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Classification with Decision Trees

Instructor: Qiang YangHong Kong University of Science and Technology

Qyang@cs.ust.hk

Thanks: Eibe Frank and Jiawei Han

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DECISION TREE [Quinlan93]

An internal node represents a test on an attribute.

A branch represents an outcome of the test, e.g., Color=red.

A leaf node represents a class label or class label distribution.

At each node, one attribute is chosen to split training examples into distinct classes as much as possible

A new case is classified by following a matching path to a leaf node.

Outlook Tempreature Humidity Windy Classsunny hot high false Nsunny hot high true Novercast hot high false Prain mild high false Prain cool normal false Prain cool normal true Novercast cool normal true Psunny mild high false Nsunny cool normal false Prain mild normal false Psunny mild normal true Povercast mild high true Povercast hot normal false Prain mild high true N

Training Set

Outlook

overcast

humidity windy

high normal falsetrue

sunny rain

N NP P

P

overcast

Example

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Building Decision Tree [Q93]

Top-down tree construction At start, all training examples are at the root. Partition the examples recursively by choosing one

attribute each time. Bottom-up tree pruning

Remove subtrees or branches, in a bottom-up manner, to improve the estimated accuracy on new cases.

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Choosing the Splitting Attribute

At each node, available attributes are evaluated on the basis of separating the classes of the training examples. A Goodness function is used for this purpose.

Typical goodness functions: information gain (ID3/C4.5) information gain ratio gini index

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Which attribute to select?

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A criterion for attribute selection

Which is the best attribute? The one which will result in the smallest tree Heuristic: choose the attribute that produces the

“purest” nodes Popular impurity criterion: information gain

Information gain increases with the average purity of the subsets that an attribute produces

Strategy: choose attribute that results in greatest information gain

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Computing information

Information is measured in bits Given a probability distribution, the info required to

predict an event is the distribution’s entropy Entropy gives the information required in bits (this

can involve fractions of bits!) Formula for computing the entropy:

Suppose a set S has n values: V1, V2, …Vn, where Vi has proportion pi,

E.g., the weather data has 2 values: Play=P and Play=N. Thus, p1=9/14, p2=5/14.

nnn ppppppppp logloglog),,,entropy( 221121

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Example: attribute “Outlook”

“ Outlook” = “Sunny”:

“Outlook” = “Overcast”:

“Outlook” = “Rainy”:

Expected information for attribute:

bits 971.0)5/3log(5/3)5/2log(5/25,3/5)entropy(2/)info([2,3]

bits 0)0log(0)1log(10)entropy(1,)info([4,0]

bits 971.0)5/2log(5/2)5/3log(5/35,2/5)entropy(3/)info([3,2]

Note: this isnormally notdefined.

971.0)14/5(0)14/4(971.0)14/5([3,2])[4,0],,info([3,2] bits 693.0

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Computing the information gain

Information gain: information before splitting – information after splitting

Information gain for attributes from weather data:

0.693-0.940[3,2])[4,0],,info([2,3]-)info([9,5])Outlook"gain("

bits 247.0

bits 247.0)Outlook"gain(" bits 029.0)e"Temperaturgain("

bits 152.0)Humidity"gain(" bits 048.0)Windy"gain("

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Continuing to split

bits 571.0)e"Temperaturgain(" bits 971.0)Humidity"gain("

bits 020.0)Windy"gain("

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The final decision tree

Note: not all leaves need to be pure; sometimes identical instances have different classes Splitting stops when data can’t be split any further

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Highly-branching attributes

Problematic: attributes with a large number of values (extreme case: ID code)

Subsets are more likely to be pure if there is a large number of values

Information gain is biased towards choosing attributes with a large number of values

This may result in overfitting (selection of an attribute that is non-optimal for prediction)

Another problem: fragmentation

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The gain ratio

Gain ratio: a modification of the information gain that reduces its bias on high-branch attributes

Gain ratio takes number and size of branches into account when choosing an attribute

It corrects the information gain by taking the intrinsic information of a split into account

Also called split ratio Intrinsic information: entropy of distribution of

instances into branches (i.e. how much info do we need to tell which branch

an instance belongs to)

