1. Constructing Decision Trees

2. A Decision Tree Example
The weather data example.
ID code Outlook Temperature Humidity Windy Play
a
b
c
d
e
f
g
h
i
j
k
l
m
n
Sunny
Sunny
Overcast
Rainy
Rainy
Rainy
Overcast
Sunny
Sunny
Rainy
Sunny
Overcast
Overcast
Rainy
Hot
Hot
Hot
Mild
Cool
Cool
Cool
Mild
Cool
Mild
Mild
Mild
Hot
Mild
High
High
High
High
Normal
Normal
Normal
High
Normal
Normal
Normal
High
Normal
High
False
True
False
False
False
True
True
False
False
False
True
True
False
True
No
No
Yes
Yes
Yes
No
Yes
No
Yes
Yes
Yes
Yes
Yes
No

3. ~continues
Outlook
humidity windy
yes
no yes
yes no
sunny overcast rainy
high normal false true
Decision tree for the weather data.

4. The Process of Constructing a
Decision Tree
• Select an attribute to place at the root of the
decision tree and make one branch for every
possible value.
• Repeat the process recursively for each
branch.

5. Which Attribute Should Be Placed
at a Certain Node
• One common approach is based on the
information gained by placing a certain
attribute at this node.

6. Information Gained by Knowing
the Result of a Decision
• In the weather data example, there are 9
instances of which the decision to play is
“yes” and there are 5 instances of which the
decision to play is “no’. Then, the
information gained by knowing the result of
the decision is
bits.
940
.
0
14
5
log
14
5
14
9
log
14
9

























7. The General Form for Calculating
the Information Gain
• Entropy of a decision =
P1, P2, …, Pn are the probabilities of the n
possible outcomes.
n
n P
P
P
P
P
P log
log
log 2
2
1
1 





 

8. Information Further Required If
“Outlook” Is Placed at the Root
Outlook
yes
yes
no
no
no
yes
yes
yes
yes
yes
yes
yes
no
no
sunny overcast rainy
.
693
.
0
971
.
0
14
5
0
14
4
971
.
0
14
5
required
further
n
Informatio
bits


























9. Information Gained by Placing
Each of the 4 Attributes
• Gain(outlook) = 0.940 bits – 0.693 bits
= 0.247 bits.
• Gain(temperature) = 0.029 bits.
• Gain(humidity) = 0.152 bits.
• Gain(windy) = 0.048 bits.

10. The Strategy for Selecting an
Attribute to Place at a Node
• Select the attribute that gives us the largest
information gain.
• In this example, it is the attribute “Outlook”.
Outlook
2 “yes”
3 “no”
4 “yes” 3 “yes”
2 “no”
sunny overcast rainy

11. The Recursive Procedure for
Constructing a Decision Tree
• The operation discussed above is applied to each
branch recursively to construct the decision tree.
• For example, for the branch “Outlook = Sunny”,
we evaluate the information gained by applying
each of the remaining 3 attributes.
• Gain(Outlook=sunny;Temperature) = 0.971 – 0.4 =
0.571
• Gain(Outlook=sunny;Humidity) = 0.971 – 0 = 0.971
• Gain(Outlook=sunny;Windy) = 0.971 – 0.951 = 0.02

12. • Similarly, we also evaluate the information
gained by applying each of the remaining 3
attributes for the branch “Outlook = rainy”.
• Gain(Outlook=rainy;Temperature) = 0.971 –
0.951 = 0.02
• Gain(Outlook=rainy;Humidity) = 0.971 – 0.951
= 0.02
• Gain(Outlook=rainy;Windy) =0.971 – 0 =
0.971

13. The Over-fitting Issue
• Over-fitting is caused by creating decision
rules that work accurately on the training set
based on insufficient quantity of samples.
• As a result, these decision rules may not
work well in more general cases.

14. Example of the Over-fitting Problem
in Decision Tree Construction
bits
848
.
0
17
9
log
17
9
17
8
log
17
8
20
17
children
at the
entropy
Average
bits
993
.
0
20
9
log
20
9
20
11
log
20
11
subroot
at the
Entropy
2
2
2
2




















11 “Yes” and 9 “No” samples;
prediction = “Yes”
8 “Yes” and 9 “No” samples;
prediction = “No”
3 “Yes” and 0 “No” samples;
prediction = “Yes” Ai=0 Ai=1

15. • Hence, with the binary split, we gain more
information.
• However, if we look at the pessimistic error rate,
i.e. the upper bound of the confidence interval of
the error rate, we may get different conclusion.
• The formula for the pessimistic error rate is
• Note that the pessimistic error rate is a function of
the confidence level used.
  user.
by the
specified
level
confidence
the
is
and
,
samples,
of
number
the
is
rate,
error
observed
the
is
where
1
4
2
1
2
2
2
2
2
c
c
z
n
r
n
z
n
z
n
r
n
r
z
n
z
r
e










16. • The pessimistic error rates under 95%
confidence are
 
6598
.
0
17
645
.
1
1
1156
706
.
2
17
17
8
17
17
8
645
.
1
34
645
.
1
17
8
4742
.
0
3
645
.
1
1
36
645
.
1
645
.
1
6
645
.
1
6278
.
0
20
645
.
1
1
1600
706
.
2
20
45
.
0
20
45
.
0
645
.
1
40
645
.
1
45
.
0
2
2
2
17
8
2
2
2
3
0
2
2
2
20
9


















e
e
e

17. • Therefore, the average pessimistic error rate
at the children is
• Since the pessimistic error rate increases
with the split, we do not want to keep the
children. This practice is called “tree
pruning”.
6278
.
0
632
.
0
6598
.
0
20
17
4742
.
0
20
3






