php - 从可变权重随机生成组合

Question

非常重要的编辑:所有 A_i 都是独一无二的。

问题

我有一个 A n unique 对象列表。每个对象 A_i 都有一个可变百分比 P_i。

我想创建一个算法，生成 k 个对象的新列表 B (k <ⁿ/₂ 并且在大多数情况下 k 明显小于 ⁿ/₂。例如 n=231，k=21)。列表 B 不应有重复项，并将填充来自列表 A 的对象，但有以下限制:

The probability that an object A_i appears in B is P_i.

我尝试过的

(这些 snipits 在 PHP 中只是为了测试目的)我首先列出了 A

$list = [
    "A" => 2.5, 
    "B" => 2.5, 
    "C" => 2.5, 
    "D" => 2.5, 
    "E" => 2.5, 
    "F" => 2.5, 
    "G" => 2.5, 
    "H" => 2.5, 
    "I" => 5,   
    "J" => 5,   
    "K" => 2.5, 
    "L" => 2.5, 
    "M" => 2.5, 
    "N" => 2.5, 
    "O" => 2.5, 
    "P" => 2.5, 
    "Q" => 2.5, 
    "R" => 2.5, 
    "S" => 2.5, 
    "T" => 2.5, 
    "U" => 5,   
    "V" => 5,   
    "W" => 5,   
    "X" => 5,   
    "Y" => 5,   
    "Z" => 20   
];

起初我尝试了以下两种算法(这些算法只是为了测试目的而用 PHP 编写的):

$result = [];

while (count($result) < 10) {
    $rnd = rand(0,10000000) / 100000;

    $sum = 0;
    foreach ($list as $key => $value) {
        $sum += $value;
        if ($rnd <= $sum) {
            if (in_array($key,$result)) {
                break;
            } else {
                $result[] = $key;
                break;
            }
        }
    }
}

和

$result = [];

while (count($result) < 10) {
    $sum = 0;
    foreach ($list as $key => $value) {
        $sum += $value;
    }

    $rnd = rand(0,$sum * 100000) / 100000;

    $sum = 0;
    foreach ($list as $key => $value) {
        $sum += $value;
        if ($rnd <= $sum) {
            $result[] = $key;
            unset($list[$key]);
            break;
        }
    }
}

这两种算法之间的唯一区别在于，一种算法在遇到重复项时会重试，而另一种算法会在对象表单列表 A 被选中时将其删除。事实证明，这两种算法具有相同的概率输出。

我将第二个算法运行了 100,000 次，并记录了每个字母被选中的次数。以下数组包含根据 100,000 次测试在任何列表 B 中选择一个字母的百分比机会。

[A] => 30.213
[B] => 29.865
[C] => 30.357
[D] => 30.198
[E] => 30.152
[F] => 30.472
[G] => 30.343
[H] => 30.011
[I] => 51.367
[J] => 51.683
[K] => 30.271
[L] => 30.197
[M] => 30.341
[N] => 30.15
[O] => 30.225
[P] => 30.135
[Q] => 30.406
[R] => 30.083
[S] => 30.251
[T] => 30.369
[U] => 51.671
[V] => 52.098
[W] => 51.772
[X] => 51.739
[Y] => 51.891
[Z] => 93.74

回顾算法时，这是有道理的。该算法错误地将原始百分比解释为对象在任何给定位置(而不是任何列表B)被选中的概率百分比。因此，例如，在现实中，Z 在列表 B 中被选中的几率是 93%，但是 Z 被选中用于索引 B_n 的几率 为 20%。这不是我想要的。我希望 Z 在列表 B 中被选中的几率为 20%。

这可能吗？怎么做到的？

编辑 1

我尝试简单地让所有 P_i 的总和 = k，如果所有 P_i 这有效是相等的，但在修改它们的值后，它开始变得越来越错误。

初始概率

$list= [
    "A" => 8.4615,
    "B" => 68.4615,
    "C" => 13.4615,
    "D" => 63.4615,
    "E" => 18.4615,
    "F" => 58.4615,
    "G" => 23.4615,
    "H" => 53.4615,
    "I" => 28.4615,
    "J" => 48.4615,
    "K" => 33.4615,
    "L" => 43.4615,
    "M" => 38.4615,
    "N" => 38.4615,
    "O" => 38.4615,
    "P" => 38.4615,
    "Q" => 38.4615,
    "R" => 38.4615,
    "S" => 38.4615,
    "T" => 38.4615,
    "U" => 38.4615,
    "V" => 38.4615,
    "W" => 38.4615,
    "X" => 38.4615,
    "Y" =>38.4615,
    "Z" => 38.4615
];

