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CSE5230/DMS/2002/7
Data Mining - CSE5230
Decision Trees
CSE5230 - Data Mining, 2002
Lecture 7.1
Lecture Outline
 Why
use Decision Trees?
 What is a Decision Tree?
 Examples
 Use as a data mining technique
 Popular Models
CART
CHAID
ID3 & C4.5
CSE5230 - Data Mining, 2002
Lecture 7.2
Why use Decision Trees? - 1
 Whereas
neural networks compute a
mathematical function of their inputs to generate
their outputs, decision trees use logical rules
Petal-length
 2.6
> 2.6
Iris setosa
Petal-width
 1.65
> 1.65
Petal-length
5
Iris virginica
>5
Iris versicolor
Sepal-length
 6.05
Iris versicolor
> 6.05
IF
Petal-length > 2.6 AND
Petal-width  1.65 AND
Petal-length > 5 AND
Sepal-length > 6.05
THEN
the flower is Iris virginica
NB. This is not the only rule for
this species. What is the other?
Iris virginica
Figure adapted from [SGI2001]
CSE5230 - Data Mining, 2002
Lecture 7.3
Why use Decision Trees? - 2
 For
some applications accuracy of classification
or prediction is sufficient, e.g.:
Direct mail firm needing to find a model for identifying
customers who will respond to mail
Predicting the stock market using past data
 In
other applications it is better (sometimes
essential) that the decision be explained, e.g.:
Rejection of a credit application
Medical diagnosis
 Humans
generally require explanations for most
decisions
CSE5230 - Data Mining, 2002
Lecture 7.4
Why use Decision Trees? - 3
 Example:
When a bank rejects a credit card
application, it is better to explain to the customer
that it was due to the fact that:
He/she is not a permanent resident of Australia AND
He/she has been residing in Australia for < 6 months AND
He/she does not have a permanent job.
 This
is better than saying:
 “We are very sorry, but our neural network thinks
that you are not a credit-worthy customer.” (In
which case the customer might become angry and
move to another bank)
CSE5230 - Data Mining, 2002
Lecture 7.5
What is a Decision Tree?
root node
 Built
Petal-length
 2.6
> 2.6
Iris setosa
Petal-width
 1.65
Petal-length
from root node (top) to leaf
test
nodes (bottom)
child node A record first enters the root node
 A test is applied to determine to
> 1.65
which child node it should go next
path
5
Iris virginica
>5
Iris versicolor
Sepal-length
 6.05
Iris versicolor
leaf nodes
CSE5230 - Data Mining, 2002
A variety of algorithms for choosing the
initial test exists. The aim is to
discriminate best between the target
classes
process is repeated until a
record arrives at a leaf node
Iris virginica
 The path from the root to a leaf
node provides an expression of a
rule
> 6.05
 The
Lecture 7.6
Building a Decision Tree - 1
 Algorithms
for building decision trees (DTs) begin
by trying to find the test which does the “best job”
of splitting the data into the desired classes
The desired classes have to be identified at the start
 Example:
we need to describe the profiles of
customers of a telephone company who “churn”
(do not renew their contracts). The DT building
algorithm examines the customer database to
find the best splitting criterion:
Phone technology
Age of customer
Time has been a customer
Gender
 The
DT algorithm may discover out that the
“Phone technology” variable is best for
separating churners from non-churners
CSE5230 - Data Mining, 2002
Lecture 7.7
Building a Decision Tree - 2
 The
process is repeated to discover the best splitting
criterion for the records assigned to each node
Phone technology
old
new
Time has been a customer
 2.3
Churners
> 2.3

Once built, the effectiveness of a decision tree can be
measured by applying it to a collection of previously
unseen records and observing the percentage of
correctly classified records
CSE5230 - Data Mining, 2002
Lecture 7.8
Phone Technology
Example - 1
Classify
customers who churn,
i.e. do not renew
their phone
contracts.
