Hi Georg,
This was an interesting challenge to me. Here's what I came up with.
The first option meets your desired result, but could get messy with
deeper nesting. The second is less code, but is not quite what you
want and requires as.data.frame() to give a reasonable result for each
list. Calling either option a "good solution" would be rather
generous. I'm iinterested to see what other people do.
Josh
## Data
list.1 <- list("I"=list("A"=c("a", "b", "c"), "B"=c("d", "e", "f")),
"II"=list("A"=c("g", "h", "i"), "B"=c("j", "k", "l")))
list.2 <- list("I"=list("A"=c("A", "B", "C"), "B"=c("D", "E", "F")),
"II"=list("A"=c("G", "H", "I"), "B"=c("J", "K", "L")))
## Try 1
list.t1 <- list.1
for(i in length(list.1)) {
for(j in length(list.1[[i]])) {
list.t1[[c(i, j)]] <- c(list.1[[c(i, j)]], list.2[[c(i, j)]])
}
}
## Try 2
list.t2 <- as.list(do.call("rbind", lapply(list(list.1, list.2),
as.data.frame, stringsAsFactors = FALSE)))
## Results
list.t1
list.t2
On Tue, Jan 11, 2011 at 7:44 AM, Georg Otto <g...@well.ox.ac.uk> wrote:
> Dear R gurus,
>
>
> first let me apologize for a question that might hve been answered
> before. I was not able to find the solution yet. I want to concatenate
> two lists of lists at their lowest level.
>
> Suppose I have two lists of lists:
>
> list.1 <- list("I"=list("A"=c("a", "b", "c"), "B"=c("d", "e", "f")),
>
> "II"=list("A"=c("g", "h", "i"), "B"=c("j", "k", "l")))
>
>
> list.2 <- list("I"=list("A"=c("A", "B", "C"), "B"=c("D", "E", "F")),
>
> "II"=list("A"=c("G", "H", "I"), "B"=c("J", "K", "L")))
>
>
>
>> list.1
> $I
> $I$A
> [1] "a" "b" "c"
>
> $I$B
> [1] "d" "e" "f"
>
>
> $II
> $II$A
> [1] "g" "h" "i"
>
> $II$B
> [1] "j" "k" "l"
>
>
>> list.2
> $I
> $I$A
> [1] "A" "B" "C"
>
> $I$B
> [1] "D" "E" "F"
>
>
> $II
> $II$A
> [1] "G" "H" "I"
>
> $II$B
> [1] "J" "K" "L"
>
>
> Now I want to concatenate list elements of the lowest levels, so the
> result looks like this:
>
>
> $I
> $I$A
> [1] "a" "b" "c" "A" "B" "C"
>
> $I$B
> [1] "d" "e" "f" "D" "E" "F"
>
>
> $II
> $II$A
> [1] "g" "h" "i" "G" "H" "I"
>
> $II$B
> [1] "j" "k" "l" "J" "K" "L"
>
>
> Has anybody a good solution for that?
>
> Best,
>
> Georg
>
> ______________________________________________
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
--
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
http://www.joshuawiley.com/
______________________________________________
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
