Hey,
I don't think this is the most efficient way to do this but here it goes:

#Create a random names vector with some repeated names
names.col <- sample(paste("name",1:4,sep =""),10,replace=T)
print(names.col)
#Now find all pairs that are the same

temp <- sapply(names.col,rep,length(names.col))
boolean.matrix <- temp==t(temp) #The entries for which the names are the
same ar
e True


# The above counts the pairs twice and includes the diagonal elements  so
to remove those
result <- boolean.matrix & upper.tri(boolean.matrix)

result.matrix <- which(result,arr.ind=T) #The result.matrix contains one
row for each pair a
nd two columns holding the indices of the matching pairs.

So you can then loop over the result.matrix rows and do whatever you need
for each of the matching pairs.

cheers,
Achilleas



On Thu, Nov 17, 2011 at 3:40 PM, Kripa R <kripa777@hotmail.com> wrote:

>
> Hello,
> I'm at an impasse in my coding and any help would be greatly appreciated.
> Below is my current code:
>
> nr=dim(fileName)[1] #number of rows
> count=0
>
> for(i in 1:nr) {
> for (j in 1:nr) {
> if(fileName$Individual[i]==fileName$Individual[j] & i != j) #look for a
> pair in the column
> {
> count=count+1
> perm<-sample(fileName$Group) #permutate sample group
> if(count==1)break;
> }
>
> }
> }
>
>
> In the matrix (fileName), first it searches for a pair in column:
> Individual, and once identified it permutes another column (Group).
>
> I would like this process to repeat itself to produce all unique outputs
> for the perm variable. If each unique permutation had its own name or was
> in a matrix that would be very helpful. ie perm1: TCTTCT perm2: CTTTCT
> perm3: TCTTTC...
>
>
> Hopefully that makes sense, thanks in advance group!
>
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