【erlang】根据奇数位的概率随机求偶数位

记录一个工作中遇到的小问题，奇数位是概率，偶数位是数值，给一个列表根据概率求数值。列表如:[50,3,30,5,20,10]

比如50%的概率返回3，30%的概率返回5，20%的概率返回10

-module(t).
-export([
    test/0,
    rand_weight/1
]).

test()->
    WeightList = [50,3,30,5,20,10],
    [
        "RandWeight:",rand_weight(WeightList),
        "Rand:1",rand_weight(1,WeightList),"Rand:50",rand_weight(50,WeightList),
        "Rand:51",rand_weight(51,WeightList),"Rand:80",rand_weight(80,WeightList),
        "Rand:81",rand_weight(81,WeightList),"Rand:100",rand_weight(100,WeightList)
    ].

%% 根据奇数位的概率随机求偶数位 列表如:[50,3,30,5,20,10]
rand_weight(WeightList)->
    OddSum = lists:sum([X || X <- WeightList,X rem 2 == 1]),
    RandNum = rand:uniform(OddSum),
    rand_weight(RandNum,WeightList).
%% 根据奇数位的概率随机求偶数位 列表如:[50,3,30,5,20,10]
rand_weight(Rand,[Perhaps,Value|T])->
    if
        Rand =< Perhaps -> Value;
        true -> rand_weight(Rand-Perhaps,T)
    end.

思路：先把奇数和加起来，然后随机1个数Rand在范围 [1,奇数和] ，然后 Rand 依次减去奇数位的概率，减不动了就是最终的范围。