Gabor Grothendieck
2006-Oct-20 20:19 UTC
[Rd] R sequence function (was: Re: [R] Recursive decreasing sequences)
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 > > ______________________________________________ > R-help at stat.math.ethz.ch mailing list > stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. >