1. Theses slides are based on the slides by
CISC 4631 Data Mining
Lecture 04:
Decision Trees
Theses slides are based on the slides by
Tan, Steinbach and Kumar (textbook authors)
Eamonn Koegh (UC Riverside)
Raymond Mooney (UT Austin)

2. 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.

3. Illustrating Classification Task

4. Classification Techniques
Decision Tree based Methods
Rule-based Methods
Memory based reasoning
Neural Networks
Naïve Bayes and Bayesian Belief Networks
Support Vector Machines

5. Example of a Decision Tree
categorical
continuous
class
Splitting Attributes
Refund
Yes No NO
MarSt
Single, Divorced
Married
TaxInc NO
< 80K
> 80K NO
YES
Training Data
Model: Decision Tree

6. Another Example of Decision Tree
categorical
categorical
continuous
class
MarSt
Single, Divorced
Married NO
Refund No
Yes NO
TaxInc
< 80K
> 80K NO
YES
There could be more than one tree that fits the same data!

7. Decision Tree Classification Task

8. Apply Model to Test Data
Start from the root of tree.
Refund
MarSt
TaxInc
YES NO
Yes No
Married
Single, Divorced
< 80K
> 80K

9. Apply Model to Test Data
Refund
MarSt
TaxInc
YES NO
Yes No
Married
Single, Divorced
< 80K
> 80K

10. Apply Model to Test Data
Refund
Yes No NO
MarSt
Single, Divorced
Married
TaxInc NO
< 80K
> 80K NO
YES

11. Apply Model to Test Data
Refund
Yes No NO
MarSt
Single, Divorced
Married
TaxInc NO
< 80K
> 80K NO
YES

12. Apply Model to Test Data
Refund
Yes No NO
MarSt
Single, Divorced
Married
TaxInc NO
< 80K
> 80K NO
YES

13. Apply Model to Test Data
Refund
Yes No NO
MarSt
Married
Assign Cheat to “No”
Single, Divorced
TaxInc NO
< 80K
> 80K NO
YES

14. Decision Tree Terminology

15. Decision Tree Classification Task

16. Decision Tree Induction
Many Algorithms:
Hunt’s Algorithm (one of the earliest)
CART
ID3, C4.5
SLIQ,SPRINT
John Ross Quinlan is a computer science researcher in data mining and decision theory. He has contributed extensively to the development of decision tree algorithms, including inventing the canonical C4.5 and ID3 algorithms. 

17. Decision Tree Classifier10 1 2 3 4 5 6 7 8 9
Ross Quinlan
Abdomen Length > 7.1?
Antenna Length no
yes
Antenna Length > 6.0?
Katydid no
yes
Grasshopper
Katydid
Abdomen Length

18. Decision trees predate computers
Antennae shorter than body?
Yes No
3 Tarsi?
Grasshopper
Yes No
Foretiba has ears?
Yes No
Cricket
Decision trees predate computers
Katydids
Camel Cricket

19. Definition
Decision tree is a classifier in the form of a tree structure
Decision node: specifies a test on a single attribute
Leaf node: indicates the value of the target attribute
Arc/edge: split of one attribute
Path: a disjunction of test to make the final decision
Decision trees classify instances or examples by starting at the root of the tree and moving through it until a leaf node.

20. Decision Tree Classification
Decision tree generation consists of two phases
Tree construction
At start, all the training examples are at the root
Partition examples recursively based on selected attributes
Tree pruning
Identify and remove branches that reflect noise or outliers
Use of decision tree: Classifying an unknown sample
Test the attribute values of the sample against the decision tree

21. Decision Tree Representation
Each internal node tests an attribute
Each branch corresponds to attribute value
Each leaf node assigns a classification
outlook
sunny
overcast
rain
humidity
yes
wind
high
normal
strong
weak no
yes no
yes

