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

下面是使用numpy的另一个版本的weighted_choice。传入weights向量,它将返回一个由0组成的数组,其中包含一个1,表示所选择的bin。该代码默认只进行一次绘制,但您可以传入绘制的数量,并且将返回每个绘制的bin的计数。

如果权重向量的和不等于1,它将被规范化,使之等于1。

import numpy as np

def weighted_choice(weights, n=1):
    if np.sum(weights)!=1:
        weights = weights/np.sum(weights)

    draws = np.random.random_sample(size=n)

    weights = np.cumsum(weights)
    weights = np.insert(weights,0,0.0)

    counts = np.histogram(draws, bins=weights)
    return(counts[0])

其他回答

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)

这取决于你想对分布进行多少次抽样。

假设要对分布进行K次抽样。当n是分布中的项数时,每次使用np.random.choice()的时间复杂度为O(K(n + log(n)))。

在我的例子中,我需要对相同的分布进行多次采样,阶数为10^3其中n阶数为10^6。我使用了下面的代码,它预先计算了累积分布,并在O(log(n))中对其进行采样。总体时间复杂度为O(n+K*log(n))。

import numpy as np

n,k = 10**6,10**3

# Create dummy distribution
a = np.array([i+1 for i in range(n)])
p = np.array([1.0/n]*n)

cfd = p.cumsum()
for _ in range(k):
    x = np.random.uniform()
    idx = cfd.searchsorted(x, side='right')
    sampled_element = a[idx]

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。

如果你有一个加权字典而不是一个列表,你可以这样写

items = { "a": 10, "b": 5, "c": 1 } 
random.choice([k for k in items for dummy in range(items[k])])

注意(k, k范围的虚拟物品(物品[k])]产生这个列表(' a ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' c ', ' b ', ' b ', ' b ', ' b ', ' b ']

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]