2

我想从 MU、MG、PU、PG 进行 70 次不替换的采样，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： matrix(sample(c("MU","MG","PU","PG"), 70*4,F), nrow = 70, byrow = TRUE) 这是不正确的，因为行可能不止有 MU、MG、PU、PG 中的每一个实例。我可以用 for 循环或更简单的方法来做到这一点吗？

4

3 回答 3

4

该replicate功能可能是您正在寻找的。我只会做 10 次复制，以免溢出屏幕。

> 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"
于 2018-05-09T15:30:17.743 回答
3

一个快速的解决方案是一个循环：

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)))
于 2018-05-09T15:30:20.647 回答
1

你也可以试试这个：

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

一些行的样本作为输出：

 #[,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"
于 2018-05-09T15:32:20.460 回答