22. How do we construct the decision tree?
Basic algorithm (a greedy algorithm)
Tree is constructed in a top-down recursive divide-and-conquer manner
At start, all the training examples are at the root
Attributes are categorical (if continuous-valued, they can be discretized in advance)
Examples are partitioned recursively based on selected attributes.
Test attributes are selected on the basis of a heuristic or statistical measure (e.g., information gain)
Conditions for stopping partitioning
All samples for a given node belong to the same class
There are no remaining attributes for further partitioning – majority voting is employed for classifying the leaf
There are no samples left

23. Top-Down Decision Tree Induction
Main loop:
A  the “best” decision attribute for next node
Assign A as decision attribute for node
For each value of A, create new descendant of node
Sort training examples to leaf nodes
If training examples perfectly classified,
Then STOP, Else iterate over new leaf nodes

24. Tree Induction Greedy strategy. Issues
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

25. How To Split Records Random Split Principled Criterion
The tree can grow huge
These trees are hard to understand.
Larger trees are typically less accurate than smaller trees.
Principled Criterion
Selection of an attribute to test at each node - choosing the most useful attribute for classifying examples.
How?
Information gain
measures how well a given attribute separates the training examples according to their target classification
This measure is used to select among the candidate attributes at each step while growing the tree

26. Tree Induction Greedy strategy:
Split the records based on an attribute test that optimizes certain criterion:
Hunt’s algorithm: recursively partition training records into successively purer subsets. How to measure purity/impurity
Entropy and information gain (covered in the lectures slides)
Gini (covered in the textbook)
Classification error

27. How to determine the Best Split
Before Splitting: 10 records of class 0, records of class 1
Gender
Which test condition is the best?
Why is student id a bad feature to use?

28. How to determine the Best Split
Greedy approach:
Nodes with homogeneous class distribution are preferred
Need a measure of node impurity:
Non-homogeneous,
High degree of impurity
Homogeneous,
Low degree of impurity

29. Picking a Good Split Feature
Goal is to have the resulting tree be as small as possible, per Occam’s razor.
Finding a minimal decision tree (nodes, leaves, or depth) is an NP-hard optimization problem.
Top-down divide-and-conquer method does a greedy search for a simple tree but does not guarantee to find the smallest.
General lesson in Machine Learning and Data Mining: “Greed is good.”
Want to pick a feature that creates subsets of examples that are relatively “pure” in a single class so they are “closer” to being leaf nodes.
There are a variety of heuristics for picking a good test, a popular one is based on information gain that originated with the ID3 system of Quinlan (1979).
R. Mooney, UT Austin

30. Information Theory
Think of playing "20 questions": I am thinking of an integer between 1 and 1, what is it? What is the first question you would ask?
What question will you ask?
Why?
Entropy measures how much more information you need before you can identify the integer.
Initially, there are 1000 possible values, which we assume are equally likely.
What is the maximum number of question you need to ask?

31. Entropy
Entropy (disorder, impurity) of a set of examples, S, relative to a binary classification is:
where p1 is the fraction of positive examples in S and p0 is the fraction of negatives.
If all examples are in one category, entropy is zero (we define 0log(0)=0)
If examples are equally mixed (p1=p0=0.5), entropy is a maximum of 1.
Entropy can be viewed as the number of bits required on average to encode the class of an example in S where data compression (e.g. Huffman coding) is used to give shorter codes to more likely cases.
For multi-class problems with c categories, entropy generalizes to:
R. Mooney, UT Austin

32. Entropy Plot for Binary Classification
The entropy is 0 if the outcome is certain.
The entropy is maximum if we have no knowledge of the system (or any outcome is equally possible).
Entropy of a 2-class problem with regard to the portion of one of the two groups

33. Information Gain
Is the expected reduction in entropy caused by partitioning the examples according to this attribute.
is the number of bits saved when encoding the target value of an arbitrary member of S, by knowing the value of attribute A.

