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)

A couple of solutions that will make this quicker using data.table

EDIT - with larger dataset showing data.table awesomeness even more.

# here are some sample data sample.list <- replicate(10000, data.frame(x = sample(1:100, 10),   y = sample(1:100, 10), capt = sample(0:1, 10, replace = TRUE)), simplify = F)

Gabor's fast solution:

# Solution Gaborbind.ith.rows <- function(i) do.call(rbind, lapply(sample.list, "[", i, TRUE))nr <- nrow(sample.list[[1]])system.time(rowbound <- lapply(1:nr, bind.ith.rows))##    user  system elapsed ##   25.87    0.01   25.92 

The data.table function rbindlist will make this even quicker even when working with data.frames)

library(data.table)fastbind.ith.rows <- function(i) rbindlist(lapply(sample.list, "[", i, TRUE))system.time(fastbound <- lapply(1:nr, fastbind.ith.rows))##    user  system elapsed ##   13.89    0.00   13.89 

A data.table solution

Here is a solution that uses data.tables - it is split solution on steroids.

# data.table solutionsystem.time({    # change each element of sample.list to a data.table (and data.frame) this    # is done instaneously by reference    invisible(lapply(sample.list, setattr, name = "class",                value = c("data.table",  "data.frame")))    # combine into a big data set    bigdata <- rbindlist(sample.list)    # add a row index column (by refere3nce)    index <- as.character(seq_len(nr))    bigdata[, `:=`(rowid, index)]    # set the key for binary searches    setkey(bigdata, rowid)    # split on this -    dt_list <- lapply(index, function(i, j, x) x[i = J(i)], x = bigdata)    # if you want to drop the `row id` column    invisible(lapply(dt_list, function(x) set(x, j = "rowid", value = NULL)))    # if you really don't want them to be data.tables run this line    # invisible(lapply(dt_list, setattr,name = 'class', value =    # c('data.frame')))})##################################    user  system elapsed    ####    0.08    0.00    0.08    ##################################

How awesome is data.table!

Caveat user with rbindlist

rbindlist is fast because it does not perform the checking that do.call(rbind,....) will. For example it assumes that any factor columns have the same levels as in the first element of the list.

Here's my attempt with plyr, but I like G. Grothendieck's approach:

library(plyr)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]])      ))