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 的列表~E12、F1~F12、G1~G12、H1~H12。像下面这样的东西...
[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"
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我尝试使用rep...或创建一个字母[1:8]的向量和一个单独的c(1:12)向量,并尝试将它们组合在一起...但我并没有很成功。
提前致谢!
附加问题有些相关......
所以在我做了这个之后,我想将其与另一个列表进行比较。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"
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不是很清楚,但是这个列表缺少“A4”。通过将此元素与我创建的包含所有 96 个元素的元素进行比较,我想知道缺少哪个元素。我尝试使用 intersect 和 setdiffer 等函数。但它们不会逐个元素地比较列表。
这是一个替代方案。关键是each参数并将rep其用于其中一个元素,而不是另一个元素:
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"
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