Splitting a dataframe by columns

Question

I want to split my dataframe by columns. Sounds trivial, but i didnt really succeed so far. Here is what i have come up with:

SG <- data.frame(num = 1:26, let = letters, LET = LETTERS)
SG <- lapply(SG, function(x) split(x, colnames(SG)))
str(SG)

List of 3
 $ num:List of 3
 $ let:List of 3
 $ LET:List of 3

I have successfully converted my dataframe into a list of lists. But i would like to have a list of dataframes, preserving the rowname info from SG , and each one of them containing one column of the initial dataframe. Is that possible?

Thank you!

Answer 1

This should work, row names are preserved. It returns a list of data frames:

SG <- lapply(SG, data.frame)

str(SG)
List of 3
 $ num:'data.frame':    26 obs. of  1 variable:
  ..$ X..i..: int [1:26] 1 2 3 4 5 6 7 8 9 10 ...
 $ let:'data.frame':    26 obs. of  1 variable:
  ..$ X..i..: Factor w/ 26 levels "a","b","c","d",..: 1 2 3 4 5 6 7 8 9 10 ...
 $ LET:'data.frame':    26 obs. of  1 variable:
  ..$ X..i..: Factor w/ 26 levels "A","B","C","D",..: 1 2 3 4 5 6 7 8 9 10 ...

Answer 2

You can use

lapply(colnames(SG), function(x) SG[,x,drop=F])

which returns an object with the structure

List of 3
 $ :'data.frame':   26 obs. of  1 variable:
  ..$ num: int [1:26] 1 2 3 4 5 6 7 8 9 10 ...
 $ :'data.frame':   26 obs. of  1 variable:
  ..$ let: Factor w/ 26 levels "a","b","c","d",..: 1 2 3 4 5 6 7 8 9 10 ...
 $ :'data.frame':   26 obs. of  1 variable:
  ..$ LET: Factor w/ 26 levels "A","B","C","D",..: 1 2 3 4 5 6 7 8 9 10 ...

In this case we are just subsetting. split() for data.frames is better when you want to separate the rows, not columns, into different groups.