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 0 和 5 个 1。 （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，哪些值应该为零，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)

我可以根据这个分布选择值，并使用 rbinom 得到一个长度为 31 的向量，但我不能准确地选择五个值。

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。失控的while() 的问题之一是，如果您一次进行一次试验，并且需要多次试验，比如 t，才能找到与 @ 的目标数量相匹配的试验。 987654323@s，您会产生 t 调用 main 函数的开销，在这种情况下为 rbinom()。

但是有一条出路，因为rbinom()，就像 R 中的所有这些（伪）随机数生成器一样，是矢量化的，我们可以一次生成 m 个试验并检查那些m 试验以符合 5 1s 的要求。如果没有找到，我们会反复进行 m 次试验，直到找到符合要求的试验。这个想法在下面的函数foo() 中实现。 chunkSize 参数是 m，即一次要绘制的试验次数。我还借此机会允许该函数查找多个保形试验；参数n 控制要返回多少保形试验。

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
}

它是这样工作的：

> 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，我在n == 1 的情况下删除了空维度。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 的大小和一次检查这么多试验的计算负担。如果要求（这里是 5 个1s）不太可能，那么增加chunkSize 以便减少对rbinom() 的调用。如果可能需要，那么如果您只想要一两个，则一次很少有点画试验和大chunkSize，因为您必须评估每个试画。

Answer 3

我认为您希望使用给定的一组概率从二项分布中重新采样，直到您达到目标值 5，对吗？如果是这样，那么我认为这可以满足您的要求。 while 循环可用于迭代直到满足条件。如果你输入非常不切实际的概率和目标值，我猜它可能会变成一个失控的函数，所以请考虑一下自己被警告:)

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

  while (sum(out) != target) {

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

FOO(概率，目标 = 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