[R] binary order combinations

[Dimitri Liakhovitski]

Sorry,

I misread it first. I tried:

library(quantreg)
 test.combos<-lapply(1:15,function(x)
 {combos(15,x)
 })

It gives me a list with an order that is somewhat different from
before, but I am not sure it helps me much:

[[1]]
     [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [,14]
[1,]    1    2    3    4    5    6    7    8    9    10    11    12    13    14
     [,15]
[1,]    15

[[2]]
     [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [,14]
[1,]    1    1    1    1    1    1    1    1    1     1     1     1     1     1
[2,]    2   15   14   13   12   11   10    9    8     7     6     5     4     3
     [,15] [,16] [,17] [,18] [,19] [,20] [,21] [,22] [,23] [,24] [,25] [,26]
[1,]     2     2     2     2     2     2     2     2     2     2     2     2
[2,]     3    15    14    13    12    11    10     9     8     7     6     5


Dimitri

On 9/5/08, roger koenker <rkoenker at uiuc.edu> wrote:
> You need to read the help file ?combos, and then use it to do indexing
> of your objects, it only knows how to construct the integer combinations
> given  a pair (n,p).
>
> url:    www.econ.uiuc.edu/~roger            Roger Koenker
> email    rkoenker at uiuc.edu            Department of Economics
> vox:     217-333-4558                University of Illinois
> fax:       217-244-6678                Champaign, IL 61820
>
>
>
> On Sep 5, 2008, at 10:23 AM, Dimitri Liakhovitski wrote:
>
> > I am not sure it can do it. Besides, I ran a test of combos from quantreg:
> >
> > library(quantreg)
> > H<-1:3
> > test.combos<-lapply(1:3,function(x)
> > {combn(H,x)
> > })
> >
> > Every time I tried it crashed my R...
> >
> > :(
> >
> > Dimitri
> >
> > On 9/5/08, roger koenker <rkoenker at uiuc.edu> wrote:
> >
> > > Does ?combos in the quantreg package do what you want?
> > >
> > >
> > > url:    www.econ.uiuc.edu/~roger            Roger Koenker
> > > email    rkoenker at uiuc.edu            Department of Economics
> > > vox:     217-333-4558                University of Illinois
> > > fax:       217-244-6678                Champaign, IL 61820
> > >
> > >
> > >
> > >
> > > On Sep 5, 2008, at 9:58 AM, Dimitri Liakhovitski wrote:
> > >
> > >
> > > >
> > > > Dear all!
> > > >
> > > > I have a vector of names
> > > > names<-("V1", "V2", "V3",....., "V15")
> > > >
> > > > I could create all possible combinations of these names (of all
> > > > lengths) using R:
> > > >
> > > > combos<-lapply(1:15,function(x)
> > > > {combn(names,x)
> > > > })
> > > >
> > > > I get a list with all possible combinations of elements of 'names'
> > > > that looks like this (just the very beginning of it):
> > > >
> > > > [[1]] - the first element contains all combinations of 1 name
> > > >  [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13]
> > > >
> > > [,14]
> > >
> > > > [1,] "V1" "V2" "V3" "V4" "V5" "V6" "V7" "V8" "V9" "V10" "V11" "V12"
> "V13"
> > > >
> > > "V14"
> > >
> > > >  [,15]
> > > > [1,] "V15"
> > > >
> > > > [[2]] - the second element contains all possible combinations of 2
> names
> > > >  [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]  [,10] [,11] [,12] [,13]
> > > > [1,] "V1" "V1" "V1" "V1" "V1" "V1" "V1" "V1" "V1"  "V1"  "V1"  "V1"
> "V1"
> > > > [2,] "V2" "V3" "V4" "V5" "V6" "V7" "V8" "V9" "V10" "V11" "V12" "V13"
> "V14"
> > > > .
> > > > .
> > > > .
> > > > etc.
> > > >
> > > > My question is: Is there any way to re-arrange all sub-elements of the
> > > > above list (i.e., all possible combinations of names such as V1,
> > > > V1:V3, V1:V2:V4:V5) in a binary system order. More specifically,
> > > > according to this system:
> > > > V1=1
> > > > V2=2
> > > > V3=4
> > > > V4=8
> > > > V5=16, etc....
> > > >
> > > > So, I'd like those combinations to be arranged in a vector in the
> > > > following order:
> > > > 1. V1 (because V1=1)
> > > > 2. V2 (because V2=2)
> > > > 3. V1:V2 (because V1=1 and V2=2 so that 1+2=3)
> > > > 4. V3 (because V3=4)
> > > > 5. V1:V3 (because V1=1 and V3=4 so that 1+4=5)
> > > > 6. V2:V3 (because V2=2 and V3=4 so that 2+4=6)
> > > > 7. V1:V2:V3 (because V1=1 and V2=2 and V3=4 so that 1+2+4=7)
> > > > 8. V4 (because V4=8)
> > > > etc.
> > > >
> > > > Is it at all possible?
> > > > Or maybe there is a way to create the name combinations in such an
> > > > order in the first place?
> > > >
> > > > Thank you very much!
> > > > Dimitri Liakhovitski
> > > >
> > > > ______________________________________________
> > > > R-help at 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 at 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.
> >
>
>