.||

||2

log||||

),(SiS

SiS

ASnfoIntrinsicI

.),(

),(),(ASnfoIntrinsicI

ASGainASGainRatio

Gain Ratio

IntrinsicInfo should be Large when data is evenly spread over all branches Small when all data belong to one branch

Gain ratio (Quinlan’86) normalizes info gain by IntrinsicInfo:

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Computing the gain ratio

Example: intrinsic information for ID code

Importance of attribute decreases as intrinsic information gets larger

Example of gain ratio:

Example:

bits 807.3)14/1log14/1(14),1[1,1,(info

)Attribute"info("intrinsic_)Attribute"gain("

)Attribute"("gain_ratio

246.0bits 3.807bits 0.940

)ID_code"("gain_ratio

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Gain ratios for weather data

Outlook Temperature

Info: 0.693 Info: 0.911

Gain: 0.940-0.693 0.247 Gain: 0.940-0.911 0.029

Split info: info([5,4,5]) 1.577 Split info: info([4,6,4]) 1.362

Gain ratio: 0.247/1.577 0.156 Gain ratio: 0.029/1.362 0.021

Humidity Windy

Info: 0.788 Info: 0.892

Gain: 0.940-0.788 0.152 Gain: 0.940-0.892 0.048

Split info: info([7,7]) 1.000 Split info: info([8,6]) 0.985

Gain ratio: 0.152/1 0.152 Gain ratio: 0.048/0.985 0.049

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More on the gain ratio

“ Outlook” still comes out top However: “ID code” has greater gain ratio

Standard fix: ad hoc test to prevent splitting on that type of attribute

Problem with gain ratio: it may overcompensate May choose an attribute just because its intrinsic

information is very low Standard fix:

First, only consider attributes with greater than average information gain

Then, compare them on gain ratio

n

jjpTgini

1

21)(

splitgini NT

NTT

Ngini

Ngini( ) ( ) ( ) 1

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Gini Index If a data set T contains examples from n classes, gini index,

gini(T) is defined as

where pj is the relative frequency of class j in T. gini(T) is

minimized if the classes in T are skewed. After splitting T into two subsets T1 and T2 with sizes N1 and

N2, the gini index of the split data is defined as

The attribute providing smallest ginisplit(T) is chosen to split the node.

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Stopping Criteria

When all cases have the same class. The leaf node is labeled by this class.

When there is no available attribute. The leaf node is labeled by the majority class.

When the number of cases is less than a specified threshold. The leaf node is labeled by the majority class.

You can make a decision at every node in a decision tree!

How?

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Pruning

Pruning simplifies a decision tree to prevent overfitting to noise in the data

Two main pruning strategies:1. Postpruning: takes a fully-grown decision tree and

discards unreliable parts2. Prepruning: stops growing a branch when

information becomes unreliable Postpruning preferred in practice because of

early stopping in prepruning

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Prepruning

Usually based on statistical significance test Stops growing the tree when there is no

statistically significant association between any attribute and the class at a particular node

Most popular test: chi-squared test ID3 used chi-squared test in addition to

information gain Only statistically significant attributes where allowed

to be selected by information gain procedure

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Postpruning

Builds full tree first and prunes it afterwards Attribute interactions are visible in fully-grown tree

Problem: identification of subtrees and nodes that are due to chance effects

Two main pruning operations: 1. Subtree replacement2. Subtree raising

Possible strategies: error estimation, significance testing, MDL principle

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Subtree replacement

Bottom-up: tree is considered for replacement once all its subtrees have been considered

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Estimating error rates

Pruning operation is performed if this does not increase the estimated error

Of course, error on the training data is not a useful estimator (would result in almost no pruning)

One possibility: using hold-out set for pruning (reduced-error pruning)

C4.5’s method: using upper limit of 25% confidence interval derived from the training data

Standard Bernoulli-process-based method

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Post-pruning in C4.5 Bottom-up pruning: at each non-leaf node v, if

merging the subtree at v into a leaf node improves accuracy, perform the merging.

Method 1: compute accuracy using examples not seen by the algorithm.

Method 2: estimate accuracy using the training examples: Consider classifying E examples incorrectly out of

N examples as observing E events in N trials in the binomial distribution.

