1. Classification II

2. Data Mining Overview Data Mining
Data warehouses and OLAP (On Line Analytical Processing.)
Association Rules Mining
Clustering: Hierarchical and Partitional approaches
Classification: Decision Trees and Bayesian classifiers

3. Setting
Given old data about customers and payments, predict new applicant’s loan eligibility.
Previous customers
Classifier
Decision rules
Age
Salary
Profession
Location
Customer type
Salary > 5 L
Good/
bad
Prof. = Exec
New applicant’s data

4. Decision trees
Tree where internal nodes are simple decision rules on one or more attributes and leaf nodes are predicted class labels.
Salary < 1 M
Prof = teaching
Age < 30
Good
Bad
Bad
Good

5. SLIQ (Supervised Learning In Quest)
Decision-tree classifier for data mining
Design goals:
Able to handle large disk-resident training sets
No restrictions on training-set size

6. Building tree GrowTree(TrainingData D) Partition(D); Partition(Data D)
if (all points in D belong to the same class) then
return;
for each attribute A do
evaluate splits on attribute A;
use best split found to partition D into D1 and D2;
Partition(D1);
Partition(D2);
2 major problems:
o How to find split points that define node tests
o How to partition the data having found split points
o CART, C4.5:
- DFS
- Repeated sorting at each node (continuous attributes)
o SLIQ
- SortOnce technique using attribute lists

7. Data Setup One list for each attribute
Entries in an Attribute List consist of:
attribute value
class list index
A list for the classes with pointers to the tree nodes
Lists for continuous attributes are in sorted order
Attribute lists may be disk-resident
Class List must be in main memory

8. Data Setup
Class list
Attribute lists N1

9. Evaluating Split Points
Gini Index
if data D contains examples from c classes
Gini(D) = 1 -  pj2
where pj is the relative frequency of class j in D
If D split into D1 & D2 with n1 & n2 tuples each
Ginisplit(D) = n1* gini(D1) + n2* gini(D2)
n n
Note: Only class frequencies are needed to compute index

10. Finding Split Points For each attribute A do
evaluate splits on attribute A using attribute list
Key idea: To evaluate a split on numerical attributes we need to sort the set at each node. But, if we have all attributes pre-sorted we don’t need to do that at the tree construction phase
Keep split with lowest GINI index
2 major problems:
o How to find split points that define node tests
o How to partition the data having found split points
o CART, C4.5:
- DFS
- Repeated sorting at each node (continuous attributes)
o SLIQ
- SortOnce technique using attribute lists

11. Finding Split Points: Continuous Attrib.
Consider splits of form: value(A) < x
Example: Age < 17
Evaluate this split-form for every value in an attribute list
To evaluate splits on attribute A for a given tree-node:
Initialize class-histograms of left and right children;
for each record in the attribute list do
find the corresponding entry in Class List and the class and Leaf node
evaluate splitting index for value(A) < record.value;
update the class histogram in the leaf

12. N1 GINI Index: und 0.33 0.22 0.5 High Low L R 4 2 1 High Low L 1 R 3 2R 4 2
und 1
High
Low L 1 R 3 2
0.33 3 1 4
High
Low L 3 R 1 2 3
0.22
1: Age < 20
3: Age < 32
Age < 32
High
Low L 3 1 R
4: Age < 43
0.5 4

13. Finding Split Points: Categorical Attrib.
Consider splits of the form: value(A)  {x1, x2, ..., xn}
Example: CarType {family, sports}
Evaluate this split-form for subsets of domain(A)
To evaluate splits on attribute A for a given tree node:
initialize class/value matrix of node to zeroes;
for each record in the attribute list do
increment appropriate count in matrix;
evaluate splitting index for various subsets using the constructed matrix;

14. class/value matrix Left Child Right Child GINI Index:
CarType in {family}
GINI = 0.444
CarType in {sports}
GINI = 0.333
CarType in {truck}
GINI = 0.267

15. Updating the Class List
Next step is to update the Class List with the new nodes
Scan the attr list that is used to split and update the corresponding leaf entry in the Class List
For each attribute A in a split traverse the attribute list
for each value u in the attr list
find the corresponding entry in the class list (e)
find the new class c to which u belongs
update the class list for e to c
update node reference in e to the node corresponding to class c

16. Preventing overfitting
A tree T overfits if there is another tree T’ that gives higher error on the training data yet gives lower error on unseen data.
An overfitted tree does not generalize to unseen instances.
Happens when data contains noise or irrelevant attributes and training size is small.
Overfitting can reduce accuracy drastically:
10-25% as reported in Minger’s 1989 Machine learning

17. Approaches to prevent overfitting
Two Approaches:
Stop growing the tree beyond a certain point
First over-fit, then post prune. (More widely used)
Tree building divided into phases:
Growth phase
Prune phase
Hard to decide when to stop growing the tree, so second appraoch more widely used.

18. Criteria for finding correct final tree size:
Three criteria:
Cross validation with separate test data
Use some criteria function to choose best size
Example: Minimum description length (MDL) criteria
Statistical bounds: use all data for training but apply statistical test to decide right size.

19. The minimum description length principle (MDL)
MDL: paradigm for statistical estimation particularly model selection
Given data D and a class of models M, our choose is to choose a model m in M such that data and model can be encoded using the smallest total length
L(D) = L(D|m) + L(m)
How to find encoding length?
Answer in Information Theory
Consider the problem of transmitting n messages where pi is probability of seeing message i
Shannon’s theorem: minimum expected length when
-log pi bits to message i

20. Encoding data Assume t records of training data D
First send tree m using L(m|M) bits
Assume all but the class labels of training data known.
Goal: transmit class labels using L(D|m)
If tree correctly predicts an instance, 0 bits
Otherwise, log k bits where k is number of classes.
Thus, if e errors on training data: total cost
e log k + L(m|M) bits.
Complex tree will have higher L(m|M) but lower e.
Question: how to encode the tree?

21. Extracting Classification Rules from Trees
Represent the knowledge in the form of IF-THEN rules
One rule is created for each path from the root to a leaf
Each attribute-value pair along a path forms a conjunction
The leaf node holds the class prediction
Rules are easier for humans to understand
Example
IF age = “<=30” AND student = “no” THEN buys_computer = “no”
IF age = “<=30” AND student = “yes” THEN buys_computer = “yes”
IF age = “31…40” THEN buys_computer = “yes”
IF age = “>40” AND credit_rating = “excellent” THEN buys_computer = “yes”
IF age = “<=30” AND credit_rating = “fair” THEN buys_computer = “no”

22. SPRINT An improvement over SLIQ
Does not need to keep a list in main memory
Parallel version is straightforward
Attribute lists are extended with class field – no Class list is needed
Uses hashing to assign records to classes and nodes

23. Pros and Cons of decision trees
Reasonable training time
Fast application
Easy to interpret
Easy to implement
Can handle large number of features
Cons
Cannot handle complicated relationship between features
simple decision boundaries
problems with lots of missing data
More information: