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。

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

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"

步骤1:生成您感兴趣的CDF F

步骤2:生成u.r.v. u

步骤3:求z=F^{-1}(u)

这种建模在概率论或随机过程课程中有描述。这是适用的，因为您有简单的CDF。

从Python 3.6开始，随机模块中有一个方法选择。

In [1]: import random

In [2]: random.choices(
...:     population=[['a','b'], ['b','a'], ['c','b']],
...:     weights=[0.2, 0.2, 0.6],
...:     k=10
...: )

Out[2]:
[['c', 'b'],
 ['c', 'b'],
 ['b', 'a'],
 ['c', 'b'],
 ['c', 'b'],
 ['b', 'a'],
 ['c', 'b'],
 ['b', 'a'],
 ['c', 'b'],
 ['c', 'b']]

注意随机。选择将与替换样本，每个文档:

返回一个k大小的元素列表，这些元素是从替换的填充中选择的。

为确保回答的完整性，请注意:

当从一个有限的总体中抽取一个抽样单位并返回时 对于该种群，在其特征被记录下来之后， 在绘制下一个单元之前，采样被称为“与” 更换”。它基本上意味着每个元素可以被选择多于 一次。

如果您需要在不替换的情况下进行采样，那么就像@ronan-paixão的精彩回答所说的那样，您可以使用numpy。Choice，其replace参数控制这种行为。

如果你碰巧有Python 3，并且害怕安装numpy或编写自己的循环，你可以这样做:

import itertools, bisect, random

def weighted_choice(choices):
   weights = list(zip(*choices))[1]
   return choices[bisect.bisect(list(itertools.accumulate(weights)),
                                random.uniform(0, sum(weights)))][0]

因为你可以用一袋管道适配器做任何东西!尽管……我必须承认，尼德的回答虽然稍长一些，但比较容易理解。