如何获得线性下降随机分布？

Question

在django中我随机实现在主页上显示人物.但现在我需要向人们展示他们的重要性(这是数字).例如:Name | Sortkey John | 5汤姆| 8

约翰必须经常线性地展示.

person_number = random.randrange(Person.objects.count())
context = { 'person': Person.objects.all()[person_number], }

我试过random.expovariate,但它让第一个人经常.那么我需要使用随机库中的哪个函数？

UPD:基本上如randrange(10)给出[6,2,8,6,4,1,9,5,3,0,]我需要算法给出一些东西[1,7,3,1,2,2, 3,4,1,5,0] 1 - 最常见(3次),2次(2次)..等等

Answer 1

我猜你的意思是加权随机选择.基本上你需要:

1. 使用累计权重总和创建列表.
2. 选择一个随机值[0,sum_of_weights-1]
3. 找到随机值下降的索引.

在Python而言,具有功能性的方法(ireduce = scanl = partial_sum = ...),它可以看看:

import operator
import bisect
import random

def ireduce(f, state, it):
    """Accumulative reduce. ireduce(operator.sum, 0, [1, 2, 3, 4]) -> 1, 3, 6, 10."""
    for x in it:
        state = f(state, x)
        yield state

people = [
  {"name": "Iain David", "weight": 1}, 
  {"name": "John Martyn", "weight": 2},
]
weights = (person["weight"] for person in people)  
accumulated_weights = list(ireduce(operator.add, 0, weights))
random_value = random.randrange(accumulated_weights[-1])
index = bisect.bisect_right(accumulated_weights, random_value)
random_person = people[index]["name"] 
# "Iain David" (p=1/3), "John Martyn" (p=2/3)