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.

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])

import numpy as np
w=np.array([ 0.4,  0.8,  1.6,  0.8,  0.4])
np.random.choice(w, p=w/sum(w))

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]

粗糙的，但可能足够:

import random
weighted_choice = lambda s : random.choice(sum(([v]*wt for v,wt in s),[]))

这有用吗?

# define choices and relative weights
choices = [("WHITE",90), ("RED",8), ("GREEN",2)]

# initialize tally dict
tally = dict.fromkeys(choices, 0)

# tally up 1000 weighted choices
for i in xrange(1000):
    tally[weighted_choice(choices)] += 1

print tally.items()

打印:

[('WHITE', 904), ('GREEN', 22), ('RED', 74)]

假设所有权重都是整数。它们的和不一定是100，我这么做只是为了让测试结果更容易理解。(如果权重是浮点数，则将它们都乘以10，直到所有权重>= 1。)

weights = [.6, .2, .001, .199]
while any(w < 1.0 for w in weights):
    weights = [w*10 for w in weights]
weights = map(int, weights)

我需要做这样的事情非常快速非常简单，从搜索的想法，我终于建立了这个模板。其思想是以json的形式从api接收加权值，这里是由dict模拟的。

然后将其转换为一个列表，其中每个值都与它的权重成比例地重复，只需使用random。选择从列表中选择一个值。

我尝试了10次、100次和1000次迭代。分布似乎很稳定。

def weighted_choice(weighted_dict):
    """Input example: dict(apples=60, oranges=30, pineapples=10)"""
    weight_list = []
    for key in weighted_dict.keys():
        weight_list += [key] * weighted_dict[key]
    return random.choice(weight_list)