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## CISC 4631 Data Mining

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### CISC 4631Data Mining

Lecture 04:

Decision Trees

Theses slides are based on the slides by

Tan, Steinbach and Kumar (textbook authors)

EamonnKoegh(UC Riverside)

Raymond Mooney (UT Austin)

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.

Classification Techniques

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

categorical

continuous

class

Example of a Decision TreeSplitting Attributes

Refund

Yes

No

NO

MarSt

Married

Single, Divorced

TaxInc

NO

< 80K

> 80K

YES

NO

Model: Decision Tree

Training Data

Another Example of Decision Tree

categorical

categorical

continuous

class

Single, Divorced

MarSt

Married

NO

Refund

No

Yes

TaxInc

< 80K

> 80K

YES

NO

There could be more than one tree that fits the same data!

Decision Tree Classification Task

Decision Tree

Yes

No

NO

MarSt

Married

Single, Divorced

TaxInc

NO

< 80K

> 80K

YES

NO

Apply Model to Test DataTest Data

Start from the root of tree.

Apply Model to Test Data

Test Data

Refund

Yes

No

NO

MarSt

Assign Cheat to “No”

Married

Single, Divorced

TaxInc

NO

< 80K

> 80K

YES

NO

Decision Tree Classification Task

Decision Tree

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.

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Ross Quinlan

Abdomen Length > 7.1?

Antenna Length

yes

no

Antenna Length > 6.0?

Katydid

yes

no

Katydid

Grasshopper

Abdomen Length

Yes

No

3 Tarsi?

Grasshopper

Yes

No

Foretiba has ears?

Yes

No

Cricket

Decision trees predate computers

Katydids

Camel Cricket

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.

- 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

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

weak

normal

strong

high

no

yes

no

yes

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

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

Tree Induction

- Greedy strategy.
- 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

How To Split Records

- Random Split
- 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

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

How to determine the Best Split

Before Splitting: 10 records of class 0, 10 records of class 1

Gender

Which test condition is the best?

Why is student id a bad feature to use?

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

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

Information Theory

- Think of playing "20 questions": I am thinking of an integer between 1 and 1,000 -- 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?

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 0log(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

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

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.

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

Examples for Computing Entropy

- NOTE: p( j | t) is computed as the relative frequency of class j at node t

P(C1) = 0/6 = 0 P(C2) = 6/6 = 1

Entropy = – 0 log20– 1 log21 = – 0 – 0 = 0

P(C1) = 1/6 P(C2) = 5/6

Entropy = – (1/6) log2 (1/6)– (5/6) log2 (5/6) = 0.65

P(C1) = 2/6 P(C2) = 4/6

Entropy = – (2/6) log2 (2/6)– (4/6) log2 (4/6) = 0.92

P(C1) = 3/6=1/2 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

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

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.

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 ?

Entropy(4F,5M) = -(4/9)log2(4/9) - (5/9)log2(5/9)

= 0.9911

no

yes

Hair Length <= 5?

Let us try splitting on Hair length

Entropy(3F,2M) = -(3/5)log2(3/5) - (2/5)log2(2/5)

= 0.9710

Entropy(1F,3M) = -(1/4)log2(1/4) - (3/4)log2(3/4)

= 0.8113

Gain(Hair Length <= 5) = 0.9911 – (4/9 * 0.8113 + 5/9 * 0.9710 ) = 0.0911

Entropy(4F,5M) = -(4/9)log2(4/9) - (5/9)log2(5/9)

= 0.9911

no

yes

Weight <= 160?

Let us try splitting on Weight

Entropy(0F,4M) = -(0/4)log2(0/4) - (4/4)log2(4/4)

= 0

Entropy(4F,1M) = -(4/5)log2(4/5) - (1/5)log2(1/5)

= 0.7219

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)

= 0.9911

no

yes

age <= 40?

Let us try splitting on Age

Entropy(1F,2M) = -(1/3)log2(1/3) - (2/3)log2(2/3)

= 0.9183

Entropy(3F,3M) = -(3/6)log2(3/6) - (3/6)log2(3/6)

= 1

Gain(Age <= 40) = 0.9911 – (6/9 * 1 + 3/9 * 0.9183 ) = 0.0183

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!

no

yes

Weight <= 160?

This time we find that we can split on Hair length, and we are done!

no

yes

Hair Length <= 2?

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

It is trivial to convert Decision Trees to rules…

Weight <= 160?

yes

no

Hair Length <= 2?

Male

no

yes

Male

Female

Rules to Classify Males/Females

IfWeightgreater than 160, classify as Male

Elseif Hair Lengthless than or equal to 2, classify as Male

Else classify as Female

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.

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?

Male

Female

For example, the rule “Wears green?” perfectly classifies the data, so does “Mothers name is Jacqueline?”, so does “Has blue shoes”…

M2

M3

M4

M1

M12

M34

How to Find the Best Split: GINIBefore Splitting:

A?

B?

Yes

No

Yes

No

Node N1

Node N2

Node N3

Node N4

Gain = M0 – M12 vs M0 – M34

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?

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).

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.

Examples for computing GINI

P(C1) = 0/6 = 0 P(C2) = 6/6 = 1

Gini = 1 – P(C1)2 – P(C2)2 = 1 – 0 – 1 = 0

P(C1) = 1/6 P(C2) = 5/6

Gini = 1 – (1/6)2 – (5/6)2 = 0.278

P(C1) = 2/6 P(C2) = 4/6

Gini = 1 – (2/6)2 – (4/6)2 = 0.444

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

Examples for Computing Error

P(C1) = 0/6 = 0 P(C2) = 6/6 = 1

Error = 1 – max (0, 1) = 1 – 1 = 0

P(C1) = 1/6 P(C2) = 5/6

Error = 1 – max (1/6, 5/6) = 1 – 5/6 = 1/6

P(C1) = 2/6 P(C2) = 4/6

Error = 1 – max (2/6, 4/6) = 1 – 4/6 = 1/3

Comparison among Splitting Criteria

For a 2-class problem:

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.

DTs in practice...

Growing to purity is bad (overfitting)

Terminate growth early

Grow to purity, then prune back

DTs in practice...

Growing to purity is bad (overfitting)

Not statistically

supportable leaf

Remove split

& merge leaves

x2: sepal width

x1: petal length

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”

Tree Induction

- Greedy strategy.
- 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

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

Family

Luxury

Sports

CarType

CarType

{Sports, Luxury}

{Family, Luxury}

{Family}

{Sports}

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.

OR

Small

Large

Medium

Size

Size

Size

{Small, Medium}

{Small, Large}

{Medium, Large}

{Medium}

{Large}

{Small}

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?

OR

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

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

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

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

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

Class = +

Class =

Oblique Decision Trees- Test condition may involve multiple attributes
- More expressive representation
- Finding optimal test condition is computationally expensive

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

Tree Replication

- Same subtree appears in multiple branches

Model Evaluation

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

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Which of the “Problems” can be solved by a Decision Tree?

- Deep Bushy Tree
- Useless
- Deep Bushy Tree

?

The Decision Tree has a hard time with correlated attributes

Advantages/Disadvantages of Decision Trees

- Advantages:
- 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

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