For a given confidence level CF, the upper limit on the error rate over the whole population is with CF% confidence. ),( NEUCF

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Usage in Statistics: Sampling error estimation Example:

population: 1,000,000 people, population mean: percentage of the left handed people sample: 100 people sample mean: 6 left-handed How to estimate the REAL population mean?

Pessimistic Estimate

U0.25(100,6)L0.25(100,6)6

Possibility(%)

2 10

25% confidence interval

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C4.5’s method

Error estimate for subtree is weighted sum of error estimates for all its leaves

Error estimate for a node:

If c = 25% then z = 0.69 (from normal distribution)

f is the error on the training data N is the number of instances covered by the

leaf

Nz

N

zNf

Nf

zN

zfe

2

2

222

142

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Example

Outlook

yes yes no

sunny

overcast

cloudy yes?

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Example cont. Consider a subtree rooted at Outlook with 3

leaf nodes: Sunny: Play = yes : (0 error, 6 instances) Overcast: Play= yes: (0 error, 9 instances) Cloudy: Play = no (0 error, 1 instance)

The estimated error for this subtree is 6*0.074+9*0.050+1*0.323=1.217

If the subtree is replaced with the leaf “yes”, the estimated error is

So no pruning is performed

323.0)0,1(,050.0)0,9(,074.0)0,6( 25.025.025.0 UUU

888.1118.0*16)1,16(*16 25.0 U

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Another Example

f=0.33 e=0.47 f=0.5

e=0.72f=0.33 e=0.47

f=5/14 e=0.46

If we split, we can compute average error usingratios 6:2:6this gives 0.51

If we have a single leaf node

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Other Trees

• Classification Trees•Current node •Children nodes (L, R):

•Decision Trees •Current node •Children nodes (L, R):

•GINI index used in CART (STD= )•Current node •Children nodes (L, R):

),min(),min( 1010RRRLLL ppppppQ

),min( 10 ppQ

1100 loglog ppppQ

RRLL QpQpQ

10 pp10 ppQ

RRLL QpQpQ

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Efforts on Scalability

Most algorithms assume data can fit in memory. Recent efforts focus on disk-resident implementation for

decision trees. Random sampling Partitioning Examples

SLIQ (EDBT’96 -- [MAR96]) SPRINT (VLDB96 -- [SAM96]) PUBLIC (VLDB98 -- [RS98]) RainForest (VLDB98 -- [GRG98])

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Questions

Consider the following variations of decision trees

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1. Apply KNN to each leaf node

Instead of choosing a class label as the majority class label, use KNN to choose a class label

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2. Apply Naïve Bayesian at each leaf node

For each leave node, use all the available information we know about the test case to make decisions

Instead of using the majority rule, use probability/likelihood to make decisions

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3. Use error rates instead of entropy

If a node has N1 positive class labels P, and N2 negative class labels N,

If N1> N2, then choose P The error rate = N2/(N1+N2) at this node The expected error at a parent node can be

calculated as weighted sum of the error rates at each child node

The weights are the proportion of training data in each child

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Cost Sensitive Decision Trees

When the FP and FN have different costs, the leaf node label is different depending on the costs:

If growing a tree has a smaller total cost, then choose an attribute with minimal total cost. Otherwise, stop and form a leaf.

Label leaf according to minimal total cost: Suppose the leaf have P positive examples and N negative

examples FP denotes the cost of a false positive example and FN false

negative If (P×FN N×FP) THEN label = positive

ELSE label = negative

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5. When there is missing value in the test data, we allow tests to be done

Attribute selection criterion: minimal total cost (Ctotal = Cmc + Ctest) instead of minimal entropy in C4.5

Typically, if there are missing values, then to obtain a value for a missing attribute (say Temperature) will incur new cost

But may increase accuracy of prediction, thus reducing the miss classification costs

In general, there is a balance between the two costs We care about the total cost

Stream Data

Suppose that you have built a decision tree from a set of training data

Now suppose that some additional training data comes at a regular interval, in fast speed

How do you adapt the existing tree to fit the new data?

Suggest an ‘online’ algorithm. Hint: find a paper online on this topic

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