[R] Variable number of loops
Bert Gunter
gunter.berton at gene.com
Sat May 16 15:28:53 CEST 2015
Are you trying to reinvent ?expand.grid ?
-- Bert
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
"Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom."
Clifford Stoll
On Fri, May 15, 2015 at 1:40 PM, WRAY NICHOLAS
<nicholas.wray at ntlworld.com> wrote:
> I am trying to build a programme which will work out the permutations of a
> number of variables, eg a=0 to 1, b=0 to 1, and c=0 to 2, so permutations
> would be (0,0,0), (1,0,0), (0,1,0)... etc In this case there would be 2 x
> 2x 3 = 12 permutations. If the number of variables are fixed it's easy to
> loop round with nesting
>
> However I don't have a fixed number of variables, so I have a variable
> number of loops. I am trying to use a recursive function to do this and
> have been building it up step-wise
>
> I want to return a list of all the permutations at the end, but the
> programme I have so ar doesn't return a full list, but just the first
> element
>
> 2 things -- 1 I don't see why this is happening, and 2 is this the right
> way to approach this problem? I cannot find find anything about this in R
> on the net
>
> recursfunc1<-function(xf,shiftvecf,vlistf){
> if(xf<4){
> xf<-xf+1
> vlistf[[length(vlistf)+1]]<-shiftvec[xf]
> #print(paste(xf,"and",shiftvec[xf]))
> print(vlistf)
> #print(shiftvec[xf])
> xf<-recursfunc1(xf,shiftvecf,vlistf)}
> return(vlistf)}
>
> shiftvec<-c(2,1,1,0)
> vlist<-list()
> perm<-recursfunc1(xf=0,shiftvecf=shiftvec,vlistf=vlist)
> perm
>
> I want perm to return the elements of shiftvec as a list so that I can then
> do all the permutations as the next stage, but it only returns the first
> element of shiftvec
>
> Thanks, Nick Wray
>
> [[alternative HTML version deleted]]
>
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