R devel - Jun 2008 - permsn incorrect when x==m (library: prob) (PR#11571)

Full_Name: Obnoxious
Version: 2.7.0
OS: Windows
Submission from: (NULL) (121.223.77.238)


Objective: 
Generate all permutations of the elements of x taken m at a time.

Library: 
prob

Function: 
permsn

Issue:
Does not appear to be working correctly when x==m.

Example:
libary(prob)
permsn(2,1) i.e., x > m
#Gives the correct result of
     [,1] [,2]
[1,]    1    2

#Yet
permsn(2,2) i.e., x==m
#Gives the incorrect? result of
[1] 1 2

#Shouldn't the result be
     [,1] [,2]???
[1]    1    2???
[2]    2    1???

#Workaround:
permsn(3,2) i.e. permsn(x+1,m)
     [,1] [,2] [,3] [,4] [,5] [,6]
[1,]    1    2    1    3    2    3
[2,]    2    1    3    1    3    2

#but then we have to remove the results with x==3 in them

#hope that all makes sense

Thankyou,

Obby

Dear Obby,

Thanks. In the future, when you have a question about something that
happens in a particular package, keep in mind that it is customary to
contact the package maintainer (me), rather than file a bug report
with the R developers.

In that case, I would have alerted you that this problem has already
been found and a fix was posted:  please see

http://tolstoy.newcastle.edu.au/R/e4/help/08/04/8178.html

This will be included in the next release of prob, on which I am
currently working.  I expect it to be ready by August.

Best,
Jay





On Wed, Jun 4, 2008 at 6:50 AM,  <obnoxious_one at hotmail.com>
wrote:
> Full_Name: Obnoxious
> Version: 2.7.0
> OS: Windows
> Submission from: (NULL) (121.223.77.238)
>
>
> Objective:
> Generate all permutations of the elements of x taken m at a time.
>
> Library:
> prob
>
> Function:
> permsn
>
> Issue:
> Does not appear to be working correctly when x==m.
>
> Example:
> libary(prob)
> permsn(2,1) i.e., x > m
> #Gives the correct result of
>     [,1] [,2]
> [1,]    1    2
>
> #Yet
> permsn(2,2) i.e., x==m
> #Gives the incorrect? result of
> [1] 1 2
>
> #Shouldn't the result be
>     [,1] [,2]???
> [1]    1    2???
> [2]    2    1???
>
> #Workaround:
> permsn(3,2) i.e. permsn(x+1,m)
>     [,1] [,2] [,3] [,4] [,5] [,6]
> [1,]    1    2    1    3    2    3
> [2,]    2    1    3    1    3    2
>
> #but then we have to remove the results with x==3 in them
>
> #hope that all makes sense
>
> Thankyou,
>
> Obby
>
> ______________________________________________
> R-devel at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-devel
>


-- 


***************************************************
G. Jay Kerns, Ph.D.
Assistant Professor / Statistics Coordinator
Department of Mathematics & Statistics
Youngstown State University
Youngstown, OH 44555-0002 USA
Office: 1035 Cushwa Hall
Phone: (330) 941-3310 Office (voice mail)
-3302 Department
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E-mail: gkerns at ysu.edu
http://www.cc.ysu.edu/~gjkerns/