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。

基于其他解决方案，您可以生成累积分布(作为整数或浮点数)，然后您可以使用平分使其更快

这是一个简单的例子(我在这里使用整数)

l=[(20, 'foo'), (60, 'banana'), (10, 'monkey'), (10, 'monkey2')]
def get_cdf(l):
    ret=[]
    c=0
    for i in l: c+=i[0]; ret.append((c, i[1]))
    return ret

def get_random_item(cdf):
    return cdf[bisect.bisect_left(cdf, (random.randint(0, cdf[-1][0]),))][1]

cdf=get_cdf(l)
for i in range(100): print get_random_item(cdf),

get_cdf函数会将20、60、10、10转换为20、20+60、20+60+10、20+60+10+10

现在我们随机选择一个20+60+10+10的随机数。然后我们用二分法快速得到实际值

你可能想看看NumPy随机抽样分布

从Python 3.6开始，Python的标准库中就有了一个解决方案，即random.choices。

示例用法:让我们建立一个与OP问题中匹配的总体和权重:

>>> from random import choices
>>> population = [1, 2, 3, 4, 5, 6]
>>> weights = [0.1, 0.05, 0.05, 0.2, 0.4, 0.2]

现在choices(population, weights)生成一个样本，包含在一个长度为1的列表中:

>>> choices(population, weights)
[4]

可选的仅关键字参数k允许一次请求多个示例。这很有价值，因为有些准备工作是随机的。在生成样本之前，每次调用choice函数都要做的事情;通过一次生成多个样本，我们只需要做一次准备工作。这里我们生成一百万个样本，并使用集合。计数器来检查我们得到的分布是否与我们给出的权重大致匹配。

>>> million_samples = choices(population, weights, k=10**6)
>>> from collections import Counter
>>> Counter(million_samples)
Counter({5: 399616, 6: 200387, 4: 200117, 1: 99636, 3: 50219, 2: 50025})

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。

我写了一个从自定义连续分布中抽取随机样本的解决方案。

我需要这个类似于你的用例(即生成随机日期与给定的概率分布)。

你只需要函数random_custDist和行samples=random_custDist(x0,x1,custDist=custDist,size=1000)。其余的都是装饰^^。

import numpy as np

#funtion
def random_custDist(x0,x1,custDist,size=None, nControl=10**6):
    #genearte a list of size random samples, obeying the distribution custDist
    #suggests random samples between x0 and x1 and accepts the suggestion with probability custDist(x)
    #custDist noes not need to be normalized. Add this condition to increase performance. 
    #Best performance for max_{x in [x0,x1]} custDist(x) = 1
    samples=[]
    nLoop=0
    while len(samples)<size and nLoop<nControl:
        x=np.random.uniform(low=x0,high=x1)
        prop=custDist(x)
        assert prop>=0 and prop<=1
        if np.random.uniform(low=0,high=1) <=prop:
            samples += [x]
        nLoop+=1
    return samples

#call
x0=2007
x1=2019
def custDist(x):
    if x<2010:
        return .3
    else:
        return (np.exp(x-2008)-1)/(np.exp(2019-2007)-1)
samples=random_custDist(x0,x1,custDist=custDist,size=1000)
print(samples)

#plot
import matplotlib.pyplot as plt
#hist
bins=np.linspace(x0,x1,int(x1-x0+1))
hist=np.histogram(samples, bins )[0]
hist=hist/np.sum(hist)
plt.bar( (bins[:-1]+bins[1:])/2, hist, width=.96, label='sample distribution')
#dist
grid=np.linspace(x0,x1,100)
discCustDist=np.array([custDist(x) for x in grid]) #distrete version
discCustDist*=1/(grid[1]-grid[0])/np.sum(discCustDist)
plt.plot(grid,discCustDist,label='custom distribustion (custDist)', color='C1', linewidth=4)
#decoration
plt.legend(loc=3,bbox_to_anchor=(1,0))
plt.show()

这个解决方案的性能肯定是可以改进的，但我更喜欢可读性。