Decision Tree Learning - ID3
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1 Decision Tree Learning - ID3 n Decision tree examples n ID3 algorithm n Occam Razor n Top-Down Induction in Decision Trees n Information Theory n gain from property 1
2 Training Examples Day Outlook Temp. Humidity Wind Play Tennis D1 Sunny Hot High Weak No D2 Sunny Hot High Strong No D3 Overcast Hot High Weak Yes D4 Rain Mild High Weak Yes D5 Rain Cool Normal Weak Yes D6 Rain Cool Normal Strong No D7 Overcast Cool Normal Weak Yes D8 Sunny Mild High Weak No D9 Sunny Cold Normal Weak Yes D10 Rain Mild Normal Strong Yes D11 Sunny Mild Normal Strong Yes D12 Overcast Mild High Strong Yes D13 Overcast Hot Normal Weak Yes D14 Rain Mild High Strong No Decision Tree for PlayTennis Outlook Sunny Overcast Rain Humidity Yes Wind High Normal Strong Weak No Yes No Yes 2
3 Decision Tree for PlayTennis Outlook Sunny Overcast Rain Humidity Each internal node tests an attribute High No Normal Yes Each branch corresponds to an attribute value node Each leaf node assigns a classification Decision Tree for PlayTennis Outlook Temperature Humidity Wind PlayTennis Sunny Hot High Weak? No Outlook Sunny Overcast Rain Humidity Yes Wind High Normal Strong Weak No Yes No Yes 3
4 Decision Tree for Conjunction Outlook=Sunny Ù Wind=Weak Outlook Sunny Overcast Rain Wind No No Strong No Weak Yes Decision Tree for Disjunction Outlook=Sunny Ú Wind=Weak Outlook Sunny Overcast Rain Yes Wind Wind Strong Weak Strong Weak No Yes No Yes 4
5 Decision Tree for XOR Outlook=Sunny XOR Wind=Weak Outlook Sunny Overcast Rain Wind Wind Wind Strong Weak Strong Weak Strong Weak Yes No No Yes No Yes Decision Tree decision trees represent disjunctions of conjunctions Outlook Sunny Overcast Rain Humidity Yes Wind High Normal Strong Weak No Yes No Yes (Outlook=Sunny Ù Humidity=Normal) Ú (Outlook=Overcast) Ú (Outlook=Rain Ù Wind=Weak) 5
6 When to consider Decision Trees n n n n n n Instances describable by attribute-value pairs Target function is discrete valued Disjunctive hypothesis may be required Possibly noisy training data Missing attribute values Examples: n Medical diagnosis n Credit risk analysis n Object classification for robot manipulator (Tan 1993) 6
7 7
8 Decision tree for credit risk assessment 8
9 n The decision tree represents the classification of the table n It can classify all the objects in the table n Each internal node represents a test on some property n Each possible value of that property corresponds to a branch of the tree n An individual of unknown type may be classified be traversing this tree n In classifying any given instance, the tree does not use all the properties in the table n Decision tree for credit risk assessment n If a person has a good credit history and low debit, we ignore her collateral income and classify her as low risk n In spite of omitting certain tests, the tree classifies all examples in the table 9
10 n In general, the size of a tree necessary to classify a given set of examples varies according to the order with which properties (=attributes) are tested n Given a set of training instances and a number of different decision trees that correctly classify the instances, we may ask which tree has the greatest likelihood of correctly classifying using instances of the population? n This is a simplified decision tree for credit risk assessment n It classifies all examples of the table correctly 10
11 n ID3 algorithm assumes that a good decision tree is the simplest decision tree n Heuristic: n Preferring simplicity and avoiding unnecessary assumptions n Known as Occam s Razor n Occam Razor was first articulated by the medieval logician William of Occam in 1324 born in the village of Ockham in Surrey (England) about 1285, believed that he died in a convent in Munich in 1349, a victim of the Black Death It is vain do with more what can be done with less.. n We should always accept the simplest answer that correctly fits our data n The smallest decision tree that correctly classifies all given examples 11
12 n The simplest decision tree that covers all examples should be the least likely to include unnecessary constraints n ID3 selects a property to test at the current node of the tree and uses this test to partition the set of examples n The algorithm then recursively constructs a sub tree for each partition n This continuous until all members of the partition are in the same class That class becomes a leaf node of the tree 12
