分配由概率分布通知的特定数量的值（在 R 中）

Question

您好，提前感谢您的帮助！

我正在尝试生成一个向量，该向量具有根据概率分布分配的特定数量的值。例如，我想要一个长度为 31 的向量，包含 26 个零和 5 个一。 （向量的总和应始终为 5。）但是，向量的位置很重要。为了确定哪些值应该为 1，哪些值应该为 0，我有一个概率向量（长度为 31），如下所示：

probs<-c(0.01,0.02,0.01,0.02,0.01,0.01,0.01,0.04,0.01,0.01,0.12,0.01,0.02,0.01,
0.14,0.06,0.01,0.01,0.01,0.01,0.01,0.14,0.01,0.07,0.01,0.01,0.04,0.08,0.01,0.02,0.01)

我可以根据此分布选择值并使用 rbinom 获得长度为 31 的向量，但是我无法恰好选择五个值。

Inv=rbinom(length(probs),1,probs)
Inv
[1] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0

有什么想法吗？

再次感谢！

Answer 1

仅使用加权 sample.int 来选择位置怎么样？

Inv<-integer(31)
Inv[sample.int(31,5,prob=probs)]<-1
Inv
[1] 0 0 0 1 0 1 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0

Answer 2

Chase 提供了一个很好的答案，并提到了 while() 迭代失控的问题。失控 while() 的问题之一是，如果您一次进行一次尝试，则需要多次尝试才能找到一个与 1 的目标数量匹配，您会产生 t 次调用主函数（在本例中为 rbinom()）的开销。

然而，还有一个出路，因为 rbinom() 与 R 中所有这些（伪）随机数生成器一样，是矢量化的，我们可以在时间并检查这些m次试验是否符合51s的要求。如果没有找到，我们会重复进行m次试验，直到找到一个符合要求的试验。这个想法在下面的函数 foo() 中实现。 chunkSize 参数是 m，即一次绘制的试验次数。我还借此机会允许该功能找到多个保形试验；参数n 控制返回多少个保形试验。

foo <- function(probs, target, n = 1, chunkSize = 100) {
    len <- length(probs)
    out <- matrix(ncol = len, nrow = 0) ## return object
    ## draw chunkSize trials
    trial <- matrix(rbinom(len * chunkSize, 1, probs),
                    ncol = len, byrow = TRUE)
    rs <- rowSums(trial)  ## How manys `1`s
    ok <- which(rs == 5L) ## which meet the `target`
    found <- length(ok)   ## how many meet the target
    if(found > 0)         ## if we found some, add them to out
        out <- rbind(out,
                     trial[ok, , drop = FALSE][seq_len(min(n,found)), , 
                                               drop = FALSE])
    ## if we haven't found enough, repeat the whole thing until we do
    while(found < n) {
        trial <- matrix(rbinom(len * chunkSize, 1, probs),
                            ncol = len, byrow = TRUE)
        rs <- rowSums(trial)
        ok <- which(rs == 5L)
        New <- length(ok)
        if(New > 0) {
            found <- found + New
            out <- rbind(out, trial[ok, , drop = FALSE][seq_len(min(n, New)), , 
                                                        drop = FALSE])
        }
    }
    if(n == 1L)           ## comment this, and
        out <- drop(out)  ## this if you don't want dimension dropping
    out
}

它的工作原理如下：

> set.seed(1)
> foo(probs, target = 5)
 [1] 1 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0
[31] 0
> foo(probs, target = 5, n = 2)
     [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11]
[1,]    0    0    0    0    0    0    0    0    0     0     0
[2,]    0    0    0    0    0    0    0    0    0     0     1
     [,12] [,13] [,14] [,15] [,16] [,17] [,18] [,19] [,20] [,21]
[1,]     0     0     0     1     1     0     0     0     0     0
[2,]     0     1     0     0     1     0     0     0     0     0
     [,22] [,23] [,24] [,25] [,26] [,27] [,28] [,29] [,30] [,31]
[1,]     1     0     1     0     0     0     1     0     0     0
[2,]     1     0     1     0     0     0     0     0     0     0

请注意，在 n == 1 的情况下，我删除了空维度。如果您不想要此功能，请将最后一个 if 代码块注释掉。

您需要平衡 chunkSize 的大小与一次检查多个试验的计算负担。如果要求（此处为 5 1s）不太可能实现，则增加 chunkSize 以便减少对 rbinom() 的调用。如果可能有这种要求，那么如果您只需要一两次，则每次抽奖试验的点数很少，而 chunkSize 则较大，因为您必须评估每次试抽。

Answer 3

我认为您想使用一组给定的概率从二项式分布中重新采样，直到达到目标值 5，对吗？如果是这样，那么我认为这就是你想要的。 while 循环可用于迭代，直到满足条件。如果你提供非常不切实际的概率和目标值，我想它可能会变成一个失控函数，所以请考虑自己被警告:)

FOO <- function(probs, target) {
  out <- rbinom(length(probs), 1, probs)

  while (sum(out) != target) {

    out <- rbinom(length(probs), 1, probs)
  }
  return(out)
}

FOO(probs, target = 5)

> FOO(probs, target = 5)  
 [1] 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 1 0 0 1 0 0 0 0 1 0