从可变权重随机生成组合
问题内容：
非常重要的编辑： 所有 _A i_都是 唯一的 。
问题
我有一个列表 一个 的 ñ 独特的 对象。每个对象 A i_具有可变的百分比 _P i。
我想，以形成产生一个新的列表的算法 乙 的 ķ 对象（ ķ < ñ / 2，并且在大多数情况下 ķ 是显著小于 ñ / 2。
例如，N = 231，K = 21 ）。列表 B 应该没有重复项，并且将填充具有以下限制的源自列表 A的 对象：
物体 A i_出现在 _B中 的概率为 P i。
我尝试过的
（这些代码段仅出于测试目的而在PHP中）我首先列出了列表 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
];
最初，我尝试了以下两个算法（这些仅在PHP中用于测试）：
$result = [];
while (count($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;
}
}
}
两种算法之间的唯一区别是，一种算法在遇到重复项时会重试，而另一种则在选择对象列表 A 时将其删除。事实证明，这两种算法具有相同的概率输出。
我运行了第二个算法100,000次，并跟踪了每个字母被选择了多少次。以下数组确定了从100,000个测试中选出任何列表 B 中的字母的机会百分比。
[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
当回顾算法时，这很有意义。该算法错误地将原始百分比解释为针对任何给定位置（而不是任何列表 B ）选择对象的机会百分比。因此，例如，实际上，在列表
B中 选择Z的机会是93％，但是在索引 B n上_选择Z的机会是20％。这不是我想要的。我希望从列表 _B中 选择Z的机会是20％。
这有可能吗？如何做呢？
编辑1
我尝试简单地让所有 P i = k的和，如果所有 _P i_都相等，则可以工作，但是修改它们的值后，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
)
问题答案：
我们必须拥有sum_i P_i = k，否则我们将无法成功。
如前所述，这个问题有些容易，但是您可能不喜欢这个答案，因为它“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
您需要使用定点算术而不是浮点算术来逼近涉及实数的计算。
缺少的条件是该分布具有称为“最大熵”的技术属性。像阿米特（Amit）一样，我想不出一个好方法。这是一种笨拙的方式。
解决这个问题的第一个（也是错误的）本能是将每个事件独立地包含A_i在B概率内，P_i然后重试，直到B正确的长度为止（不会有太多的重试，因为您可以向math.SE询问有关的原因）。问题在于，条件弄乱了概率。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,
条件概率实际上1/5是A_1和4/5的A_2。
相反，我们应该替换Q_i产生适当条件分布的新概率。我不知道封闭形式Q_i，所以我建议用像梯度下降这样的数值优化算法找到它们。初始化Q_i = P_i（为什么不呢？）。使用动态编程，对于当前设置，可以找到Q_i给定包含l元素的结果即A_i那些元素之一的概率。（我们只关心l = k条目，但是我们需要其他人来使递归起作用。）再多做一点，我们就可以得到整个梯度。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（q_i同时将每个更新到其边际正确值并进行归一化）：
#!/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])
