Smarter way of doing this?

[Max M]

Max M wrote:

> Jim Jewett wrote:

> I finally chose a combination of the suggested approaches, and ended up 
> with something I find is both more functional and nicer to look at than 
> my first try.


Well, that version had a problem. Whenever a series of values are the 
same bisect() allways selects the first one. So I never got a 
statistical valid sample of elements.

I solved it by using acumulated probabilities instead. So the final 
version is here, if anybody cares.

I used it in a Markov chain class, that is now pretty fast.

regards Max M



#########################

from random import random, randint
from bisect import bisect


class Selector:

     "Returns random elements by probability"

     def __init__(self, probabilities, elements):
         self.sum = sum(probabilities)
         acumulated_probabilities = []
         acumulated_probability = 0
         for probability in probabilities:
             acumulated_probability += probability
             acumulated_probabilities.append(acumulated_probability)
         self.decorated = zip(acumulated_probabilities, elements)
         self.decorated.sort()

     def get(self):
         "Randomly returns an element by probability"
         rnd = random() * self.sum
         index = bisect(self.decorated, (rnd, 0))
         self.decorated[index][-1]
         return self.decorated[index][-1]

     def get_range(self, n):
         "Randomly returns a range of elements by probability"
         return [self.get() for itm in range(n)]





if __name__ == '__main__':


     probabilities = [1.0, 0.5, 0.25, 0.125, 0.0625]
#    probabilities = [0.1, 0.1, 1.0, 0.1, 0.1, ]
     elements = ['a', 'b', 'c', 'd', 'e']
     s = Selector(probabilities, elements)
     r = s.get_range(10000)
     r.sort()

     for element in elements:
         print element, r.count(element)


##########
 >>a 5148
 >>b 2594
 >>c 1301
 >>d 654
 >>e 303