Another possibility is
mapply(rbind,list1,list2,SIMPLIFY=FALSE)
- Phil Spector
Statistical Computing Facility
Department of Statistics
UC Berkeley
spec...@stat.berkeley.edu
On Sun, 6 Feb 2011, jim holtman wrote:
will this do it for you:
lapply(seq(length(list1)), function(i)rbind(list1[[i]], list2[[i]]))
[[1]]
[,1] [,2] [,3] [,4] [,5]
[1,] 1 6 11 16 21
[2,] 2 7 12 17 22
[3,] 3 8 13 18 23
[4,] 4 9 14 19 24
[5,] 5 10 15 20 25
[6,] 10 11 12 13 14
[[2]]
[,1] [,2] [,3] [,4] [,5]
[1,] 2 7 12 17 22
[2,] 3 8 13 18 23
[3,] 4 9 14 19 24
[4,] 5 10 15 20 25
[5,] 6 11 16 21 26
[6,] 11 12 13 14 15
On Sun, Feb 6, 2011 at 1:32 PM, B. Jonathan B. Jonathan
<bkheijonat...@gmail.com> wrote:
Hi, I am wondering whether we can apply 'cbind/rbind' on many **equivalent**
list objects. For example please consider following:
list1 <- list2 <- vector("list", length=2); names(list1) <- names(list2)
<- c("a", "b")
list1[[1]] <- matrix(1:25, 5)
list1[[2]] <- matrix(2:26, 5)
list2[[1]] <- 10:14
list2[[2]] <- 11:15
list1
$a
[,1] [,2] [,3] [,4] [,5]
[1,] 1 6 11 16 21
[2,] 2 7 12 17 22
[3,] 3 8 13 18 23
[4,] 4 9 14 19 24
[5,] 5 10 15 20 25
$b
[,1] [,2] [,3] [,4] [,5]
[1,] 2 7 12 17 22
[2,] 3 8 13 18 23
[3,] 4 9 14 19 24
[4,] 5 10 15 20 25
[5,] 6 11 16 21 26
list2
$a
[1] 10 11 12 13 14
$b
[1] 11 12 13 14 15
Here I want to "rbind" these 2 list-s according to "a" and "b" i.e. I want
to get another list of length 2 where each element will be matrix with (6x5)
dimension. How can I do that?
Thanks for your time
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--
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
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______________________________________________
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.