Unfortunately Reduce is just as slow as the slowest of
the solutions below:
Reduce(c, 0:10000, list())
On Tue, May 27, 2008 at 11:18 PM, <[EMAIL PROTECTED]> wrote:
> 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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