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

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

其他回答

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"

我看了指向的其他线程,并在我的编码风格中提出了这种变化,这返回了用于计数的索引,但返回字符串很简单(注释返回替代):

import random
import bisect

try:
    range = xrange
except:
    pass

def weighted_choice(choices):
    total, cumulative = 0, []
    for c,w in choices:
        total += w
        cumulative.append((total, c))
    r = random.uniform(0, total)
    # return index
    return bisect.bisect(cumulative, (r,))
    # return item string
    #return choices[bisect.bisect(cumulative, (r,))][0]

# define choices and relative weights
choices = [("WHITE",90), ("RED",8), ("GREEN",2)]

tally = [0 for item in choices]

n = 100000
# tally up n weighted choices
for i in range(n):
    tally[weighted_choice(choices)] += 1

print([t/sum(tally)*100 for t in tally])

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。

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。

粗糙的,但可能足够:

import random
weighted_choice = lambda s : random.choice(sum(([v]*wt for v,wt in s),[]))

这有用吗?

# define choices and relative weights
choices = [("WHITE",90), ("RED",8), ("GREEN",2)]

# initialize tally dict
tally = dict.fromkeys(choices, 0)

# tally up 1000 weighted choices
for i in xrange(1000):
    tally[weighted_choice(choices)] += 1

print tally.items()

打印:

[('WHITE', 904), ('GREEN', 22), ('RED', 74)]

假设所有权重都是整数。它们的和不一定是100,我这么做只是为了让测试结果更容易理解。(如果权重是浮点数,则将它们都乘以10,直到所有权重>= 1。)

weights = [.6, .2, .001, .199]
while any(w < 1.0 for w in weights):
    weights = [w*10 for w in weights]
weights = map(int, weights)