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

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。

scipy.stats。Rv_discrete可能是您想要的。您可以通过values参数提供您的概率。然后，您可以使用分布对象的rvs()方法来生成随机数。

正如Eugene Pakhomov在评论中指出的那样，你也可以将p关键字参数传递给numpy.random.choice()，例如:

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

如果你使用的是Python 3.6或更高版本，你可以使用标准库中的random.choices() -请参阅Mark Dickinson的回答。

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  

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})

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。