我是R的新手，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和H1~H12的列表。像下面的东西......
[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"
我尝试使用rep ...或者创建一个LETTERS [1：8]的矢量和一个单独的c（1:12）矢量并尝试将它们组合在一起......但我并没有非常成功。 / p>
提前致谢！
附加问题有些相关......
所以在我做完之后，我想把它与另一个列表进行比较。另一个列表可能如下所示：
[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"
目前还不是很清楚但是这个清单缺少“A4”。 通过将这个与我创建的具有所有96个元素的那个进行比较，我想知道缺少哪个元素。我尝试使用像intersect和setdiffer这样的函数..但是他们不会逐个元素地比较列表。
答案 0 :(得分：3)
> 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"
答案 1 :(得分：3)
这是另一种选择。关键是each的{{1}}参数，并将其用于其中一个元素而不是另一个元素：
rep