34. Information Gain in Decision Tree Induction
Assume that using attribute A, a current set will be partitioned into some number of child sets
The encoding information that would be gained by branching on A
Note: entropy is at its minimum if the collection of objects is completely uniform

35. Examples for Computing Entropy
NOTE: p( j | t) is computed as the relative frequency of class j at node t
P(C1) = 0/6 = P(C2) = 6/6 = 1
Entropy = – 0 log2 0 – 1 log2 1 = – 0 – 0 = 0
P(C1) = 1/ P(C2) = 5/6
Entropy = – (1/6) log2 (1/6) – (5/6) log2 (5/6) = 0.65
P(C1) = 2/ P(C2) = 4/6
Entropy = – (2/6) log2 (2/6) – (4/6) log2 (4/6) = 0.92
P(C1) = 3/6=1/ P(C2) = 3/6 = 1/2
Entropy = – (1/2) log2 (1/2) – (1/2) log2 (1/2)
= -(1/2)(-1) – (1/2)(-1) = ½ + ½ = 1

36. How to Calculate log2x
Many calculators only have a button for log10x and logex (note log typically means log10)
You can calculate the log for any base b as follows:
logb(x) = logk(x) / logk(b)
Thus log2(x) = log10(x) / log10(2)
Since log10(2) = .301, just calculate the log base 10 and divide by .301 to get log base 2.
You can use this for HW if needed

37. Splitting Based on INFO...
Information Gain:
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.

38. Continuous Attribute? (more on it later)
Each non-leaf node is a test, its edge partitioning the attribute into subsets (easy for discrete attribute).
For continuous attribute
Partition the continuous value of attribute A into a discrete set of intervals
Create a new boolean attribute Ac , looking for a threshold c,
How to choose c ?

39. Person Homer 0” 250 36 M Marge 10” 150 34 F Bart 2” 90 10 Lisa 6” 78 8
Hair Length
Weight
Age
Class
Homer 0”
250 36 M
Marge
10”
150 34 F
Bart 2” 90 10
Lisa 6” 78 8
Maggie 4” 20 1
Abe 1”
170 70
Selma 8”
160 41
Otto
180 38
Krusty
200 45
Comic 8”
290 38 ?

40. Entropy(4F,5M) = -(4/9)log2(4/9) - (5/9)log2(5/9)=
yes no
Hair Length <= 5?
Let us try splitting on Hair length
Entropy(1F,3M) = -(1/4)log2(1/4) - (3/4)log2(3/4) =
Entropy(3F,2M) = -(3/5)log2(3/5) - (2/5)log2(2/5) =
Gain(Hair Length <= 5) = – (4/9 * /9 * ) =

41. Gain(Weight <= 160) = 0.9911 – (5/9 * 0.7219 + 4/9 * 0 ) = 0.5900
Entropy(4F,5M) = -(4/9)log2(4/9) - (5/9)log2(5/9) =
yes no
Weight <= 160?
Let us try splitting on Weight
Entropy(4F,1M) = -(4/5)log2(4/5) - (1/5)log2(1/5) =
Entropy(0F,4M) = -(0/4)log2(0/4) - (4/4)log2(4/4)
= 0
Gain(Weight <= 160) = – (5/9 * /9 * 0 ) =

42. Gain(Age <= 40) = 0.9911 – (6/9 * 1 + 3/9 * 0.9183 ) = 0.0183
Entropy(4F,5M) = -(4/9)log2(4/9) - (5/9)log2(5/9) =
yes no
age <= 40?
Let us try splitting on Age
Entropy(3F,3M) = -(3/6)log2(3/6) - (3/6)log2(3/6)
= 1
Entropy(1F,2M) = -(1/3)log2(1/3) - (2/3)log2(2/3) =
Gain(Age <= 40) = – (6/9 * 1 + 3/9 * ) =

43. This time we find that we can split on Hair length, and we are done!
Of the 3 features we had, Weight was best. But while people who weigh over 160 are perfectly classified (as males), the under 160 people are not perfectly classified… So we simply recurse!
yes no
Weight <= 160?
This time we find that we can split on Hair length, and we are done!
yes no
Hair Length <= 2?

