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

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.

其他回答

使用numpy

def choice(items, weights):
    return items[np.argmin((np.cumsum(weights) / sum(weights)) < np.random.rand())]
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"

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())]

通解:

import random
def weighted_choice(choices, weights):
    total = sum(weights)
    treshold = random.uniform(0, total)
    for k, weight in enumerate(weights):
        total -= weight
        if total < treshold:
            return choices[k]

如果你没有提前定义你想要选择多少项(所以,你没有做k=10这样的事情),你只有概率,你可以做下面的事情。注意,你的概率加起来不需要等于1,它们可以相互独立:

soup_items = ['pepper', 'onion', 'tomato', 'celery'] 
items_probability = [0.2, 0.3, 0.9, 0.1]

selected_items = [item for item,p in zip(soup_items,items_probability) if random.random()<p]
print(selected_items)
>>>['pepper','tomato']