13 n Because the order of tests is critical to constructing a simple tree, ID3 relies heavily on its criteria for selecting the test at the root of each sub tree Top-Down Induction of Decision Trees ID3 1. A the best decision attribute for next node 2. Assign A as decision attribute (=property) for node 3. For each value of A create new descendant 4. Sort training examples to leaf node according to the attribute value of the branch 5. If all training examples are perfectly classified (same value of target attribute) stop, else iterate over new leaf nodes 13
14 n ID3 constructs the tree for credit rist assessment n Beginning with the full table of examples, ID3 selects income as the root property using function selecting best property (attribute) n The examples are divided, listed by their number in the list 14
15 n ID3 applies the method recursively for each partition n The partition {1,4,7,11} consists entirely of high-risk individuals, a node is created n ID3 selects credit history property as the root of the subtree for the partition {2,3,12,14} n Credit history further divides this partition into {2,3},{14} and {12} n ID3 implements a form of hill climbing in the space of all possible trees 15
16 Hypothesis Space Search ID3 two classes: +,- A A A2 A A A4 Two classes: +,- 16
17 Information Theory n Information theory (Shannon 1948) provides the information content of a message n We may think of a message as an instance in a universe of possible messages n The act of transmitting a message is the same as selecting one of these possible messages n Define information content of a message as depending upon both the size of this universe and the frequency with which each possible message occurs Issues with MDL n What is the right model family? n This determines the kind of solutions that we can have E.g., polynomials Clusterings n What is the encoding cost? n Determines the function that we optimize n Information theory 17
18 II Entropy n Let be F an experiment (e.g. : two dice) n Before we perform the experiment, we do not know what will be the results... n We are uncertain about the outcome n How can we measure the uncertainty n Instead of uncertainty we use the word Entropy of the experiment 0 H(F) Entropy - Information n Experiments starts at t 0 and ends at t 1 n At t 0 we have no information about the results of the experiment n At t 1 we have all information, so the Entropy of the experiment is 0 n From t 1 to t 0 we have wone information 18
19 Time Entropy Information t 0 (before) H(F) 0 t 1 (after) 0 H(F) n We can describe an experiment by probabilities n Experiment, outcome of the flip of a honest coin n Head or Tail, both probability 0.5, the outcome can be either heat or tail, p=(0.5,0.5) n H(F)=H(p 1,p 2 )=(0.5,0.5) 19
20 Interpretation of H(F) n The experiment F was done n Person A knows the outcome, person B not n How to define H? n H = number of questions to A, B has to pose to know the result of the experiment n Questions of the form yes/no n Example coin, p=(0.5,0.5) n We can pose the question, is it tail? n H=1 20
21 n Example cards, p=(1/2,1/4,1/4) n red, clubs, spade n We can ask, is the card red, if the answer is no, we have only to ask is it spade... n If the card is red, we need only one question, else we need two questions 21
22 n We have to speak about the mean number of questions n H(F)=1/2*1+1/4*2+1/4*2=1.5 n If the card is red, we need only one question, for clubs and spade we need 2 questions... Interpretation of H(F) n The experiment F was done n Person A knows the outcome, person B not n How to define H? n H = mean number of optimal questions to A, B has to pose to know the result of the experiment n Questions of the form yes/no 22
23 n For four cards of which one is the joker the probability of a joker is 0.25 and of other cards =0.75, p=(0.25,0.75) n In the mean we have to ask n 1* *0.75=1 n questions to determine to determine if the card is a joker or not. n Given n cards of which one is the joker the probability of a joker is 1/n and of other cards is 1-1/n n In the mean we have to ask 1 * 1/n + 1 * (1-1/n) n questions to determine if the card is a joker or not. n Its results in one question independent of the size of n. 23
