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

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)

其他回答

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。

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"

我可能已经来不及提供任何有用的东西了,但这里有一个简单,简短,非常有效的片段:

def choose_index(probabilies):
    cmf = probabilies[0]
    choice = random.random()
    for k in xrange(len(probabilies)):
        if choice <= cmf:
            return k
        else:
            cmf += probabilies[k+1]

不需要排序你的概率或用你的cmf创建一个向量,它一旦找到它的选择就会终止。内存:O(1),时间:O(N),平均运行时间~ N/2。

如果你有权重,只需添加一行:

def choose_index(weights):
    probabilities = weights / sum(weights)
    cmf = probabilies[0]
    choice = random.random()
    for k in xrange(len(probabilies)):
        if choice <= cmf:
            return k
        else:
            cmf += probabilies[k+1]

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])

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。