Fast vectorized merge of list of data.frames by row

Question

Most of the questions about merging data.frame in lists on SO don't quite relate to what I'm trying to get across here, but feel free to prove me wrong.

I have a list of data.frames. I would like to "rbind" rows into another data.frame by row. In essence, all first rows form one data.frame, second rows second data.frame and so on. Result would be a list of the same length as the number of rows in my original data.frame(s). So far, the data.frames are identical in dimensions.

Here's some data to play around with.

sample.list <- list(data.frame(x = sample(1:100, 10), y = sample(1:100, 10), capt = sample(0:1, 10, replace = TRUE)),         data.frame(x = sample(1:100, 10), y = sample(1:100, 10), capt = sample(0:1, 10, replace = TRUE)),         data.frame(x = sample(1:100, 10), y = sample(1:100, 10), capt = sample(0:1, 10, replace = TRUE)),         data.frame(x = sample(1:100, 10), y = sample(1:100, 10), capt = sample(0:1, 10, replace = TRUE)),         data.frame(x = sample(1:100, 10), y = sample(1:100, 10), capt = sample(0:1, 10, replace = TRUE)),         data.frame(x = sample(1:100, 10), y = sample(1:100, 10), capt = sample(0:1, 10, replace = TRUE)),         data.frame(x = sample(1:100, 10), y = sample(1:100, 10), capt = sample(0:1, 10, replace = TRUE))) 

Here's what I've come up with with the good ol' for loop.

#solution 1 my.list <- vector("list", nrow(sample.list[[1]])) for (i in 1:nrow(sample.list[[1]])) {     for (j in 1:length(sample.list)) {         my.list[[i]] <- rbind(my.list[[i]], sample.list[[j]][i, ])     } }  #solution 2 (so far my favorite) sample.list2 <- do.call("rbind", sample.list) my.list2 <- vector("list", nrow(sample.list[[1]]))  for (i in 1:nrow(sample.list[[1]])) {     my.list2[[i]] <- sample.list2[seq(from = i, to = nrow(sample.list2), by = nrow(sample.list[[1]])), ] } 

Can this be improved using vectorization without much brainhurt? Correct answer will contain a snippet of code, of course. "Yes" as an answer doesn't count.

EDIT

#solution 3 (a variant of solution 2 above) ind <- rep(1:nrow(sample.list[[1]]), times = length(sample.list)) my.list3 <- split(x = sample.list2, f = ind) 

BENCHMARKING

I've made my list larger with more rows per data.frame. I've benchmarked the results which are as follows:

#solution 1 system.time(for (i in 1:nrow(sample.list[[1]])) {     for (j in 1:length(sample.list)) {         my.list[[i]] <- rbind(my.list[[i]], sample.list[[j]][i, ])     } })    user  system elapsed   80.989   0.004  81.210   # solution 2 system.time(for (i in 1:nrow(sample.list[[1]])) {     my.list2[[i]] <- sample.list2[seq(from = i, to = nrow(sample.list2), by = nrow(sample.list[[1]])), ] })    user  system elapsed    0.957   0.160   1.126   # solution 3 system.time(split(x = sample.list2, f = ind))    user  system elapsed    1.104   0.204   1.332   # solution Gabor system.time(lapply(1:nr, bind.ith.rows))    user  system elapsed    0.484   0.000   0.485   # solution ncray system.time(alply(do.call("cbind",sample.list), 1,                 .fun=matrix, ncol=ncol(sample.list[[1]]), byrow=TRUE,                 dimnames=list(1:length(sample.list),names(sample.list[[1]]))))    user  system elapsed   11.296   0.016  11.365 

Answer 1

Try this:

bind.ith.rows <- function(i) do.call(rbind, lapply(sample.list, "[", i, TRUE)) nr <- nrow(sample.list[[1]]) lapply(1:nr, bind.ith.rows)