10,000 次运行后的结果

Array
(
    [A] => 10.324
    [B] => 59.298
    [C] => 15.902
    [D] => 56.299
    [E] => 21.16
    [F] => 53.621
    [G] => 25.907
    [H] => 50.163
    [I] => 30.932
    [J] => 47.114
    [K] => 35.344
    [L] => 43.175
    [M] => 39.141
    [N] => 39.127
    [O] => 39.346
    [P] => 39.364
    [Q] => 39.501
    [R] => 39.05
    [S] => 39.555
    [T] => 39.239
    [U] => 39.283
    [V] => 39.408
    [W] => 39.317
    [X] => 39.339
    [Y] => 39.569
    [Z] => 39.522
)

Answer 1

我们必须有sum_i P_i = k，否则我们无法成功。

如前所述，问题有点简单，但您可能不喜欢这个答案，理由是它“不够随机”。

Sample a uniform random permutation Perm on the integers [0, n)
Sample X uniformly at random from [0, 1)
For i in Perm
    If X < P_i, then append A_i to B and update X := X + (1 - P_i)
    Else, update X := X - P_i
End

您需要使用定点运算而不是浮点来近似涉及实数的计算。

缺少的条件是分布具有称为“最大熵”的技术属性。像阿米特一样，我想不出一个好的方法来做到这一点。这是一个笨拙的方法。

我解决这个问题的第一个(也是错误的)直觉是将每个 A_i 独立地包含在 B 中，概率为 P_i 并重试直到 B 是正确的长度(不会重试太多次，原因你可以问 math.SE)。问题是条件反射打乱了概率。如果 P_1 = 1/3 和 P_2 = 2/3 和 k = 1，则结果为

{}: probability 2/9
{A_1}: probability 1/9
{A_2}: probability 4/9
{A_1, A_2}: probability 2/9,

A_1 的条件概率实际上是 1/5，A_2 的条件概率是 4/5。

相反，我们应该替换产生适当条件分布的新概率 Q_i。我不知道 Q_i 的封闭形式，所以我建议使用像 gradient descent 这样的数值优化算法来找到它们。 .初始化 Q_i = P_i(为什么不呢？)。使用动态规划，对于 Q_i 的当前设置，有可能找到在给定具有 l 元素的结果的情况下，A_i 的概率是这些元素之一。 (我们只关心 l = k 条目，但我们需要其他条目来使递归工作。)再做一点工作，我们可以获得整个梯度。抱歉，这太粗略了。

在 Python 3 中，使用似乎总是收敛的非线性求解方法(将每个 q_i 同时更新为其边缘正确的值并归一化):

#!/usr/bin/env python3
import collections
import operator
import random


def constrained_sample(qs):
    k = round(sum(qs))
    while True:
        sample = [i for i, q in enumerate(qs) if random.random() < q]
        if len(sample) == k:
            return sample


def size_distribution(qs):
    size_dist = [1]
    for q in qs:
        size_dist.append(0)
        for j in range(len(size_dist) - 1, 0, -1):
            size_dist[j] += size_dist[j - 1] * q
            size_dist[j - 1] *= 1 - q
    assert abs(sum(size_dist) - 1) <= 1e-10
    return size_dist


def size_distribution_without(size_dist, q):
    size_dist = size_dist[:]
    if q >= 0.5:
        for j in range(len(size_dist) - 1, 0, -1):
            size_dist[j] /= q
            size_dist[j - 1] -= size_dist[j] * (1 - q)
        del size_dist[0]
    else:
        for j in range(1, len(size_dist)):
            size_dist[j - 1] /= 1 - q
            size_dist[j] -= size_dist[j - 1] * q
        del size_dist[-1]
    assert abs(sum(size_dist) - 1) <= 1e-10
    return size_dist


def test_size_distribution(qs):
    d = size_distribution(qs)
    for i, q in enumerate(qs):
        d1a = size_distribution_without(d, q)
        d1b = size_distribution(qs[:i] + qs[i + 1 :])
        assert len(d1a) == len(d1b)
        assert max(map(abs, map(operator.sub, d1a, d1b))) <= 1e-10


def normalized(qs, k):
    sum_qs = sum(qs)
    qs = [q * k / sum_qs for q in qs]
    assert abs(sum(qs) / k - 1) <= 1e-10
    return qs


def approximate_qs(ps, reps=100):
    k = round(sum(ps))
    qs = ps[:]
    for j in range(reps):
        size_dist = size_distribution(qs)
        for i, p in enumerate(ps):
            d = size_distribution_without(size_dist, qs[i])
            d.append(0)
            qs[i] = p * d[k] / ((1 - p) * d[k - 1] + p * d[k])
        qs = normalized(qs, k)
    return qs


def test(ps, reps=100000):
    print(ps)
    qs = approximate_qs(ps)
    print(qs)
    counter = collections.Counter()
    for j in range(reps):
        counter.update(constrained_sample(qs))
    test_size_distribution(qs)
    print("p", "Actual", sep="\t")
    for i, p in enumerate(ps):
        print(p, counter[i] / reps, sep="\t")


if __name__ == "__main__":
    test([2 / 3, 1 / 2, 1 / 2, 1 / 3])