50 Churners
50 Non-churners
 Requirement:
(adapted from [BeS1997])
new
Time has been a Customer
30 Churners
50 Non-churners
<= 2.3 years
5 Churners
40 Non-churners
25 Churners
10 Non-churners
20 Churners
0 Non-churners
CSE5230 - Data Mining, 2002
20 Churners
0 Non-churners
> 2.3 years
Age
<= 35
old
> 35
5 Churners
10 Non-churners
Lecture 7.9
Example - 2
 The
number of records in a given parent node
equals the sum of the records contained in the
child nodes
 Quite easy to understand how the model is being
built (unlike NNs)
 Easy use the model
say for a targeted marketing campaign aimed at
customers likely to churn
 Provides
intuitive ideas about the customer base
e.g: “Customers who have been with the company for a
couple of years and have new phones are pretty loyal”
CSE5230 - Data Mining, 2002
Lecture 7.10
Use as a data mining technique - 1
 Exploration
Analyzing the predictors and splitting criteria selected
by the algorithm may provide interesting insights which
can be acted upon
e.g. if the following rule was identified:
IF
time a customer < 1.1 years AND
sales channel = telesales
THEN chance of churn is 65%
It might be worthwhile conducting a study on the way
the telesales operators are making their calls
CSE5230 - Data Mining, 2002
Lecture 7.11
Use as a data mining technique - 2
 Exploration
(continued)
Gleaning information from rules that fail
e.g. from the phone example we obtained the rule:
IF
Phone technology = old AND
Time has been a customer  2.3 years AND
Age > 35
THEN there are only 15 customers (15% of total)
Can this rule be useful?
» Perhaps we can attempt to build up this small market
segment. If this is possible then we have the edge over
competitors since we have a head start in this knowledge
» We can remove these customers from our direct
marketing campaign since there are so few of them
CSE5230 - Data Mining, 2002
Lecture 7.12
Use as a data mining technique - 3
 Exploration
(continued)
Again from the phone company example we noticed
that:
» There was no combination of rules to reliably
discriminate between churners and non-churners
for the small market segment mentioned on the
previous slide (5 churners, 10 non-churners).
Do we consider this as an occasion where it was not
possible to achieve our objective?
From this failure we have learnt that age is not all that
important for this category churners (unlike those under
35).
Perhaps we were asking the wrong questions all along this warrants further analysis
CSE5230 - Data Mining, 2002
Lecture 7.13
Use as a data mining technique - 4
 Data
Pre-processing
Decision trees are very robust at handling different
predictor types (number/categorical), and run quickly.
Therefore the can be good for a first pass over the data
in a data mining operation
This will create a subset of the possibly useful
predictors which can then be fed into another model,
say a neural network
 Prediction
Once the decision tree is built it can be then be used as
a prediction tool, by using it on a new set of data
CSE5230 - Data Mining, 2002
Lecture 7.14
Popular Decision Tree Models:
CART
 CART:
Classification And Regression Trees,
developed in 1984 by a team of researchers (Leo
Breiman et al.) from Stanford University
Used in the DM software Darwin - from Thinking
Machines Corporation (recently bought by Oracle)
 Often
uses an entropy measure to determine the
split point (Shannon’s Information theory).
measure of disorder (MOD) =
  p log 2 ( p)
where p is is the probability of that prediction
value occurring in a particular node of the tree.
Other measures used include Gini and twoing.
 CART produces a binary tree
CSE5230 - Data Mining, 2002
Lecture 7.15
CART - 2
 Consider
the “Churn” problem from slide 7.9
At the first node there are 100 customers to split, 50 who
churn and 50 who don’t churn
The MOD of this node is:
MOD = -0.5*log2(0.5) + -0.5*log2(0.5) = 1.00
The algorithm will try each predictor
For each predictor the algorithm will calculate the MOD of the
split produced by several values to identify the optimum
splitting on “Phone technology” produces two nodes, one with
50 churners and 30 non-churners, the other with 20 churners
and 0 non-churners. The first of these has:
MOD = -5/8*log2(5/8) + -3/8log2(3/8) = 0.95
and the second has a MOD of 0.
