Is this what you want:
> l <- list(a=c(1), b=c(2,3), c=c(4,5,6))
> l
$a
[1] 1
$b
[1] 2 3
$c
[1] 4 5 6
> l$b <- c(l$b, 99)
> l
$a
[1] 1
$b
[1] 2 3 99
$c
[1] 4 5 6
>
Or is you want to dynamically specify the list element:
> x <- 'a'
> l[[x]] <- c(l[[x]], -99)
> l
$a
[1] 1 -99
$b
[1] 2 3 99
$c
[1] 4 5 6
On Sat, Sep 27, 2008 at 2:44 PM, Bodea, Tudor D <gtg757i at
mail.gatech.edu> wrote:> Dear R users:
>
> Is there a way to append selectively to components of a list (if possible,
loops are to be avoided)? To illustrate the point, in the example below, I would
like to append 99 to vector b of the list l.
>
>> l <- list(a=c(1), b=c(2,3), c=c(4,5,6))
>> l
> $a
> [1] 1
>
> $b
> [1] 2 3
>
> $c
> [1] 4 5 6
>
> As you may expect, the result should look like:
>
>> l
> $a
> [1] 1
>
> $b
> [1] 2 3 99
>
> $c
> [1] 4 5 6
>
> I believe that a combination of match and lapply will do it, but,
unfortunately, it seems that I cannot get it right. lapply alone will append
the 99 to all vectors in l. Trying to somehow subset with match gives me the
result expected but in a form independent of the original list (see below). I
think that match should appear inside the lapply function but, for now, I just
cannot make it work.
>
>> lapply(l, function(x, y) {x<-c(x,y)}, y=99)
> $a
> [1] 1 99
>
> $b
> [1] 2 3 99
>
> $c
> [1] 4 5 6 99
>
>> lapply(l[match("b", names(l))], function(x,y) x <- c(x,y),
y=99)
> $b
> [1] 2 3 99
>
> I use R2.7.1 on a Windows machine. Thank you so much.
>
> Tudor
>
> ______________________________________________
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http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?