From ggrothendieck at gmail.com  Fri Oct 20 22:19:45 2006
From: ggrothendieck at gmail.com (Gabor Grothendieck)
Date: Fri, 20 Oct 2006 16:19:45 -0400
Subject: [Rd] R sequence function (was: Re: [R] Recursive decreasing
	sequences)
Message-ID: <971536df0610201319r3caad88ehabcf82ab077fc39a@mail.gmail.com>

This seems a lot faster than using the R sequence function.  Suggest
that sequence be rewritten.

On 10/20/06, Marc Schwartz <MSchwartz at mn.rr.com> wrote:
> On Fri, 2006-10-20 at 12:51 -0700, Julian Burgos wrote:
> > Hello fellow R's,
> >
> > I'm sure there must be an easy way to do this.  But after digging in the
> > documentation and thinking about it for a while I couldn't figure it
> > out.  I need to get a decreasing recursive vector in.  I mean something
> > like this: if starting at 2, and ending at 6, the vector should be
> >
> >  2 3 4 5 6 3 4 5 6 4 5 6 5 6 6
> >
> > An easy way would be to do this
> >
> >     x <- integer(0)
> >     for (i in 5) x <- c(x, i:5)
> >
> > But I need to create really long vectors (where the ending value is in
> > the order of 6500) , and using loops is way to slow.  I'm looking for a
> > vectorized method.  Any help will be welcomed.
>
> How about this:
>
> Range <- c(2:6)
>
> > unlist(sapply(seq(along = Range), function(x) Range[x]:max(Range)))
>  [1] 2 3 4 5 6 3 4 5 6 4 5 6 5 6 6
>
>
> Then:
>
> Range2 <- 2:6500
>
> > str(Range2)
>  int [1:6499] 2 3 4 5 6 7 8 9 10 11 ...
>
>
> system.time(res <- unlist(sapply(seq(along = Range2),
>                                 function(x) Range2[x]:max(Range2))))
> [1] 0.492 0.136 0.647 0.000 0.000
>
>
> > str(res)
>  int [1:21121750] 2 3 4 5 6 7 8 9 10 11 ...
>
>
> HTH,
>
> Marc Schwartz
>
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>




[Rd] R sequence function (was: Re: [R] Recursive decreasing sequences)
Gabor Grothendieck