CART will select the predictor producing nodes with the lowest
MOD as the split point
CSE5230 - Data Mining, 2002
Lecture 7.16
Node splitting
An ideally good split
Name
Churned?
Name
Churned?
Jim
Yes
Bob
No
Sally
Yes
Betty
No
Steve
Yes
Sue
No
Joe
Yes
Alex
No
An ideally bad split
Name
Churned?
Name
Churned?
Jim
Yes
Bob
No
Sally
Yes
Betty
No
Steve
No
Sue
Yes
Joe
No
Alex
Yes
CSE5230 - Data Mining, 2002
Lecture 7.17
Popular Decision Tree Models:
CHAID
 CHAID:
Chi-squared Automatic Interaction
Detector, developed by J. A. Hartigan in 1975.
Widely used since it is distributed as part of the popular
statistical packages SAS and SPSS
 Differs
from CART in the way it identifies the split
points. Instead of the information measure, it
uses chi-squared test to identify the split points
(a statistical measure used for identifying
independent variables)
 All predictors must be categorical or put into
categorical form by binning
 The accuracy of the two methods CHAID and
CART have been found to be similar
CSE5230 - Data Mining, 2002
Lecture 7.18
Popular Decision Tree Models:
ID3 & C4.5
 ID3:
Iterative Dichtomiser, developed by the
Australian researcher Ross Quinlan in 1979
Used in the data mining software Clementine of Integral
Solutions Ltd. (taken over by SPSS)
 ID3
picks predictors and their splitting values on
the basis of the information gain provided
Gain is the difference between the amount of
information that is needed to make a correct prediction
both before and after the split has been made
If the amount of information required is much lower after
the split is made, then the split is said to have
decreased the disorder of the original data
CSE5230 - Data Mining, 2002
Lecture 7.19
ID3 & C4.5 - 2
A
B
+ ----
++++ By using the entropy
left
A ++++-
+++++----
-
  p log( p )
right left entropy
right entropy
start entropy
+---- -4/5log(4/5)+
-1/5log(1/5)=.72
-4/5log(4/5)+
-1/5log(1/5)=.72
-5/10log(5/10)+
-5/10(log(5/10)
=1
B +++++ ----
CSE5230 - Data Mining, 2002
-5/9log(5/9)+
-4/9log(4/9)=.99
-1/1log(1/1)
=0
Lecture 7.20
ID3 & C4.5 - 3
Weighted Entropy
Gain
A
(5/10)*0.72+(5/10)*0.72
= 0.72
0.28
B
(9/10)*0.99+(1/10)*0
= 0.89
0.11
 Split
A will be selected
 C4.5 introduces a number of extensions to ID3:
Handles unknown field values in training set
Tree pruning method
Automated rule generation
CSE5230 - Data Mining, 2002
Lecture 7.21
Strengths and Weaknesses
 Strengths
of decision trees
Able to generate understandable rules
Classify with very little computation
Handle both continuous and categorical data
Provides a clear indication of which variables are most
important for prediction or classification
 Weaknesses
Not appropriate for estimation or prediction tasks
(income, interest rates, etc.)
Problematic with time series data (much pre-processing
required), can be computationally expensive
CSE5230 - Data Mining, 2002
Lecture 7.22
References
 [SGI2001]
Silicon Graphics Inc. MLC++ Utilities
Manual, 2001
http://www.sgi.com/tech/mlc/utils.html
 [BeL1997] J. A. Berry and G. Linoff, Data Mining
Techniques: For Marketing, Sales, and Customer
Support, John Wiley & Sons Inc.,1997
 [BeS1997] A. Berson and S. J. Smith, Data
Warehousing, Data Mining and OLAP, McGraw
Hill, 1997
CSE5230 - Data Mining, 2002
Lecture 7.23