Python加权随机

Question

我需要根据加权循环法返回不同的值,以便20中的1得到A,20中的1得到B,其余的转到C.

所以:

A => 5%
B => 5%
C => 90%

这是一个似乎有用的基本版本:

import random

x = random.randint(1, 100)

if x <= 5:
    return 'A'
elif x > 5 and x <= 10:
    return 'B'
else:
    return 'C'

这个算法是否正确？如果是这样,可以改进吗？

Answer 1

你的算法是正确的,如何更优雅:

import random
my_list = ['A'] * 5 + ['B'] * 5 + ['C'] * 90
random.choice(my_list)

Answer 2

没关系.更一般地说,您可以定义如下内容:

from collections import Counter
from random import randint

def weighted_random(pairs):
    total = sum(pair[0] for pair in pairs)
    r = randint(1, total)
    for (weight, value) in pairs:
        r -= weight
        if r <= 0: return value

results = Counter(weighted_random([(1,'a'),(1,'b'),(18,'c')])
                  for _ in range(20000))
print(results)

这使

Counter({'c': 17954, 'b': 1039, 'a': 1007})

如你所料,它接近18:1:1.

Answer 3

如果你想使用加权随机而不是百分位随机,你可以创建自己的Randomizer类:

import random

class WeightedRandomizer:
    def __init__ (self, weights):
        self.__max = .0
        self.__weights = []
        for value, weight in weights.items ():
            self.__max += weight
            self.__weights.append ( (self.__max, value) )

    def random (self):
        r = random.random () * self.__max
        for ceil, value in self.__weights:
            if ceil > r: return value

w = {'A': 1.0, 'B': 1.0, 'C': 18.0}
#or w = {'A': 5, 'B': 5, 'C': 90}
#or w = {'A': 1.0/18, 'B': 1.0/18, 'C': 1.0}
#or or or

wr = WeightedRandomizer (w)

results = {'A': 0, 'B': 0, 'C': 0}
for i in range (10000):
    results [wr.random () ] += 1

print ('After 10000 rounds the distribution is:')
print (results)