which have a less variations in their values.
 Such a partition appears to be useless for classification.
 This type of problem is called overfitting problem.
Note:
Decision Tree Induction Algorithm ID3 may suffer from overfitting problem.
103

Algorithm: C 4.5 : Introduction
 The overfitting problem in ID3 is due to the measurement of information
gain.
 In order to reduce the effect of the use of the bias due to the use of
information gain, C4.5 uses a different measure called Gain Ratio, denoted
as .
 Gain Ratio is a kind of normalization to information gain using a split
information.
104

Algorithm: C 4.5 : Gain Ratio
The gain ratio can be defined as follows. We first define split information
as
Here, m is the number of distinct values in A.
The gain ratio is then defined as where denotes the information gain on splitting
the attribute A in the dataset D.
Definition 9.8: Gain Ratio
105

Physical Interpretation of (D)
(D)
 The value of split information depends on
 the number of (distinct) values an attribute has and
 how uniformly those values are distributed.
 In other words, it represents the potential information generated by splitting a
data set D into m partitions, corresponding to the m outcomes of on attribute
A.
 Note that for each outcome, it considers the number of tuples having that
outcome with respect to the total number of tuples in D.
106

Physical Interpretation of (D )
Example 9.18 : (D)
𝐒𝐩𝐥𝐢𝐭 𝐢𝐧𝐟𝐨𝐫𝐦𝐚𝐭𝐢𝐨𝐧
 To illustrate (D), let us examine the case where there are 32 instances and
splitting an attribute A which has , , and sets of distinct values.
 Distribution 1 : Highly non-uniform distribution of attribute values
 Distribution 2
Frequenc
y
32 0 0 0
(D) = () 1 = 0
Frequenc
y
16 16 0 0
(D) = ()() = 1
107

 Distribution 3
 Distribution 4
 Distribution 5: Uniform distribution of attribute values
Frequenc
y
16 8 8 0
(D) = ()() 1.5
Frequenc
y
16 8 4 4
(D) = 1.75
Frequenc
y
8 8 8 8
(D) = (())*4 = ().0
108

 In general, if there are m attribute values, each occurring equally frequently, then the split
information is .
 Based on the Example 9.18, we can summarize our observation on split information as
under:
 Split information is 0 when there is a single attribute value. It is a trivial case and implies the
minimum possible value of split information.
 For a given data set, when instances are uniformly distributed with respect to the attribute values,
split information increases as the number of different attribute values increases.
 The maximum value of split information occur when there are many possible attribute values, all
are equally frequent.
Note:
 Split information varies between 0 and log2m (both inclusive)
109

Physical Interpretation of
 Information gain signifies how much information will be gained on partitioning the
values of attribute A
 Higher information gain means splitting of A is more desirable.
•
 On the other hand, split information forms the denominator in the gain ratio formula.
 This implies that higher the value of split information is, lower the gain ratio.
 In turns, it decreases the information gain.
 Further, information gain is large when there are many distinct attribute values.
 When many distinct values, split information is also a large value.
 This way split information reduces the value of gain ratio, thus resulting a balanced value
for information gain.
 Like information gain (in ID3), the attribute with the maximum gain ratio is selected
as the splitting attribute in C4.5.
110

Calculation of using Frequency Table
 The frequency table can be used to calculate the gain ratio for a given data set and an attribute.
•
 We have already learned the calculation of information gain using Frequency Table.
 To calculate gain ratio, in addition to information gain, we are also to calculate split information.
 This split information can be calculated from frequency table as follows.
 For each non-zero column sum say contribute || for the j-th column (i.e., the j-th value of the
attribute). Thus the split information is
(D) =
If there are m-columns in the frequency table.
Practice:
Using Gain ratio as the measurement of splitting attributes, draw the decision trees for
OPTH and EMP data sets. Give calculation of gain ratio at each node.
111

Summary of Decision Tree Induction Algorithms
 We have learned the building of a decision tree given a training data.
 The decision tree is then used to classify a test data.
 For a given training data D, the important task is to build the decision tree so
that:
 All test data can be classified accurately
 The tree is balanced and with as minimum depth as possible, thus the
classification can be done at a faster rate.
 In order to build a decision tree, several algorithms have been proposed.
These algorithms differ from the chosen splitting criteria, so that they satisfy
the above mentioned objectives as well as the decision tree can be induced
with minimum time complexity. We have studied three decision tree
induction algorithms namely ID3, CART and C4.5. A summary of these three
algorithms is presented in the following table. 112