18. Tree Pruning based on 2 Test of
Independence
• We construct the corresponding
contingency table
Ai=
0
Ai=
1
Yes 3 8 11
No 0 9 9
3 17 20
11 “Yes” and
9 “No” samples;
8 “Yes” and
9 “No” samples;
3 “Yes” and
0 “No samples; Ai=0 Ai=1
15
.
1
20
7
1
9
20
7
1
9
-
9
20
9
3
20
9
3
-
0
20
1
1
17
20
1
1
17
-
8
20
3
11
20
3
11
-
3
statistic
The
2
2
2
2
2







 







 







 







 



19. • Therefore, we should not split the subroot
node, if we require that the 2 statistic must
be larger than 2
k,0.05 , where k is the degree
of freedom of the corresponding
contingency table.

20. Constructing Decision Trees based
on 2 test of Independence
• Using the following example, we can
construct a contingency table accordingly.
75 “Yes”s out of
100 samples;
Prediction = “Yes”
45 “Yes”s out of
50 samples;
20 “Yes”s out of
25 samples;
10 “Yes”s out of
25 samples; 100
100
50
100
25
100
25
100
25
5
15
5
100
75
45
10
20
2
1
0
No
Yes
Ai
Ai=0 Ai=1 Ai=2

21. • Therefore, we may say that the split is
statistically robust.
991
.
5
67
.
22
100
4
1
2
1
100
4
1
2
1
5
100
4
1
4
1
100
4
1
4
1
15
100
4
1
4
1
100
4
1
4
1
5
100
4
3
2
1
100
4
3
2
1
45
100
4
3
4
1
100
4
3
4
1
10
100
4
3
4
1
100
4
3
4
1
20
2
05
.
0
,
2
2
2
2
2
2
2
2













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





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



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


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







22. Assume that we have another attribute
Aj to consider
Aj=0 Aj=1
Yes 25 50 75
No 0 25 25
25 75 100
75 “Yes” out of
100 samples;
50 “Yes” out of
75 samples;
25 “Yes” out
of 25 samples; Aj=0 Aj=1
841
.
3
11
.
11
100
5
7
5
2
100
5
7
5
2
-
25
100
5
7
5
7
100
5
7
5
7
-
50
100
25
5
2
100
25
5
2
-
0
100
5
7
5
2
100
5
7
5
2
-
25
2
05
.
0
,
1
2
2
2
2
2









 
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
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

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

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



 







 




23. • Now, both Ai and Aj pass our criterion. How
should we make our selection?
• We can make our selection based on the
significance levels of the two contingency tables.
 
   
  .
10
8
0008
.
0
33
.
3
)
1
,
0
(
Prob
2
33
.
3
)
1
,
0
(
Prob
33
.
3
)
1
,
0
(
Prob
'
11
.
11
)
1
,
0
(
Prob
)
11
.
11
(
1
'
11
.
11
4
2
2
'
,
1 2
1



















N
N
N
N
F


 


24. • Therefore, Ai is preferred over Aj.
.
10
19
.
1
1
1
)
67
.
22
(
1
"
67
.
22
5
)
67
.
22
(
2
1
2
"
,
2 2
2



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

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









e
F
 


25. • If a subtree is as follows
• 2 = 4.543 < 5.991
• In this case, we do not want to carry out the split.
15 “Yes”s out of
20 samples;
9 “Yes”s out of
10 samples;
4 “Yes”s out of
5 samples;
2 “Yes”s out of
5 samples;
Termination of Split due to Low
Significance level

26. A More Realistic Example and
Some Remarks
• In the following example, a bank wants to
derive a credit evaluation tree for future use
based on the records of existing customers.
• As the data set shows, it is highly likely that
the training data set contains inconsistencies.
• Furthermore, some values may be missing.
• Therefore, for most cases, it is impossible to
derive perfect decision trees, i.e. decision
trees with 100% accuracy.

27. ~continues
Attributes Class
Education Annual Income Age Own House Sex Credit ranking
College High Old Yes Male Good
High school ----- Middle Yes Male Good
High school Middle Young No Female Good
College High Old Yes Male Poor
College High Old Yes Male Good
College Middle Young No Female Good
High school High Old Yes Male Poor
College Middle Middle ----- Female Good
High school Middle Young No Male Poor

28. ~continues
• A quality measure of decision trees can be
based on the accuracy. There are alternative
measures depending on the nature of
applications.
• Overfitting is a problem caused by making
the derived decision tree work accurately
for the training set. As a result, the decision
tree may work less accurately in the real
world.

29. ~continues
• There are two situations in which
overfitting may occur:
• insufficient number of samples at the subroot.
• some attributes are highly branched.
• A conventional practice for handling
missing values is to treat them as possible
attribute values. That is, each attribute has
one additional attribute value corresponding
to the missing value.

30. Alternative Measures of Quality of
Decision Trees
• The recall rate and precision are two widely used
measures.
• where C is the set of samples in the class and C’ is
the set of samples which the decision tree puts into
the class.
'
Precision
Rate
Recall
C
C
C
C
C
C
'
'





31. ~continues
• A situation in which the recall rate is the
main concern:
• “A bank wants to find all the potential credit
card customers”.
• A situation in which precision is the main
concern:
• “A bank wants to find a decision tree for credit
approval.”