循环R的采样

Question

我想从MU、MG、PU、PG中取样70次而不替换，以创建一个矩阵（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），这是不正确的，因为行可能有不止一个MU，MG，PU，PG的实例。我可以用for循环或更简单的方法来完成吗？

Answer 1

您也可以尝试以下方法：

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

作为输出的某些行的示例：

 #[,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"

Answer 2

快速解决方案是循环：

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

对于那些对循环过敏的人：

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

Answer 3

replicate函数可能就是您想要的。我只做10次复制，以免屏幕溢出。

> 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"