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样本in for循环R.

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我想在没有替换MU，MG，PU，PG 70次的情况下进行采样以创建矩阵（ncol = 4，nrow = 70），例如，

sample(c("MU","MG","PU","PG"), 4,F)
sample(c("MU","MG","PU","PG"), 4,F)
sample(c("MU","MG","PU","PG"), 4,F)
sample(c("MU","MG","PU","PG"), 4,F)
sample(c("MU","MG","PU","PG"), 4,F)
#etc

到目前为止，我有：矩阵（样本（c（“MU”，“MG”，“PU”，“PG”），70 * 4，F），nrow = 70，byrow = TRUE）这是不对的，因为行可以不仅仅具有MU，MG，PU，PG中的每一个 . 我可以使用for循环或更简单的方法吗？

r sample rep

3 回答

  • 4

    replicate 函数可能是您只进行10次重复而不会溢出屏幕的函数 .

    > sample(c("MU","MG","PU","PG"), 4,F)
[1] "MG" "MU" "PU" "PG"
> replicate(10, sample(c("MU","MG","PU","PG"), 4,F))
     [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] "MG" "PG" "MU" "MU" "PU" "MG" "PU" "MG" "MG" "MU" 
[2,] "PU" "MG" "PU" "PU" "MU" "MU" "MU" "PG" "MU" "PG" 
[3,] "MU" "PU" "MG" "PG" "MG" "PG" "MG" "MU" "PG" "MG" 
[4,] "PG" "MU" "PG" "MG" "PG" "PU" "PG" "PU" "PU" "PU" 
> # Output is transposed from how we would want it so we'll just transpose it back
> t(replicate(10, sample(c("MU","MG","PU","PG"), 4,F)))
      [,1] [,2] [,3] [,4]
 [1,] "MG" "PU" "PG" "MU"
 [2,] "PU" "MG" "MU" "PG"
 [3,] "MG" "MU" "PU" "PG"
 [4,] "PU" "PG" "MU" "MG"
 [5,] "MG" "PU" "PG" "MU"
 [6,] "PU" "PG" "MU" "MG"
 [7,] "PG" "MG" "PU" "MU"
 [8,] "MU" "PU" "MG" "PG"
 [9,] "PG" "MU" "PU" "MG"
[10,] "MU" "MG" "PG" "PU"
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  • 1

    一个快速的解决方案是循环：

    mat <- matrix(NA_character_, nrow = 70, ncol = 4)
for (i in 1:70) {
  mat[i, ] <- sample(c("MU","MG","PU","PG"), 4, replace = FALSE)
}

    对于那些对环过敏的人：

    t(sapply(1:70, function(x) sample(c("MU","MG","PU","PG"), 4, replace = FALSE)))
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  • 2

    您也可以尝试这样做：

    lyst <- lapply( 1:70, function(x) {set.seed(x); sample(c("PU", "MU", "MG", "PG"), 4, replace=F)})
do.call("rbind", lyst)

    Sample of some rows as an output ：

    #[,1] [,2] [,3] [,4]
 #[1,] "MU" "PG" "MG" "PU"
 #[2,] "PU" "MG" "MU" "PG"
 #[3,] "PU" "MG" "PG" "MU"
 #[4,] "MG" "PU" "PG" "MU"
 #[5,] "PU" "MG" "MU" "PG"
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