Jeff,
Thanks! This one is much faster, no doubt!
system.time(directionless_circular_permutations2(12))
user system elapsed
5.331 0.745 6.089
system.time(directionless_circular_permutations(12))
user system elapsed
173.659 0.890 174.946
However, from n = 13, things start becoming slow.
system.time(directionless_circular_permutations2(13))
user system elapsed
60.588 14.933 75.864
Perhaps the loop can be parallelized using doMC or doSNOW, etc. but most likely
it is best to do a .Call or .C or Rcpp as you suggested earlier. This may be a
consequence of the permutations function itself being slow.
Thanks again!
Ranjan
On Fri, 30 Mar 2018 13:49:01 -0700 Jeff Newmiller <jdnew...@dcn.davis.ca.us>
wrote:
> New function below is a bit faster due to more efficent memory handling.
>
> for-loop FTW!
>
> directionless_circular_permutations2 <- function( n ) {
> n1 <- n - 1L
> v <- seq.int( n1 )
> ix <- combinations( n1, 2L )
> jx <- permutations( n-3L, n-3L )
> jxrows <- nrow( jx )
> jxoffsets <- seq.int( jxrows )
> result <- matrix( n, nrow = factorial( n1 )/2L, ncol = n )
> k <- seq( 2L, n - 2L )
> for ( i in seq.int( nrow( ix ) ) ) {
> result[ ( i - 1L ) * jxrows + jxoffsets, k ] <-
> matrix( v[ -ix[ i, ] ][ jx ], nrow = jxrows )
> }
> result[ , 1L ] <- rep( ix[ , 1L ], each = jxrows )
> result[ , n1 ] <- rep( ix[ , 2L ], each = jxrows )
> result
> }
>
>
> On Fri, 30 Mar 2018, Ranjan Maitra wrote:
>
> > Jeff,
> >
> > I wanted to let you know that your function is faster than generating
> > the directional circular permutations and weeding.
> >
> > Here is the time for n = 10. I compared with just doing the
> > permutations, there is no point in proceeding further with the weeding
> > since it is slower at the start itself.
> >
> >
> > system.time(directionless_circular_permutations(10))
> > user system elapsed
> > 1.576 0.000 1.579
> >
> > system.time(permutations(9,9))
> > user system elapsed
> > 3.030 0.000 3.037
> >
> > Repeats keep the values similar. All computations on a 10-core processor
> > @3.1GHz with 20 threads (using only one thread) and running the Fedora
> > 27 with 4.15.9 kernel.
> >
> > I have to say that I was surprised to see the difference at n = 10 itself.
> >
> > Thanks again!
> >
> > Best wishes,
> > Ranjan
> >
> > On Thu, 29 Mar 2018 23:26:12 -0700 Jeff Newmiller
> > <jdnew...@dcn.davis.ca.us> wrote:
> >
> >> I don't know if this is more efficient than enumerating with distinct
> >> directions and weeding... it seems kind of heavyweight to me:
> >>
> >> #######
> >> library(gtools)
> >> directionless_circular_permutations <- function( n ) {
> >> v <- seq.int( n-1 )
> >> ix <- combinations( n-1, 2 )
> >> jx <- permutations( n-3, n-3 )
> >> x <- lapply( seq.int( nrow( ix ) )
> >> , function( i ) {
> >> cbind( ix[ i, 1 ]
> >> , permutations( n-3
> >> , n-3
> >> , v[ -ix[ i, ] ]
> >> )
> >> , ix[ i, 2 ]
> >> )
> >> }
> >> )
> >> cbind( do.call( rbind, x ), n )
> >> }
> >> #######
> >>
> >> For more speed, perhaps use Rcpp with [1]?
> >>
> >> [1] http://howardhinnant.github.io/combinations.html
> >>
> >> On Thu, 29 Mar 2018, Ranjan Maitra wrote:
> >>
> >>> Thanks!
> >>>
> >>> Yes, however, this seems a bit wasteful. Just wondering if there are
> >>> other, more efficient options possible.
> >>>
> >>> Best wishes,
> >>> Ranjan
> >>>
> >>>
> >>> On Thu, 29 Mar 2018 22:20:19 -0400 Boris Steipe
> >>> <boris.ste...@utoronto.ca> wrote:
> >>>
> >>>> If one is equal to the reverse of another, keep only one of the pair.
> >>>>
> >>>> B.
> >>>>
> >>>>
> >>>>
> >>>>> On Mar 29, 2018, at 9:48 PM, Ranjan Maitra <mai...@email.com> wrote:
> >>>>>
> >>>>> Dear friends,
> >>>>>
> >>>>> I would like to get all possible arrangements of n objects listed 1:n
> >>>>> on a circle.
> >>>>>
> >>>>> Now this is easy to do in R. Keep the last spot fixed at n and fill in
> >>>>> the rest using permuations(n-1, n-1) from the gtools package.
> >>>>>
> >>>>> However, what if clockwise or counterclockwise arrangements are the
> >>>>> same? I know that half of the above (n - 1)! arrangements are redundant.
> >>>>>
> >>>>> Is there an easy way to list these (n-1)!/2 arrangements?
> >>>>>
> >>>>> I thought of only listing the first half from a call to permuations(n -
> >>>>> 1, n - 1), but while this holds for n = 4, it does not for n = 5. So, I
> >>>>> am wondering if there is another function or tweak which would easily
> >>>>> do this.
> >>>>>
> >>>>> Many thanks in advance for any help. and best wishes,
> >>>>> Ranjan
> >>>>>
> >>>>> ______________________________________________
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> >>
> >> ---------------------------------------------------------------------------
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> ---------------------------------------------------------------------------
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