使用for循环在R中创建数据帧列表

Question

I have a list of 26 data frames. I'm using the code below to create a vector. Is there a way I can use the for loop to apply the same unique function on all the 26 data frames and have a new list of 26 data frames with the vector in each data frame? Each data frame would create a different vector depending on that data frame.

(unique(paste(list[[1]]$row, list[[1]]$col, sep=""))

> (unique(paste(list[[1]]$row, list[[1]]$col, sep="")))
 [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"

So the result I want is something like...

[[1]]
[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"

[[2]]
[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"

[[3]] 
[1] "E10" "E11" "E12" "F1"  "F2"  "F3"  "F4"  "F5"  "F6"  "F7"  "F8"  "F9"  "F10" "F11" "F12" "G1"  "G2"  "G3"  "G4"  "G5" 
[21] "G6"  "G7"  "G8"  "G9"  "G10" "G11" "G12" "H1"  "H2"  "H3"  "H4"  "H5"  "H6"  "H7"  "H8"  "H9"  "H10" "H11" "H12"

etc etc until [[26]]

Clarification

Each data frame contains any or all of the elements that the following list contains (the following list has 96 elements):

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

So I need something that would go through my list of 26 data frames and tell me the elements that each data frame contains. Because some of the data frames may not have all 96 elements, I showed that [[3]] in my example of result I want only has only like.. 40 elements. Hope that's more clear now..

Answer 1

something like:

lapply(list, function(x) unique(do.call(paste0, x)))

should do it. do.call calls its first argument (in this case the function paste0 ) with the arguments set to do.call 's second argument. Since x is a data frame, which is a list, this works nicely. Here is the sample output:

[[1]]
[1] "G18" "J20" "N12" "U11" "E1" 

[[2]]
[1] "F10" "E14" "Q18" "I7"  "X13"

[[3]]
[1] "Y8"  "F1"  "P7"  "C15" "Z6" 

[[4]]
[1] "M14" "O16" "L2"  "E13" "S7" 

[[5]]
[1] "V16" "Q1"  "S9"  "M13" "L12"

And the data I used:

set.seed(1)
list <- replicate(5, data.frame(row=sample(LETTERS, 5), col=sample(1:20, 5)), s=F)
lapply(list, function(x) do.call(paste0, x))

Answer 2

I didn't understand the structure of your data, so please pardon my fake data sets.

You can also use transform within the lapply function.

df1 <- data.frame(a = c(1, 2, 3),
                  b = c(4, 5, 6))

df2 <- data.frame(a = c(5, 6, 7),
                  b = c(8, 9, 10))

dfList <- list(df1 = df1, df2 = df2)

lapply(dfList, transform, c = a - b)