Table 11.6
Algorithm Splitting Criteria Remark
ID3 Information Gain
(D)
Where
= Entropy of D (a measure of
uncertainty) =
where D is with set of k classes
, , … , and = ; Here, is the
set of tuples with class in D.
(D) = Weighted average
entropy when D is partitioned
on the values of attribute A = )
Here, m denotes the distinct
values of attribute A.
• The algorithm
calculates ,D) for all in
D and choose that
attribute which has
maximum ,D).
• The algorithm can handle
both categorical and
numerical attributes.
• It favors splitting those
attributes, which has a
large number of distinct
values.
113

CART Gini Index
(D)
where
= Gini index (a measure of
impurity)
=
Here, = and D is with k
number of classes
GA(D) = ) + ),
when D is partitioned into two
data sets and based on some
values of attribute A.
• The algorithm calculates
all binary partitions for
all possible values of
attribute A and choose
that binary partition
which has the maximum
• The algorithm is
computationally very
expensive when the
attribute A has a large
number of values.
114

C4.5 Gain Ratio
where
= Information gain of D (same
as in ID3, and
= splitting information
=
when D is partitioned into , ,
… , partitions corresponding
to m distinct attribute values of
A.
• The attribute A with
maximum value of is
selected for splitting.
• Splitting information is a
kind of normalization, so
that it can check the
biasness of information
gain towards the
choosing attributes with
a large number of distinct
values.
In addition to this, we also highlight few important characteristics
of decision tree induction algorithms in the following.
115

Notes on Decision Tree Induction algorithms
1. Optimal Decision Tree: Finding an optimal decision tree is an NP-complete
problem. Hence, decision tree induction algorithms employ a heuristic based
approach to search for the best in a large search space. Majority of the algorithms
follow a greedy, top-down recursive divide-and-conquer strategy to build decision
trees.
2. Missing data and noise: Decision tree induction algorithms are quite robust to the
data set with missing values and presence of noise. However, proper data pre-
processing can be followed to nullify these discrepancies.
3. Redundant Attributes: The presence of redundant attributes does not adversely
affect the accuracy of decision trees. It is observed that if an attribute is chosen for
splitting, then another attribute which is redundant is unlikely to chosen for splitting.
4. Computational complexity: Decision tree induction algorithms are computationally
inexpensive, in particular, when the sizes of training sets are large, Moreover, once a
decision tree is known, classifying a test record is extremely fast, with a worst-case
time complexity of O(d), where d is the maximum depth of the tree. 116

5. Data Fragmentation Problem: Since the decision tree induction algorithms
employ a top-down, recursive partitioning approach, the number of tuples
becomes smaller as we traverse down the tree. At a time, the number of
tuples may be too small to make a decision about the class representation,
such a problem is known as the data fragmentation. To deal with this
problem, further splitting can be stopped when the number of records falls
below a certain threshold.
6. Tree Pruning: A sub-tree can replicate two or more times in a decision tree
(see figure below). This makes a decision tree unambiguous to classify a test
record. To avoid such a sub-tree replication problem, all sub-trees except one
can be pruned from the tree.
A
B
C
C
D
D
1
1 1
1
0
0
0
117

7. Decision tree equivalence: The different splitting criteria followed in
different decision tree induction algorithms have little effect on the
performance of the algorithms. This is because the different heuristic
measures (such as information gain (), Gini index () and Gain ratio () are
quite consistent with each other); also see the figure below.
118

Reference
 The detail material related to this lecture can be found in
Data Mining: Concepts and Techniques, (3rd
Edn.), Jiawei Han, Micheline Kamber,
Morgan Kaufmann, 2015.
Introduction to Data Mining, Pang-Ning Tan, Michael Steinbach, and Vipin Kumar,
Addison-Wesley, 2014
119

Any question?
You may post your question(s) at the “Discussion Forum”
maintained in the course Web page!
120

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