2

我有以下类型的数据

ntrt = paste ("EL", 1:4, sep= "")
repl = 3

我想从 ntrt 中采样，但是 3 次（rep = 3），输出如下所示：

nsam <- c(sample(ntrt),sample(ntrt), sample(ntrt)) 
repl <- rep (1:3, each = length (ntrt))
newd <- data.frame (nsam, repl)
newd 
 nsam repl
1   EL3    1
2   EL1    1
3   EL4    1
4   EL2    1

5   EL2    2
6   EL4    2
7   EL1    2
8   EL3    2

9   EL1    3
10  EL3    3
11  EL4    3
12  EL2    3

这是我的循环试验：

nsam <- rep (NULL, ntrt)
for (i in 1:rep){
         nsam[i] <- sample(ntrt)
         }

编辑：只是澄清

ntrt
[1] "EL1" "EL2" "EL3" "EL4"

> sample(ntrt,4)
[1] "EL4" "EL3" "EL2" "EL1" 

# is equal to:
sample(ntrt)

but what I need:
c(sample(ntrt), sample(ntrt), sample(ntrt))

[1] "EL4" "EL3" "EL1" "EL2" "EL1" "EL2" "EL3" "EL4" "EL3" "EL1" "EL2" "EL4"

which is equal to 
c(sample(ntrt,4), sample(ntrt,4), sample(ntrt,4))

在此处输入图像描述

因此过程是：

# for repl = 1
sample from ntrt  size = length(ntrt)

# repeat the same process again
# for repl = 2
sample from ntrt  size = length(ntrt)
# note: I only concerned with order of names within ntrt (randomization process)

# repeat the same process again
# for repl = 3
sample from ntrt  size = length(ntrt)

same process for n levels of repl

我收到错误，对不起一个简单的问题

4

2 回答 2

4

尝试使用lapply：

set.seed(1) # Just so you can compare -- remove for your actual purposes
ntrt = paste ("EL", 1:4, sep= "")
repl <- rep (1:3, each = length (ntrt))
nsam = unlist(lapply(1:3, FUN=function(i) sample(ntrt)))
newd <- data.frame (nsam, repl)
# > newd
# nsam repl
# 1   EL2    1
# 2   EL4    1
# 3   EL3    1
# 4   EL1    1
# 5   EL1    2
# 6   EL3    2
# 7   EL2    2
# 8   EL4    2
# 9   EL3    3
# 10  EL1    3
# 11  EL4    3
# 12  EL2    3

更新

我注意到@joran 已经replicate在评论中推荐了。这是复制方法：

data.frame(nsam = as.vector(replicate(3, sample(ntrt))), repl)
于 2012-06-23T17:15:23.380 回答
1

看来您需要学习使用 R 中的帮助工具。该sample函数应该执行您所要求的操作，即您只需提交第二个参数：

?sample      # to get the help page
nsam <- sample(ntrt, 3)
nsam
#[1] "EL2" "EL1" "EL4"

重复排列 3 次：

replicate(3, sample(ntrt, length(ntrt)))
于 2012-06-23T13:36:01.623 回答