使用 paste 和 rep 函数指定序列的顺序

Question

我正在尝试对以下代码进行逆向工程

paste("a", rep(1:4, each=4), 1:4, sep="")`

给出以下结果：

[1] "a11" "a12" "a13" "a14" "a21" "a22" "a23" "a24" "a31" "a32" "a33" "a34" "a41" "a42" "a43" "a44" 

作为参考，此代码来自 LTRE{popbio} 帮助文件的示例部分。

对于我的数据，我需要重复以下序列 14 次：“a11 a12 a21 a22”。当我尝试修改原始代码时，

paste("a", rep(1:2, each=14), 1:2, sep="") 

我反而得到了

[1] "a11" "a12" "a11" "a12" "a11" "a12" "a11" "a12" "a11" "a12" "a11" "a12" "a11" "a12" "a21" "a22" "a21" "a22" "a21" "a22" "a21" "a22" "a21" "a22" "a21" "a22" "a21" "a22"`. 

从技术上讲，这些是正确的组合，但我需要序列为“a11、a12、a21、a22”、“a11、a12、a21、a22”等，而不是在切换之前重复“a11 a12”7 次到“a21 a22”7 次。这看起来应该很简单，但是在尝试了各种代码修改之后，我想不通。任何建议将不胜感激。

Answer 1

我们也可以

rep(paste0("a", rep(1:2, each = 2), 1:2), 7)
#[1] "a11" "a12" "a21" "a22" "a11" "a12" "a21" "a22" "a11" "a12" "a21" "a22" "a11" "a12" "a21" "a22" "a11" "a12" "a21" "a22" "a11" "a12"
#[23] "a21" "a22" "a11" "a12" "a21" "a22"

Answer 2

尝试rep 与each 和times 参数一起使用

paste0("a", rep(1:2,times = 7, each = 2), 1:2)

#[1] "a11" "a12" "a21" "a22" "a11" "a12" "a21" "a22" "a11" "a12" "a21" "a22" "a11"
#[14] "a12" "a21" "a22" "a11" "a12" "a21" "a22" "a11" "a12" "a21" "a22" "a11" "a12"
#[27] "a21" "a22"

Answer 3

尝试将paste 与replicate 一起使用：

paste(replicate(14, "a11 a12 a21 a22"), collapse = " ")

[1] "a11 a12 a21 a22 a11 a12 a21 a22 a11 a12 a21 a22 a11 a12 a21 a22 ...