这里有一个更有效的方法:

只需使用'weights'数组调用以下函数(假设索引为相应的项)和no。所需样品的数量。可以很容易地修改此函数以处理有序对。

使用它们各自的概率返回采样/选择(带有替换)的索引(或项):

def resample(weights, n):
    beta = 0

    # Caveat: Assign max weight to max*2 for best results
    max_w = max(weights)*2

    # Pick an item uniformly at random, to start with
    current_item = random.randint(0,n-1)
    result = []

    for i in range(n):
        beta += random.uniform(0,max_w)

        while weights[current_item] < beta:
            beta -= weights[current_item]
            current_item = (current_item + 1) % n   # cyclic
        else:
            result.append(current_item)
    return result

关于while循环中使用的概念的简短说明。 我们从累积beta(均匀随机构造的累积值)中减少当前项的权重，并增加当前索引以找到与beta值匹配的项。

另一个答案，可能更快:)

distribution = [(1, 0.2), (2, 0.3), (3, 0.5)]  
# init distribution  
dlist = []  
sumchance = 0  
for value, chance in distribution:  
    sumchance += chance  
    dlist.append((value, sumchance))  
assert sumchance == 1.0 # not good assert because of float equality  

# get random value  
r = random.random()  
# for small distributions use lineair search  
if len(distribution) < 64: # don't know exact speed limit  
    for value, sumchance in dlist:  
        if r < sumchance:  
            return value  
else:  
    # else (not implemented) binary search algorithm  

你可能想看看NumPy随机抽样分布

根据物品的重量列出一个清单:

items = [1, 2, 3, 4, 5, 6]
probabilities= [0.1, 0.05, 0.05, 0.2, 0.4, 0.2]
# if the list of probs is normalized (sum(probs) == 1), omit this part
prob = sum(probabilities) # find sum of probs, to normalize them
c = (1.0)/prob # a multiplier to make a list of normalized probs
probabilities = map(lambda x: c*x, probabilities)
print probabilities

ml = max(probabilities, key=lambda x: len(str(x)) - str(x).find('.'))
ml = len(str(ml)) - str(ml).find('.') -1
amounts = [ int(x*(10**ml)) for x in probabilities]
itemsList = list()
for i in range(0, len(items)): # iterate through original items
  itemsList += items[i:i+1]*amounts[i]

# choose from itemsList randomly
print itemsList

优化可能是用最大公约数归一化，使目标列表更小。

另外，这可能会很有趣。

也许有点晚了。但是你可以使用numpy.random.choice()，传递p参数:

val = numpy.random.choice(numpy.arange(1, 7), p=[0.1, 0.05, 0.05, 0.2, 0.4, 0.2])

from __future__ import division
import random
from collections import Counter


def num_gen(num_probs):
    # calculate minimum probability to normalize
    min_prob = min(prob for num, prob in num_probs)
    lst = []
    for num, prob in num_probs:
        # keep appending num to lst, proportional to its probability in the distribution
        for _ in range(int(prob/min_prob)):
            lst.append(num)
    # all elems in lst occur proportional to their distribution probablities
    while True:
        # pick a random index from lst
        ind = random.randint(0, len(lst)-1)
        yield lst[ind]

验证:

gen = num_gen([(1, 0.1),
               (2, 0.05),
               (3, 0.05),
               (4, 0.2),
               (5, 0.4),
               (6, 0.2)])
lst = []
times = 10000
for _ in range(times):
    lst.append(next(gen))
# Verify the created distribution:
for item, count in Counter(lst).iteritems():
    print '%d has %f probability' % (item, count/times)

1 has 0.099737 probability
2 has 0.050022 probability
3 has 0.049996 probability 
4 has 0.200154 probability
5 has 0.399791 probability
6 has 0.200300 probability