制作r中的字符列表

Question

我是R新手，这是一个非常简单的问题，我似乎无法解决... 所以我想制作一个包含A1〜A12，B1〜 B12，C1〜C12，D1〜D12，E1〜E12，F1〜F12，G1〜G12，H1〜H12。类似下面...

[1] "A1" "A2" "A3" "A4" "A5" "A6" "A7" "A8" "A9" "A10" "A11" "A12" "B1" "B2" "B3" "B4" "B5" "B6" "B7" "B8" 
[21] "B9" "B10" "B11" "B12" "C1" "C2" "C3" "C4" "C5" "C6" "C7" "C8" "C9" "C10" "C11" "C12" "D1" "D2" "D3" "D4" 
[41] "D5" "D6" "D7" "D8" "D9" "D10" "D11" "D12" "E1" "E2" "E3" "E4" "E5" "E6" "E7" "E8" "E9" "E10" "E11" "E12" 
[61] "F1" "F2" "F3" "F4" "F5" "F6" "F7" "F8" "F9" "F10" "F11" "F12" "G1" "G2" "G3" "G4" "G5" "G6" "G7" "G8" 
[81] "G9" "G10" "G11" "G12" "H1" "H2" "H3" "H4" "H5" "H6" "H7" "H8" "H9" "H10" "H11" "H12" 

我试图使用Rep ...或创建LETTERS的矢量[1：8]和c（1:12）的一个单独的载体，并试图它们结合在一起...但我并不是很成功。

提前致谢！

其他问题有点关系...

所以我做这个之后，我想这个比较另一个列表。另一个列表可能看起来像这样：

[1] "A1" "A2" "A3" "A5" "A6" "A7" "A8" "A9" "A10" "A11" "A12" "B1" "B2" "B3" "B4" "B5" "B6" "B7" "B8" "B9" 
[21] "B10" "B11" "B12" "C1" "C2" "C3" "C4" "C5" "C6" "C7" "C8" "C9" "C10" "C11" "C12" "D1" "D2" "D3" "D4" "D5" 
[41] "D6" "D7" "D8" "D9" "D10" "D11" "D12" "E1" "E2" "E3" "E4" "E5" "E6" "E7" "E8" "E9" "E10" "E11" "E12" "F1" 
[61] "F2" "F3" "F4" "F5" "F6" "F7" "F8" "F9" "F10" "F11" "F12" "G1" "G2" "G3" "G4" "G5" "G6" "G7" "G8" "G9" 
[81] "G10" "G11" "G12" "H1" "H2" "H3" "H4" "H5" "H6" "H7" "H8" "H9" "H10" "H11" "H12" 

这不是很清楚，但此列表缺少“A4”。 通过比较这一个和我创建的具有所有96个元素的人之间的关系，我想知道哪个元素丢失了。我尝试过使用像intersect和setdiffer这样的函数，但是它们不会比较列表中的元素。

Answer 1

下面是另一种选择。关键是each参数rep和使用它的内容之一，但不是其他：

paste0(rep(LETTERS[1:8], each=12), rep(1:12, 8)) 

[1] "A1" "A2" "A3" "A4" "A5" "A6" "A7" "A8" "A9" "A10" "A11" "A12" "B1" "B2" 
[15] "B3" "B4" "B5" "B6" "B7" "B8" "B9" "B10" "B11" "B12" "C1" "C2" "C3" "C4" 
[29] "C5" "C6" "C7" "C8" "C9" "C10" "C11" "C12" "D1" "D2" "D3" "D4" "D5" "D6" 
[43] "D7" "D8" "D9" "D10" "D11" "D12" "E1" "E2" "E3" "E4" "E5" "E6" "E7" "E8" 
[57] "E9" "E10" "E11" "E12" "F1" "F2" "F3" "F4" "F5" "F6" "F7" "F8" "F9" "F10" 
[71] "F11" "F12" "G1" "G2" "G3" "G4" "G5" "G6" "G7" "G8" "G9" "G10" "G11" "G12" 
[85] "H1" "H2" "H3" "H4" "H5" "H6" "H7" "H8" "H9" "H10" "H11" "H12" 

Answer 2

> e <- expand.grid(1:12,LETTERS[1:8]) 
> paste0(e[,2],e[,1]) 
[1] "A1" "A2" "A3" "A4" "A5" "A6" "A7" "A8" "A9" "A10" "A11" "A12" 
[13] "B1" "B2" "B3" "B4" "B5" "B6" "B7" "B8" "B9" "B10" "B11" "B12" 
[25] "C1" "C2" "C3" "C4" "C5" "C6" "C7" "C8" "C9" "C10" "C11" "C12" 
[37] "D1" "D2" "D3" "D4" "D5" "D6" "D7" "D8" "D9" "D10" "D11" "D12" 
[49] "E1" "E2" "E3" "E4" "E5" "E6" "E7" "E8" "E9" "E10" "E11" "E12" 
[61] "F1" "F2" "F3" "F4" "F5" "F6" "F7" "F8" "F9" "F10" "F11" "F12" 
[73] "G1" "G2" "G3" "G4" "G5" "G6" "G7" "G8" "G9" "G10" "G11" "G12" 
[85] "H1" "H2" "H3" "H4" "H5" "H6" "H7" "H8" "H9" "H10" "H11" "H12"