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

我不喜欢它们的语法。我只想具体说明这些项目是什么以及每项的权重是多少。我意识到我可以用随机。选项,但我很快就写了下面的类。

import random, string
from numpy import cumsum

class randomChoiceWithProportions:
    '''
    Accepts a dictionary of choices as keys and weights as values. Example if you want a unfair dice:


    choiceWeightDic = {"1":0.16666666666666666, "2": 0.16666666666666666, "3": 0.16666666666666666
    , "4": 0.16666666666666666, "5": .06666666666666666, "6": 0.26666666666666666}
    dice = randomChoiceWithProportions(choiceWeightDic)

    samples = []
    for i in range(100000):
        samples.append(dice.sample())

    # Should be close to .26666
    samples.count("6")/len(samples)

    # Should be close to .16666
    samples.count("1")/len(samples)
    '''
    def __init__(self, choiceWeightDic):
        self.choiceWeightDic = choiceWeightDic
        weightSum = sum(self.choiceWeightDic.values())
        assert weightSum == 1, 'Weights sum to ' + str(weightSum) + ', not 1.'
        self.valWeightDict = self._compute_valWeights()

    def _compute_valWeights(self):
        valWeights = list(cumsum(list(self.choiceWeightDic.values())))
        valWeightDict = dict(zip(list(self.choiceWeightDic.keys()), valWeights))
        return valWeightDict

    def sample(self):
        num = random.uniform(0,1)
        for key, val in self.valWeightDict.items():
            if val >= num:
                return key

其他回答

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])

另一种方法是,假设我们的权重与元素数组中的元素的下标相同。

import numpy as np
weights = [0.1, 0.3, 0.5] #weights for the item at index 0,1,2
# sum of weights should be <=1, you can also divide each weight by sum of all weights to standardise it to <=1 constraint.
trials = 1 #number of trials
num_item = 1 #number of items that can be picked in each trial
selected_item_arr = np.random.multinomial(num_item, weights, trials)
# gives number of times an item was selected at a particular index
# this assumes selection with replacement
# one possible output
# selected_item_arr
# array([[0, 0, 1]])
# say if trials = 5, the the possible output could be 
# selected_item_arr
# array([[1, 0, 0],
#   [0, 0, 1],
#   [0, 0, 1],
#   [0, 1, 0],
#   [0, 0, 1]])

现在我们假设,我们要在一次试验中抽取3个项目。你可以假设有三个球R、G、B大量存在,它们的权重由权重数组给定,可能的结果如下:

num_item = 3
trials = 1
selected_item_arr = np.random.multinomial(num_item, weights, trials)
# selected_item_arr can give output like :
# array([[1, 0, 2]])

您还可以将要选择的项目数量视为一组中二项/多项试验的数量。所以,上面的例子仍然可以作为工作

num_binomial_trial = 5
weights = [0.1,0.9] #say an unfair coin weights for H/T
num_experiment_set = 1
selected_item_arr = np.random.multinomial(num_binomial_trial, weights, num_experiment_set)
# possible output
# selected_item_arr
# array([[1, 4]])
# i.e H came 1 time and T came 4 times in 5 binomial trials. And one set contains 5 binomial trails.

假设你有

items = [11, 23, 43, 91] 
probability = [0.2, 0.3, 0.4, 0.1]

你有一个函数,它生成一个介于[0,1)之间的随机数(我们可以在这里使用random.random())。 现在求概率的前缀和

prefix_probability=[0.2,0.5,0.9,1]

现在,我们只需取一个0-1之间的随机数,然后使用二分搜索来查找该数字在prefix_probability中的位置。这个索引就是你的答案

代码是这样的

return items[bisect.bisect(prefix_probability,random.random())]

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。

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]