如何基于加权概率从python字典中选择键? [英] How to choose keys from a python dictionary based on weighted probability?

问题描述

我有一个Python字典，其中的键代表某些项目，值代表该项目的某些(标准化)权重.例如:

I have a Python dictionary where keys represent some item and values represent some (normalized) weighting for said item. For example:

d = {'a': 0.0625, 'c': 0.625, 'b': 0.3125}
# Note that sum([v for k,v in d.iteritems()]) == 1 for all `d`

鉴于项目与权重之间的这种相关性，我如何从d中选择一个键，使得结果的6.25％的时间为'a'，结果的32.25％的时间为'b'，以及62.5％的时间结果是"c"?

Given this correlation of items to weights, how can I choose a key from d such that 6.25% of the time the result is 'a', 32.25% of the time the result is 'b', and 62.5% of the result is 'c'?

推荐答案

def weighted_random_by_dct(dct):
    rand_val = random.random()
    total = 0
    for k, v in dct.items():
        total += v
        if rand_val <= total:
            return k
    assert False, 'unreachable'

应该做到这一点.仔细检查每个密钥并保持连续的总和，如果随机值(介于0和1之间)落在插槽中，它将返回该密钥

Should do the trick. Goes through each key and keeps a running sum and if the random value (between 0 and 1) falls in the slot it returns that key

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