Data Mining: Practical Machine Learning Tools and Techniques, Second Edition
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- Table 4.2 The weather data with counts and probabilities.
- Table 4.3 A new day.
makes real-life datasets interesting is that the attributes are certainly not equally important or independent. But it leads to a simple scheme that again works sur- prisingly well in practice. Table 4.2 shows a summary of the weather data obtained by counting how many times each attribute–value pair occurs with each value (yes and no) for
ples, two of which have play = yes and three of which have play = no. The cells in the first row of the new table simply count these occurrences for all possible values of each attribute, and the play figure in the final column counts the total number of occurrences of yes and no. In the lower part of the table, we rewrote the same information in the form of fractions, or observed probabilities. For example, of the nine days that play is yes, outlook is sunny for two, yielding a fraction of 2/9. For play the fractions are different: they are the proportion of days that play is yes and no, respectively. Now suppose we encounter a new example with the values that are shown in Table 4.3. We treat the five features in Table 4.2—outlook, temperature, humid- ity, windy, and the overall likelihood that play is yes or no—as equally impor- tant, independent pieces of evidence and multiply the corresponding fractions. Looking at the outcome yes gives: The fractions are taken from the yes entries in the table according to the values of the attributes for the new day, and the final 9/14 is the overall fraction likelihood of yes = ¥
¥ ¥ = 2 9 3 9 3 9 3 9 9 14 0 0053
. . 4 . 2 S TAT I S T I C A L M O D E L I N G 8 9
Table 4.2 The weather data with counts and probabilities. Outlook
Temperature Humidity
Windy Play
yes no yes no yes no yes no yes no sunny
2 3 hot 2 2 high 3 4 false 6 2 9 5 overcast
4 0 mild 4 2 normal 6 1 true 3 3 rainy 3 2 cool 3 1 sunny 2/9 3/5
hot 2/9
2/5 high
3/9 4/5
false 6/9
2/5 9/14
5/14 overcast
4/9 0/5
mild 4/9
2/5 normal
6/9 1/5
true 3/9
3/5 rainy
3/9 2/5
cool 3/9
1/5 Table 4.3 A new day. Outlook
Temperature Humidity
Windy Play
sunny cool
high true
? P088407-Ch004.qxd 4/30/05 11:13 AM Page 89 representing the proportion of days on which play is yes. A similar calculation for the outcome no leads to This indicates that for the new day, no is more likely than yes—four times more likely. The numbers can be turned into probabilities by normalizing them so that they sum to 1: This simple and intuitive method is based on Bayes’s rule of conditional prob- ability. Bayes’s rule says that if you have a hypothesis H and evidence E that bears on that hypothesis, then We use the notation that Pr[A] denotes the probability of an event A and that Pr[A |B] denotes the probability of A conditional on another event B. The hypothesis H is that play will be, say, yes, and Pr[H |E] is going to turn out to be 20.5%, just as determined previously. The evidence E is the particular combi- nation of attribute values for the new day, outlook = sunny, temperature = cool, humidity = high, and windy = true. Let’s call these four pieces of evidence E 1 , E 2 ,
3 , and E 4 , respectively. Assuming that these pieces of evidence are independent (given the class), their combined probability is obtained by multiplying the probabilities: Don’t worry about the denominator: we will ignore it and eliminate it in the final normalizing step when we make the probabilities of yes and no sum to 1, just as we did previously. The Pr[yes] at the end is the probability of a yes outcome without knowing any of the evidence E, that is, without knowing any- thing about the particular day referenced—it’s called the prior probability of the hypothesis H. In this case, it’s just 9/14, because 9 of the 14 training examples had a yes value for play. Substituting the fractions in Table 4.2 for the appro- priate evidence probabilities leads to Pr Pr
E [ ] = ¥ ¥ ¥ ¥ [ ] 2 9 3 9 3 9 3 9 9 14 , Pr Pr Pr Pr Pr Pr Pr yes E E yes E yes E yes E yes yes E [ ] = [ ] ¥ [ ] ¥ [ ] ¥ [ ] ¥ [ ] [ ] 1 2 3 4 . Pr Pr Pr Pr H E E H H E [ ] = [ ] [ ] [ ] . Probability of no = + = 0 0206 0 0053 0 0206 79 5 .
. . %.
Probability of yes = + = 0 0053
0 0053 0 0206 20 5
. . . . %, likelihood of no = ¥
¥ ¥ = 3 5 1 5 4 5 3 5 5 14 0 0206 . .
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