This is somewhat subtle.
Rolf's solution (here corrected to...)
a <- list()
for(i in 0:10000) a[[i+1]] <- i
is the best of the loop solutions (or at least the best I know of). The
apparently similar
a <- list(0)
for(i in 1:10000) a <- c(a, list(i))
will take a lot longer, although the result is the same. For example:
> system.time({
a <- list()
for(i in 0:10000) a[[i+1]] <- i
})
user system elapsed
0.59 0.00 0.59
> system.time({
a <- list(0)
for(i in 1:10000) a <- c(a, list(i))
})
user system elapsed
6.87 0.00 6.89
>
That's a factor of about 11 times as long. The best of the lot is
a <- as.list(0:10000)
of course, but this has problems with generalisation, (which everyone
suspects is going to be needed here...).
Bill Venables.
-----Original Message-----
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of Rolf Turner
Sent: Wednesday, 28 May 2008 1:02 PM
To: Daniel Yang
Cc: r-help@r-project.org
Subject: Re: [R] how to bind lists recursively
On 28/05/2008, at 2:43 PM, Daniel Yang wrote:
> Dear all,
>
> I want to create a list that contains 0,1,2,3, ..., 10000 as its
> elements. I used the following code, which apparently doesn't work
> very well.
>
> a <- 0
> for(i in 1:10000) {
> a <- list(a, i)
> }
>
> The result is not what I wanted. So how to create the bind lists
> recursively so that the last element would be the newly added one
> while the previous elements all remain the same?
a <- list()
for(i in 1:10000) a[[i]] <- i
(The word ``bind'' is inappropriate.)
cheers,
Rolf Turner
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