Generating weighted random numbers

问题:

Hi I'm doing some code for a genomics class and I am having difficulty on a certain part.

I have a set of mutually exclusive events with probabilities

I want to simulate randomly sampling an event n times with the given probability.

input: probabilities = {0.3, 0.2, 0.5} events{e1,e2,e3} n=100

output: there should be ~50 results for e3, ~20 for e2 and ~30 for e1. Note that these are probably not exactly 50, 20, 30 because empirical values are different from theoretical values...

回答1:

Python doesn't have any weighted sampling functionality built in (NumPy/SciPy does), but for a really simple case like this, it's pretty easy:

import itertools
import random

probabilities = [0.3, 0.2, 0.5]
totals = list(itertools.accumulate(probabilities))

def sample():
    n = random.uniform(0, totals[-1])
    for i, total in enumerate(totals):
        if n <= total:
            return i

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If you don't have Python 3.2+, you don't have the accumulate function; you can fake it with an inefficient one-liner if the list really is this short:

totals = [sum(probabilities[:i+1]) for i in range(len(probabilities))]

… or you can write an explicit loop, or an ugly reduce call, or copy the equivalent Python function from the docs.

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Also, note that random.uniform(0, totals[-1]) is just a more complicated way of writing random.random() if you can be sure that your numbers add up to 1.0.

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A quick way to test this:

>>> samples = [sample() for _ in range(100000)]
>>> samples.count(0)
29878
>>> samples.count(1)
19908
>>> samples.count(2)
50214

Those are pretty close to 30%, 20%, and 50% of 100000, respectively.

回答2:

Let's assume that we have three events, each with probabilities .3, .2 and .5, respectively. Then for each sample generated, we generate a number in the range [0,1), let's call this "rand." If "rand" < .3, we generate event 1, if .3 <= "rand" < .5, we generate even 2, otherwise we generate event 3. This can be accomplished using random(), which indeed generates a number in the range [0,1).