Re: [R] ? extended rep()
On 20-Oct-08 21:17:22, Gabor Grothendieck wrote: > Try this: > with(data.frame(x = 0:1, times = 3:6), rep(x, times)) > > or even shorter: > do.call(rep, data.frame(x = 0:1, times = 3:6)) That is sneaky! data.frame(x = 0:1, times = 3:6) # x times # 1 0 3 # 2 1 4 # 3 0 5 # 4 1 6 (Which is why it won't work with list(x=0:1,times=3:6)) Ted. > On Mon, Oct 20, 2008 at 4:57 PM, Ted Harding > <[EMAIL PROTECTED]> wrote: >> Hi Folks, >> I'm wondering if there's a compact way to achieve the >> following. The "dream" is that, by analogy with >> >> rep(c(0,1),times=c(3,4)) >> # [1] 0 0 0 1 1 1 1 >> >> one could write >> >> rep(c(0,1),times=c(3,4,5,6)) >> >> which would produce >> >> # [1] 0 0 0 1 1 1 1 0 0 0 0 0 1 1 1 1 1 1 >> >> in effect "recycling" x through 'times'. >> >> The objective is to produce a vector of alternating runs of >> 0s and 1s, with the lengths of the runs supplied as a vector. >> Indeed, more generally, something like >> >> rep(c(0,1,2), times=c(1,2,3,2,3,4)) >> # [1] 0 1 1 2 2 2 0 0 1 1 1 2 2 2 2 >> >> Suggestions appreciated! With thanks, >> Ted. >> >> >> E-Mail: (Ted Harding) <[EMAIL PROTECTED]> >> Fax-to-email: +44 (0)870 094 0861 >> Date: 20-Oct-08 Time: 21:57:15 >> -- XFMail -- >> >> __ >> R-help@r-project.org mailing list >> https://stat.ethz.ch/mailman/listinfo/r-help >> PLEASE do read the posting guide >> http://www.R-project.org/posting-guide.html >> and provide commented, minimal, self-contained, reproducible code. >> > > __ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. E-Mail: (Ted Harding) <[EMAIL PROTECTED]> Fax-to-email: +44 (0)870 094 0861 Date: 20-Oct-08 Time: 23:02:00 -- XFMail -- __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] ? extended rep()
On 20-Oct-08 21:19:21, Stefan Evert wrote: > On 20 Oct 2008, at 22:57, (Ted Harding) wrote: >> I'm wondering if there's a compact way to achieve the >> following. The "dream" is that one could write >> >> rep(c(0,1),times=c(3,4,5,6)) >> >> which would produce >> >> # [1] 0 0 0 1 1 1 1 0 0 0 0 0 1 1 1 1 1 1 >> >> in effect "recycling" x through 'times'. > > rep2 <- function (x, times) rep(rep(x, length.out=length(times)), > times) > > rep2(c(0,1),times=c(3,4,5,6)) > [1] 0 0 0 1 1 1 1 0 0 0 0 0 1 1 1 1 1 1 > > Any prizes for shortest solution? ;-) > > Best, > Stefan If ever we are both within reach of 'en øl', then yes. But Gabor came up with a shorter one. I tried to shorten Gabor's but failed. However, all competitors are entitled to a consolation prize! (And that includes me ... ) Ted. E-Mail: (Ted Harding) <[EMAIL PROTECTED]> Fax-to-email: +44 (0)870 094 0861 Date: 20-Oct-08 Time: 22:59:16 -- XFMail -- __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] ? extended rep()
Here is one other solution: x <- 0:1 times <- 3:6 rep(x + 0*times, times) This solution also works if the length of times is not a whole number of lengths of x but in that case it does give a warning which seems reasonable since that is the way recycling in R works elsewhere too. On Mon, Oct 20, 2008 at 5:17 PM, Gabor Grothendieck <[EMAIL PROTECTED]> wrote: > Try this: > > with(data.frame(x = 0:1, times = 3:6), rep(x, times)) > > or even shorter: > > do.call(rep, data.frame(x = 0:1, times = 3:6)) > > > On Mon, Oct 20, 2008 at 4:57 PM, Ted Harding > <[EMAIL PROTECTED]> wrote: >> Hi Folks, >> I'm wondering if there's a compact way to achieve the >> following. The "dream" is that, by analogy with >> >> rep(c(0,1),times=c(3,4)) >> # [1] 0 0 0 1 1 1 1 >> >> one could write >> >> rep(c(0,1),times=c(3,4,5,6)) >> >> which would produce >> >> # [1] 0 0 0 1 1 1 1 0 0 0 0 0 1 1 1 1 1 1 >> >> in effect "recycling" x through 'times'. >> >> The objective is to produce a vector of alternating runs of >> 0s and 1s, with the lengths of the runs supplied as a vector. >> Indeed, more generally, something like >> >> rep(c(0,1,2), times=c(1,2,3,2,3,4)) >> # [1] 0 1 1 2 2 2 0 0 1 1 1 2 2 2 2 >> >> Suggestions appreciated! With thanks, >> Ted. >> >> >> E-Mail: (Ted Harding) <[EMAIL PROTECTED]> >> Fax-to-email: +44 (0)870 094 0861 >> Date: 20-Oct-08 Time: 21:57:15 >> -- XFMail -- >> >> __ >> R-help@r-project.org mailing list >> https://stat.ethz.ch/mailman/listinfo/r-help >> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html >> and provide commented, minimal, self-contained, reproducible code. >> > __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] ? extended rep()
