A weighted version of random.choice

Question

I needed to write a weighted version of random.choice (each element in the list has a different probability for being selected). This is what I came up with:

Answer 1

Crude, but may be sufficient:

import random
weighted_choice = lambda s : random.choice(sum(([v]*wt for v,wt in s),[]))

Does it work?

# define choices and relative weights
choices = [("WHITE",90), ("RED",8), ("GREEN",2)]

# initialize tally dict
tally = dict.fromkeys(choices, 0)

# tally up 1000 weighted choices
for i in xrange(1000):
    tally[weighted_choice(choices)] += 1

print tally.items()

Prints:

[('WHITE', 904), ('GREEN', 22), ('RED', 74)]

Assumes that all weights are integers. They don't have to add up to 100, I just did that to make the test results easier to interpret. (If weights are floating point numbers, multiply them all by 10 repeatedly until all weights >= 1.)

weights = [.6, .2, .001, .199]
while any(w < 1.0 for w in weights):
    weights = [w*10 for w in weights]
weights = map(int, weights)