random.choice的加权版本

我需要写一个加权版的random。选择(列表中的每个元素有不同的被选择的概率)。这是我想到的:

def weightedChoice(choices):
    """Like random.choice, but each element can have a different chance of
    being selected.

    choices can be any iterable containing iterables with two items each.
    Technically, they can have more than two items, the rest will just be
    ignored.  The first item is the thing being chosen, the second item is
    its weight.  The weights can be any numeric values, what matters is the
    relative differences between them.
    """
    space = {}
    current = 0
    for choice, weight in choices:
        if weight > 0:
            space[current] = choice
            current += weight
    rand = random.uniform(0, current)
    for key in sorted(space.keys() + [current]):
        if rand < key:
            return choice
        choice = space[key]
    return None

这个函数对我来说太复杂了，而且很丑。我希望这里的每个人都能提供一些改进的建议或其他方法。对我来说，效率没有代码的整洁和可读性重要。

当前回答

import numpy as np
w=np.array([ 0.4,  0.8,  1.6,  0.8,  0.4])
np.random.choice(w, p=w/sum(w))

2013-12-11 16:38:41

其他回答

通解:

import random
def weighted_choice(choices, weights):
    total = sum(weights)
    treshold = random.uniform(0, total)
    for k, weight in enumerate(weights):
        total -= weight
        if total < treshold:
            return choices[k]

2013-06-09 15:26:44

为random.choice()提供一个预先加权的列表:

解决方案和测试:

import random

options = ['a', 'b', 'c', 'd']
weights = [1, 2, 5, 2]

weighted_options = [[opt]*wgt for opt, wgt in zip(options, weights)]
weighted_options = [opt for sublist in weighted_options for opt in sublist]
print(weighted_options)

# test

counts = {c: 0 for c in options}
for x in range(10000):
    counts[random.choice(weighted_options)] += 1

for opt, wgt in zip(options, weights):
    wgt_r = counts[opt] / 10000 * sum(weights)
    print(opt, counts[opt], wgt, wgt_r)

输出:

['a', 'b', 'b', 'c', 'c', 'c', 'c', 'c', 'd', 'd']
a 1025 1 1.025
b 1948 2 1.948
c 5019 5 5.019
d 2008 2 2.008

2019-10-02 18:37:14

使用numpy

def choice(items, weights):
    return items[np.argmin((np.cumsum(weights) / sum(weights)) < np.random.rand())]

2018-08-31 13:07:59

def weighted_choice(choices):
   total = sum(w for c, w in choices)
   r = random.uniform(0, total)
   upto = 0
   for c, w in choices:
      if upto + w >= r:
         return c
      upto += w
   assert False, "Shouldn't get here"

2010-09-09 19:08:40

如果您的加权选项列表是相对静态的，并且您希望频繁采样，则可以执行一个O(N)预处理步骤，然后使用相关答案中的函数在O(1)中进行选择。

# run only when `choices` changes.
preprocessed_data = prep(weight for _,weight in choices)

# O(1) selection
value = choices[sample(preprocessed_data)][0]

2015-04-21 17:42:55

random.choice的加权版本

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