Re: [R] getting all circular arrangements without accounting for order

2018-03-30 Thread Ranjan Maitra 
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  
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 
> >  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 
> >>>  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  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
> >
> > __
> > R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see

Re: [R] getting all circular arrangements without accounting for order

2018-03-30 Thread Jeff Newmiller 

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  
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  
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  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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Re: [R] getting all circular arrangements without accounting for order

2018-03-30 Thread Ranjan Maitra 
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  
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  
> > 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  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
> >>>
> >>> __
> >>> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> >>> 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 -- To UNSUBSCRIBE and more, see
> >> 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.
> >>
> >
> >
> > -- 
> > Important Notice: This mailbox is ignored: e-mails are set to be deleted on 
> > receipt. Please respond to the mailing list if appropriate. For those 
> > needing to send personal or professional e-mail, please use appropriate 
> > addresses.
> >
> > __
> > R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> > 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.
> >
> 
> ---
> Jeff NewmillerThe .   .  Go Live...
> DCN:Basics: ##.#.   ##.#.  Live Go...
>Live:   OO#.. Dead: OO#..  Playing
> Research Engineer (Solar/BatteriesO.O#.   #.O#.  with
> /Software/Embedded Controllers)   .OO#.   .OO#.  rocks...1k
> 
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE

Re: [R] getting all circular arrangements without accounting for order

2018-03-30 Thread Berry, Charles 


> On Mar 29, 2018, at 6:48 PM, Ranjan Maitra  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? 
> 

Well half of these arrangements will be of the form 

`k, ... , j, n' 

and half will be of the form 

`j, ...,  k, n'

So fix n in position n, select (k,j), and require that the first position is 
min(k,j) and position n-1 is max(k,j). 

There are choose(n-1,2) choices for {(k,j):k

Re: [R] getting all circular arrangements without accounting for order

2018-03-30 Thread Ranjan Maitra 
Thanks! 

It is not clear to me that the weeding step is straightforward to do 
efficiently. Comparing using rev appears to me to be an operation on the order 
of O(n^3).

I guess one way would be to include everything with all 1's in the first slot 
(taking them out of consideration for future operations) and eliminate 
everything with 1's in the last slot. Then go for the 2's and so on, perhaps 
ending until nothing left. This looks like an O(n) operation, however I do not 
know if I will be missing something.

Many thanks again and best wishes,
Ranjan


On Thu, 29 Mar 2018 23:26:12 -0700 Jeff Newmiller  
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  
> > 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  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
> >>>
> >>> __
> >>> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> >>> 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 -- To UNSUBSCRIBE and more, see
> >> 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.
> >>
> >
> >
> > -- 
> > Important Notice: This mailbox is ignored: e-mails are set to be deleted on 
> > receipt. Please respond to the mailing list if appropriate. For those 
> > needing to send personal or professional e-mail, please use appropriate 
> > addresses.
> >
> > __
> > R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> > 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.
> >
> 
> ---
> Jeff NewmillerThe .   .  Go Live...
> DCN:Basics: ##.#.   ##.#.  Live Go...
>Live:   OO#.. Dead: OO#..  Playing
> Research Engineer (Solar/BatteriesO.O#.   #.O#.  with
> /Software/Embedded Controllers)   .OO#.   .OO#.  rocks...1k
> 
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> 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.
> 


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Re: [R] getting all circular arrangements without accounting for order

2018-03-30 Thread Jeff Newmiller 
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  
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  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

__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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 -- To UNSUBSCRIBE and more, see
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.




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Important Notice: This mailbox is ignored: e-mails are set to be deleted on 
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---
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DCN:Basics: ##.#.   ##.#.  Live Go...
  Live:   OO#.. Dead: OO#..  Playing
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/Software/Embedded Controllers)   .OO#.   .OO#.  rocks...1k

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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] getting all circular arrangements without accounting for order

2018-03-29 Thread Ranjan Maitra 
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  
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  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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> > and provide commented, minimal, self-contained, reproducible code.
> 
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Re: [R] getting all circular arrangements without accounting for order

2018-03-29 Thread Boris Steipe 
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  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
> 
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> 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.

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