Classification with Decision Trees
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Transcript of Classification with Decision Trees
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Classification with Decision Trees
Instructor: Qiang YangHong Kong University of Science and Technology
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.
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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
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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)
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.||
||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
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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
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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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