On 20 Oct 2008, at 22:57, (Ted Harding) wrote: I'm wondering if there's a compact way to achieve the following. The "dream" is that one could write rep(c(0,1),times=c(3,4,5,6)) which would produce # [1] 0 0 0 1 1 1 1 0 0 0 0 0 1 1 1 1 1 1 in effect "recycling" x through 'times'. rep2 <- function (x, times) rep(rep(x, length.out=length(times)), times) rep2(c(0,1),times=c(3,4,5,6)) [1] 0 0 0 1 1 1 1 0 0 0 0 0 1 1 1 1 1 1 Any prizes for shortest solution? ;-) Best, Stefan __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] ? extended rep()
Try this: with(data.frame(x = 0:1, times = 3:6), rep(x, times)) or even shorter: do.call(rep, data.frame(x = 0:1, times = 3:6)) On Mon, Oct 20, 2008 at 4:57 PM, Ted Harding <[EMAIL PROTECTED]> wrote: > Hi Folks, > I'm wondering if there's a compact way to achieve the > following. The "dream" is that, by analogy with > > rep(c(0,1),times=c(3,4)) > # [1] 0 0 0 1 1 1 1 > > one could write > > rep(c(0,1),times=c(3,4,5,6)) > > which would produce > > # [1] 0 0 0 1 1 1 1 0 0 0 0 0 1 1 1 1 1 1 > > in effect "recycling" x through 'times'. > > The objective is to produce a vector of alternating runs of > 0s and 1s, with the lengths of the runs supplied as a vector. > Indeed, more generally, something like > > rep(c(0,1,2), times=c(1,2,3,2,3,4)) > # [1] 0 1 1 2 2 2 0 0 1 1 1 2 2 2 2 > > Suggestions appreciated! With thanks, > Ted. > > > E-Mail: (Ted Harding) <[EMAIL PROTECTED]> > Fax-to-email: +44 (0)870 094 0861 > Date: 20-Oct-08 Time: 21:57:15 > -- XFMail -- > > __ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] ? extended rep()
will this do what you want:
> f.rep <- function(x, times){
+ # make sure the 'x' is long enough
+ x <- head(rep(x, length(times)), length(times))
+ rep(x, times)
+ }
>
> f.rep(c(0,1), c(3,4,5,6,7))
[1] 0 0 0 1 1 1 1 0 0 0 0 0 1 1 1 1 1 1 0 0 0 0 0 0 0
>
On Mon, Oct 20, 2008 at 4:57 PM, Ted Harding
<[EMAIL PROTECTED]> wrote:
> Hi Folks,
> I'm wondering if there's a compact way to achieve the
> following. The "dream" is that, by analogy with
>
> rep(c(0,1),times=c(3,4))
> # [1] 0 0 0 1 1 1 1
>
> one could write
>
> rep(c(0,1),times=c(3,4,5,6))
>
> which would produce
>
> # [1] 0 0 0 1 1 1 1 0 0 0 0 0 1 1 1 1 1 1
>
> in effect "recycling" x through 'times'.
>
> The objective is to produce a vector of alternating runs of
> 0s and 1s, with the lengths of the runs supplied as a vector.
> Indeed, more generally, something like
>
> rep(c(0,1,2), times=c(1,2,3,2,3,4))
> # [1] 0 1 1 2 2 2 0 0 1 1 1 2 2 2 2
>
> Suggestions appreciated! With thanks,
> Ted.
>
>
> E-Mail: (Ted Harding) <[EMAIL PROTECTED]>
> Fax-to-email: +44 (0)870 094 0861
> Date: 20-Oct-08 Time: 21:57:15
> -- XFMail --
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide 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?
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] ? extended rep()
Hi Folks, I'm wondering if there's a compact way to achieve the following. The "dream" is that, by analogy with rep(c(0,1),times=c(3,4)) # [1] 0 0 0 1 1 1 1 one could write rep(c(0,1),times=c(3,4,5,6)) which would produce # [1] 0 0 0 1 1 1 1 0 0 0 0 0 1 1 1 1 1 1 in effect "recycling" x through 'times'. The objective is to produce a vector of alternating runs of 0s and 1s, with the lengths of the runs supplied as a vector. Indeed, more generally, something like rep(c(0,1,2), times=c(1,2,3,2,3,4)) # [1] 0 1 1 2 2 2 0 0 1 1 1 2 2 2 2 Suggestions appreciated! With thanks, Ted. E-Mail: (Ted Harding) <[EMAIL PROTECTED]> Fax-to-email: +44 (0)870 094 0861 Date: 20-Oct-08 Time: 21:57:15 -- XFMail -- __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