24 n It seems some thing is missing in our definition n Our result is correct for one independent experiment n For several experiments the mean number of questions is lower n We define the real entropy: n for one experiment as H 0 (F 1 ) n for two experiments as H 0 (F 2 ) n.. n For k experiments as H 0 (F k ) 24
25 n The mean number of question for one experiment in the sequence of k experiments is n 1/k *H 0 (F k ) n For four cards of which one is the joker the probability of a joker is 0.25 and of other cards =0.75 n H 0 (F 1 )=1 n H 0 (F 1 )=1= 1* *0.25=1 n k=1, 1/k *H 0 (F k )=1/1*H 0 (F 1 )=1 n 25
26 n What is the size ofh 0 (F 2 )? 26
27 27
28 Ideal Entropy 28
29 n An experiment is described by probabilities p=(p 1,p 2,...,p n ) n Does the distribution of these probabilities have an effect on the ideal entropy? n It turns out that the ideal entropy is maximal in the case all probabilities are equal, means p=(1/n,1/n...,1/n) n In this case the maximal ideal Entropy is 29
30 Only two probabilities n Shannon formalized these intuitions n Given a universe of messages M={m 1,m 2,...,m n } and a probability p(m i ) for the occurrence of each message, the information content (also called entropy)of a message M is given n I(M) = p(m i )log 2 ( p(m i )) i=1 30
31 n Information content of a message telling the outcome of the flip of a honest coin I(Coin _ toss) = p(heads)log 2 (p(heads)) p(tails)log 2 ( p(tails)) I(Coin _ toss) = p(0.5)log 2 ( p(0.5)) p(0.5)log 2 (p(0.5)) I(Coin _ toss) =1 bit n However if the coin has been rigged to come up heads 75 percent I(Coin _ toss) = p(heads)log 2 (p(heads)) p(tails)log 2 ( p(tails)) I(Coin _ toss) = p(0.75)log 2 ( p(0.75)) p(0.25)log 2 (p(0.25)) I(Coin _ toss) = bits 31
32 n We may think of a decision tree as conveying information about the classification of examples in the decision table n The information content of the tree is computed from the probabilities of different classifications n The credit history loan table has following information n p(risk is high)=6/14 n p(risk is moderate)=3/14 n p(risk is low)=5/14 I(credit _ table) = 6 14 log log log I(credit _ table) = bits 32
33 n For a given test, the information gain provided by making that test at the root of the current tree is equal to n Total information of the table - the amount of information needed to complete the classification after performing the test n The amount of information needed to complete the tree is defined as weighted average of the information content of each sub tree n The amount of information needed to complete the tree is defined as weighted average of the information content of each sub tree by the percentage of the examples present n C a set of training instances. If property (for example income) with n values, C will be divided into the subsets {C 1,C 2,..,C n } n Expected information needed to complete the tree after making P root n C E(P) = i I(C i ) C i=1 33
34 n C E(P) = i I(C i ) C i=1 n The gain from the property P is computed by subtracting the expected information to complete E(P) fro the total information gain(p) = I(C) E(P) n In the credit history loan table we make income the property tested at the root n This makes the division into C 1 ={1,4,7,11},C 2 ={2,3,12,14},C 3 ={5,6,8,9,10,13} E(income) = 4 14 I(C 1) I(C 2) I(C 3) E(income) = E(income) = bits 34
35 gain(income)=i(credit_table)-e(income) gain(income)= gain(income)=0.967 bits gain(credit history)=0.266 gain(debt)=0.581 gain(collateral)=0.756 n Because income provides the greatest information gain, ID will select it as the root of the tree n The algorithm continues to apply this analysis recursively to each subtree, until it has completed the tree. 35
36 Day Outlook Temp. Humidity Wind Play Tennis D1 Sunny Hot High Weak No D2 Sunny Hot High Strong No D3 Overcast Hot High Weak Yes D4 Rain Mild High Weak Yes D5 Rain Cool Normal Weak Yes D6 Rain Cool Normal Strong No D7 Overcast Cool Normal Weak Yes D8 Sunny Mild High Weak No D9 Sunny Cold Normal Weak Yes D10 Rain Mild Normal Strong Yes D11 Sunny Mild Normal Strong Yes D12 Overcast Mild High Strong Yes D13 Overcast Hot Normal Weak Yes D14 Rain Mild High Strong No 36
37 37
38 n Enhancements to basic decision tree induction n C4.5, CART algorithm 38
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