how abount a more complicated one?
outer( 1:5, 1:5, '-')[ outer( 1:5, 1:5, '>')]
[1] 1 2 3 4 1 2 3 1 2 1
On Thu, Sep 17, 2015 at 11:52:27AM -0700, David Winsemius wrote:
> You can add this to the list of options to be tested, although my bet would
> be placed on `sequence(5:1)`:
>
> > Reduce( function(x,y){c( 1:y, x)}, 1:5)
> [1] 1 2 3 4 5 1 2 3 4 1 2 3 1 2 1
>
>
> On Sep 17, 2015, at 11:40 AM, Achim Zeileis wrote:
>
> > On Thu, 17 Sep 2015, Peter Langfelder wrote:
> >
> >> Not sure if this is slicker or easier to follow than your solution,
> >> but it is shorter :)
> >>
> >> do.call(c, lapply(n:1, function(n1) 1:n1))
> >
> > Also not sure about efficiency but somewhat shorter...
> > unlist(lapply(5:1, seq))
> >
> >> Peter
> >>
> >> On Thu, Sep 17, 2015 at 11:19 AM, Dan D <ddalth...@usgs.gov> wrote:
> >>> Can anyone think of a slick way to create an array that looks like c(1:n,
> >>> 1:(n-1), 1:(n-2), ... , 1)?
> >>>
> >>> The following works, but it's inefficient and a little hard to follow:
> >>> n<-5
> >>> junk<-array(1:n,dim=c(n,n))
> >>> junk[((lower.tri(t(junk),diag=T)))[n:1,]]
> >>>
> >>> Any help would be greatly appreciated!
> >>>
> >>> -Dan
> >>>
> >>>
>
> David Winsemius
> Alameda, CA, USA
>
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