我有以下类型的数据
ntrt = paste ("EL", 1:4, sep= "")
repl = 3
我想从ntrt中抽样,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))
因此,流程为:
# 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,抱歉一个简单的问题
答案 0 :(得分: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
我注意到@joran已在评论中推荐replicate
。这是复制方法:
data.frame(nsam = as.vector(replicate(3, sample(ntrt))), repl)
答案 1 :(得分:1)
看来你需要学习使用R中的帮助工具。sample
函数应该做你看起来要问的只是提交第二个参数:
?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)))