问题描述
有没有人有任何 Prolog 代码可以从列表中非均匀地选择一个随机元素?
Does anyone have any Prolog code to non-uniformly select a random element from a list?
我想复制 numpy.random.choice 当给定与输入列表中的每个条目相关联的概率时.
I want to replicate the functionality of numpy.random.choice when given the probabilities associated with each entry in the input list.
推荐答案
我在 library(random) 中没有发现任何有用的东西.
I found nothing useful in library(random).
这是我的实现choice(Xs, Ps, Y):
choice([X|_], [P|_], Cumul, Rand, X) :-
Rand < Cumul + P.
choice([_|Xs], [P|Ps], Cumul, Rand, Y) :-
Cumul1 is Cumul + P,
Rand >= Cumul1,
choice(Xs, Ps, Cumul1, Rand, Y).
choice([X], [P], Cumul, Rand, X) :-
Rand < Cumul + P.
choice(Xs, Ps, Y) :- random(R), choice(Xs, Ps, 0, R, Y).
它的工作原理是根据 中给出的概率递归构建 累积概率分布Ps，并检查随机数 R 是否低于该值.
It works by recursively building the cumulative probability distribution from the probabilities given in Ps, and checking if the random number R is below that.
注意:要正确运行，概率 Ps 的总和必须为 1，如果情况并非如此，则不会进行检查以警告您.
Note: to function correctly the probabilities Ps must sum to 1, no check is made to warn you if that is not the case.
示例:
?- choice([1,2,3], [0.1,0.2,0.7], Y).
Y = 3 .
?- choice([1,2,3], [0.1,0.2,0.7], Y).
Y = 1 .
?- choice([1,2,3], [0.1,0.2,0.7], Y).
Y = 2 .
?- choice([1,2,3], [0.1,0.2,0.7], Y).
Y = 3 .
?- choice([1,2,3], [0.1,0.2,0.7], Y).
Y = 2 .
?- choice([1,2,3], [0.1,0.2,0.7], Y).
Y = 3 .
?- choice([1,2,3], [0.1,0.2,0.7], Y).
Y = 3 .
?- choice([1,2,3], [0.1,0.2,0.7], Y).
Y = 3 .
?- choice([1,2,3], [0.1,0.2,0.7], Y).
Y = 3 .
?- choice([1,2,3], [0.1,0.2,0.7], Y).
Y = 3 .
?- choice([1,2,3], [0.1,0.2,0.7], Y).
Y = 3 .
...
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