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Arbitrary categorisation - learning decision trees:
Through visualising a set of boxes with some balls in. There if all the balls were in a single box so this would be nicely ordered but it would be extremely easy to find a particular ball. Moreover If the balls were distributed amongst the boxes then this would not be so nicely ordered but it might take rather a whereas to find a particular ball. It means if we were going to define a measure based at this notion of purity then we would want to be able to calculate a value for each box based on the number of balls in it so then take the sum of these as the overall measure. Thus we would want to reward two situations: nearly empty boxes as very neat and boxes just with nearly all the balls in as also very neat. However this is the basis for the general entropy measure that is defined follows like:
Now next here instantly an arbitrary categorisation like C into categories c1, ..., cn and a set of examples, S, for that the proportion of examples in ci is pi, then the entropy of S is as:
Here measure satisfies our criteria that is of the -p*log2(p) construction: where p gets close to zero that is the category has only a few examples in it so then the log(p) becomes a big negative number and the p part dominates the calculation then the entropy works out to be nearly zero. However make it sure that entropy calculates the disorder in the data in this low score is good and as it reflects our desire to reward categories with few examples in. Such of similarly if p gets close to 1 then that's the category has most of the examples in so then the log(p) part gets very close to zero but it is this that dominates the calculation thus the overall value gets close to zero. Thus we see that both where the category is nearly - or completely - empty and when the category nearly contains as - or completely contains as - all the examples and the score for the category gets close to zero that models what we wanted it to. But note that 0*ln(0) is taken to be zero by convention them.
De Morgan's Laws : With continuing the relationship between ^ and ? , we can also needs De Morgan's Law to rearrange sentences involving negation in conjunction with these con
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