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"

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.

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]

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])

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"

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]