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

将权重排列成a 累积分布。 使用random.random()来选择一个随机的 浮点0.0 <= x < total。 搜索 用等分法进行分布。二等分的 如http://docs.python.org/dev/library/bisect.html#other-examples中的示例所示。

from random import random
from bisect import bisect

def weighted_choice(choices):
    values, weights = zip(*choices)
    total = 0
    cum_weights = []
    for w in weights:
        total += w
        cum_weights.append(total)
    x = random() * total
    i = bisect(cum_weights, x)
    return values[i]

>>> weighted_choice([("WHITE",90), ("RED",8), ("GREEN",2)])
'WHITE'

如果需要做出多个选择,可以将其分成两个函数,一个用于构建累积权重,另一个用于对随机点进行等分。

其他回答

从版本1.7.0开始,NumPy有一个支持概率分布的选择函数。

from numpy.random import choice
draw = choice(list_of_candidates, number_of_items_to_pick,
              p=probability_distribution)

注意,probability_distribution是一个与list_of_candidate顺序相同的序列。您还可以使用关键字replace=False来更改行为,这样绘制的项就不会被替换。

如果不介意使用numpy,可以使用numpy.random.choice。

例如:

import numpy

items  = [["item1", 0.2], ["item2", 0.3], ["item3", 0.45], ["item4", 0.05]
elems = [i[0] for i in items]
probs = [i[1] for i in items]

trials = 1000
results = [0] * len(items)
for i in range(trials):
    res = numpy.random.choice(items, p=probs)  #This is where the item is selected!
    results[items.index(res)] += 1
results = [r / float(trials) for r in results]
print "item\texpected\tactual"
for i in range(len(probs)):
    print "%s\t%0.4f\t%0.4f" % (items[i], probs[i], results[i])

如果你知道你需要提前做多少选择,你可以不像这样循环:

numpy.random.choice(items, trials, p=probs)

下面是Python 3.6标准库中包含的版本:

import itertools as _itertools
import bisect as _bisect

class Random36(random.Random):
    "Show the code included in the Python 3.6 version of the Random class"

    def choices(self, population, weights=None, *, cum_weights=None, k=1):
        """Return a k sized list of population elements chosen with replacement.

        If the relative weights or cumulative weights are not specified,
        the selections are made with equal probability.

        """
        random = self.random
        if cum_weights is None:
            if weights is None:
                _int = int
                total = len(population)
                return [population[_int(random() * total)] for i in range(k)]
            cum_weights = list(_itertools.accumulate(weights))
        elif weights is not None:
            raise TypeError('Cannot specify both weights and cumulative weights')
        if len(cum_weights) != len(population):
            raise ValueError('The number of weights does not match the population')
        bisect = _bisect.bisect
        total = cum_weights[-1]
        return [population[bisect(cum_weights, random() * total)] for i in range(k)]

来源:https://hg.python.org/cpython/file/tip/Lib/random.py l340

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]

加权选择的一个非常基本和简单的方法如下:

np.random.choice(['A', 'B', 'C'], p=[0.3, 0.4, 0.3])