Classification II

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

Setting

Given old data about customers and payments, predict new applicant’s loan eligibility.

AgeSalaryProfessionLocationCustomer type

Previous customers Classifier Decision rules

Salary > 5 L

Prof. = Exec

New applicant’s data

Good/bad

Tree where internal nodes are simple decision rules on one or more attributes and leaf nodes are predicted class labels.

Decision trees

Salary < 1 M

Prof = teaching

Good

Age < 30

BadBad Good

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

Building treeGrowTree(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);

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

Data Setup

Age Car Type Risk 23 family High 17 sports High 43 sports High 68 family Low

32 truck Low

20 family High

Age CLI 23 0 17 1 43 2 68 3

32 4

20 5

Car Type CLI family 0 sports 1 sports 2 family 3

truck 4

family 5

Risk Leaf High N1 High N1

High N1

Low N1

Low N1

High N1

Attribute lists Class list

N1

Age CLI 17 1 20 5

23 0

32 4

43 2

68 3

Risk Leaf 0 High N1 1 High N1

2 High N1

3 Low N1

4 Low N1

5 High N1

Car Type CLI family 0 sports 1

sports 2

family 3

truck 4

family 5

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

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

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 dofind the corresponding entry in Class List and the class and Leaf nodeevaluate splitting index for value(A) < record.value;update the class histogram in the leaf

Age CLI 17 1 20 5

23 0

32 4

43 2

68 3

Class Leaf 0 High N1 1 High N1

2 High N1

3 Low N1

4 Low N1

5 High N1

High Low

L 0 0

R 4 2

GINI Index:

0

N1

1

4

High Low

L 1 0

R 3 2

High Low

L 3 0

R 1 2

3 1

3

und

High Low

L 3 1

R 1 1

4

1: Age < 20

3: Age < 32

4: Age < 43

0

0.33

0.22

0.5

Age < 32

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 doincrement appropriate count in matrix;

evaluate splitting index for various subsets using the constructed matrix;

High Low

family 2 1

sports 2 0

truck 0 1

class/value matrix

CarType in {family}High Low2 1

High Low2 1

Left Child Right Child GINI Index:

High Low0 1

High Low4 1

CarType in {truck}

GINI = 0.444

GINI = 0.267

High Low2 0

High Low2 2

CarType in {sports} GINI = 0.333

Car Type CLI family 0 sports 1

sports 2

family 3

truck 4

family 5

Risk Leaf High N1 High N1

High N1

Low N1

Low N1

High N1

Updating 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

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

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

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.

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.

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

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?

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”

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

Pros and Cons of decision trees

• Cons– Cannot handle complicated relationship between features– simple decision boundaries– problems with lots of missing data

• Pros+ Reasonable training time+ Fast application+ Easy to interpret+ Easy to implement+ Can handle large number of features

More information: http://www.recursive-partitioning.com/