44. Male Male Female Weight <= 160?
We don’t need to keep the data around, just the test conditions.
Weight <= 160?
yes no
How would these people be classified?
Hair Length <= 2?
Male
yes no
Male
Female

45. Male Male Female Weight <= 160? Rules to Classify Males/Females
It is trivial to convert Decision Trees to rules…
Weight <= 160?
yes no
Hair Length <= 2?
Male
yes no
Male
Female
Rules to Classify Males/Females
If Weight greater than 160, classify as Male
Elseif Hair Length less than or equal to 2, classify as Male
Else classify as Female

46. Once we have learned the decision tree, we don’t even need a computer!
This decision tree is attached to a medical machine, and is designed to help nurses make decisions about what type of doctor to call.
Decision tree for a typical shared-care setting applying the system for the diagnosis of prostatic obstructions.

47. The worked examples we have seen were performed on small datasets
The worked examples we have seen were performed on small datasets. However with small datasets there is a great danger of overfitting the data…
When you have few datapoints, there are many possible splitting rules that perfectly classify the data, but will not generalize to future datasets.
Yes No
Wears green?
Female
Male
For example, the rule “Wears green?” perfectly classifies the data, so does “Mothers name is Jacqueline?”, so does “Has blue shoes”…

48. How to Find the Best Split: GINI
Before Splitting: M0 A? B?
Yes No
Yes No
Node N1
Node N2
Node N3
Node N4 M1 M2 M3 M4
M12
M34
Gain = M0 – M12 vs M0 – M34

49. Measure of Impurity: GINI (at node t)
Gini Index for a given node t with classes j
NOTE: p( j | t) is computed as the relative frequency of class j at node t
Example: Two classes C1 & C2 and node t has 5 C1 and 5 C2 examples. Compute Gini(t)
1 – [p(C1|t) + p(C2|t)] = 1 – [(5/10)2 + [(5/10)2 ]
1 – [¼ + ¼] = ½.
Do you think this Gini value indicates a good split or bad split? Is it an extreme value?

50. More on Gini
Worst Gini corresponds to probabilities of 1/nc, where nc is the number of classes.
For 2-class problems the worst Gini will be ½
How do we get the best Gini? Come up with an example for node t with 10 examples for classes C1 and C2
10 C1 and 0 C2
Now what is the Gini?
1 – [(10/10)2 + (0/10)2 = 1 – [1 + 0] = 0
So 0 is the best Gini
So for 2-class problems:
Gini varies from 0 (best) to ½ (worst).

51. Some More Examples
Below we see the Gini values for 4 nodes with different distributions. They are ordered from best to worst. See next slide for details
Note that thus far we are only computing GINI for one node. We need to compute it for a split and then compute the change in Gini from the parent node.

52. Examples for computing GINI
P(C1) = 0/6 = P(C2) = 6/6 = 1
Gini = 1 – P(C1)2 – P(C2)2 = 1 – 0 – 1 = 0
P(C1) = 1/ P(C2) = 5/6
Gini = 1 – (1/6)2 – (5/6)2 = 0.278
P(C1) = 2/ P(C2) = 4/6
Gini = 1 – (2/6)2 – (4/6)2 = 0.444

53. Splitting Criteria based on Classification Error
Classification error at a node t :
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

54. Examples for Computing Error
P(C1) = 0/6 = P(C2) = 6/6 = 1
Error = 1 – max (0, 1) = 1 – 1 = 0
P(C1) = 1/ P(C2) = 5/6
Error = 1 – max (1/6, 5/6) = 1 – 5/6 = 1/6
P(C1) = 2/ P(C2) = 4/6
Error = 1 – max (2/6, 4/6) = 1 – 4/6 = 1/3

55. Comparison among Splitting Criteria
For a 2-class problem:

56. Discussion
Error rate is often the metric used to evaluate a classifier (but not always)
So it seems reasonable to use error rate to determine the best split
That is, why not just use a splitting metric that matches the ultimate evaluation metric?
But this is wrong!
The reason is related to the fact that decision trees use a greedy strategy, so we need to use a splitting metric that leads to globally better results
The other metrics will empirically outperform error rate, although there is no proof for this.

57. DTs in practice... Growing to purity is bad (overfitting)
x2: sepal width
x1: petal length

58. DTs in practice... Growing to purity is bad (overfitting)
x2: sepal width
x1: petal length

59. DTs in practice... Growing to purity is bad (overfitting)
Terminate growth early
Grow to purity, then prune back

60. DTs in practice... Growing to purity is bad (overfitting)
Not statistically
supportable leaf
Remove split
& merge leaves
x2: sepal width
x1: petal length

61. (more on overfitting later)
Avoid Overfitting in Classification
(more on overfitting later)
The generated tree may overfit the training data
Too many branches, some may reflect anomalies due to noise or outliers
Result is in poor accuracy for unseen samples
Two approaches to avoid overfitting
Prepruning: Halt tree construction early—do not split a node if this would result in the goodness measure falling below a threshold
Difficult to choose an appropriate threshold
Postpruning: Remove branches from a “fully grown” tree—get a sequence of progressively pruned trees
Use a set of data different from the training data to decide which is the “best pruned tree”

62. Tree Induction Greedy strategy. Issues
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

63. 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

64. Splitting Based on Nominal Attributes
Multi-way split: Use as many partitions as distinct values.
Binary split: Divides values into two subsets Need to find optimal partitioning.
CarType
Family
Sports
Luxury
CarType
{Sports, Luxury}
{Family}
CarType
{Family, Luxury}
{Sports} OR

65. Splitting Based on Ordinal Attributes
Multi-way split: Use as many partitions as distinct values.
Binary split: Divides values into two subsets Need to find optimal partitioning.
What about this split?
Size
Small
Medium
Large
Size
{Small, Medium}
{Large}
Size
{Medium, Large}
{Small} OR
Size
{Small, Large}
{Medium}

66. 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

67. Splitting Based on Continuous Attributes

68. Data Fragmentation
Number of instances gets smaller as you traverse down the tree
Number of instances at the leaf nodes could be too small to make any statistically significant decision

69. Search Strategy Finding an optimal decision tree is NP-hard
The algorithm presented so far uses a greedy, top-down, recursive partitioning strategy to induce a reasonable solution

70. Expressiveness
Decision tree provides expressive representation for learning discrete-valued function
But they do not generalize well to certain types of Boolean functions
Example: parity function:
Class = 1 if there is an even number of Boolean attributes with truth value = True
Class = 0 if there is an odd number of Boolean attributes with truth value = True
For accurate modeling, must have a complete tree
Not expressive enough for modeling continuous variables
Particularly when test condition involves only a single attribute at-a-time

71. Decision Boundary
Border line between two neighboring regions of different classes is known as decision boundary
Decision boundary is parallel to axes because test condition involves a single attribute at-a-time

72. Oblique Decision Trees
x + y < 1
Class = +
Class =
Test condition may involve multiple attributes
More expressive representation
Finding optimal test condition is computationally expensive

73. Vertical/Horizontal Boundaries
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

74. Tree Replication
Same subtree appears in multiple branches

75. Model Evaluation Metrics for Performance Evaluation
How to evaluate the performance of a model?
Methods for Performance Evaluation
How to obtain reliable estimates?

76. ? Deep Bushy Tree Useless
Which of the “Problems” can be solved by a Decision Tree? 10 1 2 3 4 5 6 7 8 9
Deep Bushy Tree
Useless ? 10 1 2 3 4 5 6 7 8 9
100 10 20 30 40 50 60 70 80 90
The Decision Tree has a hard time with correlated attributes

77. Advantages/Disadvantages of Decision Trees
Easy to understand (Doctors love them!)
Easy to generate rules
Disadvantages:
May suffer from overfitting.
Classifies by rectangular partitioning (so does not handle correlated features very well).
Can be quite large – pruning is necessary.
Does not handle streaming data easily