CODEWITHSUNDEEP
stringrlist
hj14
hj14

Reputation: 169

making a list of characters in r

I'm new to R and this is a very simple problem that I can't seem to solve... So I want to make a list which contains A1~A12, B1~B12, C1~C12, D1~D12, E1~E12, F1~F12, G1~G12, and H1~H12. Something like below...

 [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"

I tried using rep... or creating a vector of LETTERS[1:8] and a separate vector of c(1:12) and tried combining them together... but I haven't been very successful.

Thanks in advance!

Additional Question somewhat related...

So after I make this, I want to compare this to another list. The other list may look something like this: 

[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"

It's not very clear but this list is missing "A4". By comparing this one to the one I have created that has all 96 elements, I want to know which element is missing. I tried using functions like intersect and setdiffer.. but they don't compare the lists, element by element.

Upvotes: 0

Views: 1512

Answers (2)

BrodieG
BrodieG

Reputation: 52687

Here is an alternative. The key is the each argument to rep and using it for one of the elements but not the other:

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"

Upvotes: 6

Thomas
Thomas

Reputation: 44565

> 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"

Upvotes: 3

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