Php 从可变权重随机生成组合
非常重要的编辑:所有Ai都是唯一的 问题 我有一个n唯一对象的列表。每个对象Ai都有一个可变百分比Pi 我想创建一个算法,生成k个对象的新列表B(kPhp 从可变权重随机生成组合,php,algorithm,probability,Php,Algorithm,Probability,非常重要的编辑:所有Ai都是唯一的 问题 我有一个n唯一对象的列表。每个对象Ai都有一个可变百分比Pi 我想创建一个算法,生成k个对象的新列表B(k
$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
];
10000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
)
让我们分析一下。 带有替换项(不是您想要的,但更易于分析) 给定大小为
k
的列表L
,以及元素a_i
,列表中a_i
的概率由值p_i
表示
让我们检查a_i
位于列表中某个索引j
的概率。让我们把概率表示为q_i,j
。请注意,对于列表中的任何索引t
,q_i,j=q_i,t
——因此我们可以简单地说q_i_1=q_i_2=…=q_i_k=q_i
a_i位于列表中任何位置的概率表示为:
1-(1-q_i)^k
但是它也是p_i,所以我们需要解这个方程
1-(1-q_i)^k = pi
1 - (1-q_i)^k -pi = 0
一种方法是
计算每个元素的概率后,检查其是否确实是一个可概率空间(总和为1,所有概率均在[0,1]中)。如果不是-对于给定的概率和k
,就不能这样做
无需更换:这更为棘手,因为现在
q_i,j!=q_i,t
(所选内容不是i.i.d.)。这里的概率计算要复杂得多,我现在不确定如何计算它们,需要在运行时完成,在创建列表的过程中
(删除了一个我几乎可以肯定是有偏见的解决方案)。除非我的数学技能比我认为的要差得多,否则在你的示例中,列表a中的一个元素在列表B中被发现的平均几率应该是10/26=0.38。
如果你降低了任何物体的这个几率,肯定还有其他物体的几率更高。 此外,列表A中的您的遗嘱认证无法计算:他们太低:您无法填写您的列表/您没有足够的元素可供选择 假设以上是正确的(或足够正确),这意味着在你的列表中,你的平均体重必须是随机挑选的平均机会。这反过来意味着列表a中的概率总和不等于100
除非我完全错了,否则…我们必须有
sum\u i p\u i=k
,否则我们不能成功
如上所述,这个问题有点简单,但你可能不喜欢这个答案,因为它“不够随机”
对于A_1
和A_2
,条件概率实际上是1/5
相反,我们应该用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,所以我建议使用一种数值优化算法来找到它们,如。初始化Q_i=P_i
(为什么不?)。使用动态规划,对于当前设置的Q_i
,有可能找到l
元素的结果,A_i
是这些元素之一的概率。(我们只关心l=k
条目,但我们需要其他条目使复发生效。)再多做一点工作,我们就可以得到整个梯度。对不起,这太粗略了
在Python 3中,使用一种似乎总是收敛的非线性求解方法(同时将每个q_i
更新为其略微正确的值并进行规范化):
#/usr/bin/env蟒蛇3
导入集合
进口经营者
随机输入
def约束_样本(qs):
k=四舍五入(总和(qs))
尽管如此:
sample=[i代表i,q在枚举(qs)中,如果为random.random()assert abs(sum(size_dist)-1)对象a出现在B中的概率是Pn。
这很棘手,我相信这不是你想要的。具体来说,如果k=n/2
,至少一半的元素应该具有B_i>=1/2
@amit,我很确定这就是我想要的
$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
];
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
)
1-(1-q_i)^k
1-(1-q_i)^k = pi
1 - (1-q_i)^k -pi = 0
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
{}: probability 2/9
{A_1}: probability 1/9
{A_2}: probability 4/9
{A_1, A_2}: probability 2/9,
#!/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])