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Split list every n elements and cbind, then rbind slices

I would like to slice every n elements of list, cbind the slice, and then rbind the slices.

I can do this with the code below (n = 10 elements, list is 30 elements long). I 'manually' select every 10 elements of the list and then cbind these 10 element slices. I then rbind those cbinded slices.

However, I think that there could be a way to this with l*ply in plyr or dplyr, or at least some of it. For starters, I do not now how to select every n elements of list, and don't seem to know the appropriate search term to find this answer.

dl <- list(c(2L, 1L, 3L, 2L, 1L, 1L, 3L), c(1L, 1L, 2L, 1L, 1L, 2L, 
1L), c(1L, 1L, 2L, 2L, 3L, 3L, 3L), c(1L, 1L, 2L, 2L, 3L, 3L, 
3L), c(1L, 1L, 2L, 2L, 3L, 3L, 3L), c(1L, 1L, 2L, 2L, 3L, 3L, 
1L), c(1L, 1L, 2L, 2L, 3L, 3L, 3L), c(1L, 3L, 2L, 1L, 3L, 2L, 
1L), c(3L, 1L, 2L, 3L, 3L, 1L, 3L), c(3L, 2L, 1L, 1L, 3L, 3L, 
1L), c(1L, 1L, 2L, 2L, 2L, NA, NA), c(1L, 1L, 2L, 2L, 3L, NA, 
NA), c(1L, 1L, 2L, 2L, 3L, NA, NA), c(1L, 1L, 2L, 2L, 3L, NA, 
NA), c(1L, 1L, 2L, 2L, 3L, NA, NA), c(1L, 1L, 2L, 2L, 3L, NA, 
NA), c(1L, 1L, 2L, 2L, 3L, NA, NA), c(1L, 1L, 2L, 2L, 3L, NA, 
NA), c(1L, 1L, 2L, 2L, 3L, NA, NA), c(2L, 1L, 2L, 1L, 1L, NA, 
NA), c(2L, 3L, 1L, 2L, 1L, 2L, NA), c(1L, 1L, 2L, 2L, 1L, 3L, 
NA), c(1L, 1L, 2L, 2L, 3L, 3L, NA), c(1L, 1L, 2L, 2L, 3L, 3L, 
NA), c(1L, 1L, 2L, 2L, 3L, 3L, NA), c(1L, 1L, 2L, 2L, 3L, 3L, 
NA), c(1L, 1L, 2L, 2L, 3L, 3L, NA), c(1L, 1L, 2L, 2L, 3L, 3L, 
NA), c(1L, 1L, 2L, 2L, 3L, 3L, NA), c(1L, 1L, 2L, 2L, 3L, 3L, 
NA)) 

# slice list 'manually' cbind those slices
dl1 <- dl[1:10]
dl1.c <- do.call("cbind", dl1)
dl2 <- dl[11:20]
dl2.c <- do.call("cbind", dl2)
dl3 <- dl[21:30]
dl3.c <- do.call("cbind", dl3)

# rbind the cbind slices for result
ans <- as.data.frame(rbind(dl1.c, dl2.c, dl3.c)) # ans as df
# ans <- rbind(dl1.c, dl2.c, dl3.c)
like image 304
nofunsally Avatar asked Jul 27 '15 15:07

nofunsally


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1 Answers

Try

do.call(mapply, c(cbind, split(dl, cut(seq_along(dl), length(dl)/10, labels = FALSE))))
like image 78
Steven Beaupré Avatar answered Sep 20 '22 10:09

Steven Beaupré