Getting the excluded elements for each of the combn(n,k) combinations

Question

Suppose we have generated a matrix A where each column contains one of the combinations of n elements in groups of k. So, its dimensions will be k,choose(n,k). Such a matrix is produced giving the command combn(n,k). What I would like to get is another matrix B with dimensions (n-k),choose(n,k), where each column B[,j] will contain the excluded n-k elements of A[,j].

Here is an example of the way I use tho get table B. Do you think it is a safe method to use? Is there another way?

n <- 5 ; k <- 3
(A <- combn(n,k))
(B <- combn(n,n-k)[,choose(n,k):1])

Another example

x<-c(0,1,0,2,0,1) ; k<- 4
(A <- combn(x,k))
(B <- combn(x,length(x)-k)[,choose(length(x),k):1])

That previous question of mine is part of this problem.
Thank you.

Answer 1

using Musa's idea

B <- apply(A,2,function(z) x[is.na(pmatch(x,z))])

as regards the first example:

B <- apply(A,2,function(z) (1:n)[is.na(pmatch((1:n),z))])

Answer 2

Use the setdiff function:

N <- 5
m <- 2    
A <- combn(N,m)
B <- apply(A,2,function(S) setdiff(1:N,S))

MODIFIED: The above works only when the vectors have unique values. For the second example, we write a replacement for setdiff that can handle duplicate values. We use rle to count the number of occurence of each element in the two sets, subtract the counts, then invert the RLE:

diffdup <- function(x,y){
  rx <- do.call(data.frame,rle(sort(x)))
  ry <- do.call(data.frame,rle(sort(y)))
  m <- merge(rx,ry,by='values',all.x=TRUE)
  m$lengths.y[is.na(m$lengths.y)] <- 0
  rz <- list(values=m$values,lengths=m$lengths.x-m$lengths.y)
  inverse.rle(rz)
}

x<-c(0,1,0,2,0,1) ; k<- 4
A <- combn(x,k)
B <- apply(A,2,function(z) diffdup(x,z))

Answer 3

Here a more general solution (you can replace X by any vector containing unique entries):

X<-1:n
B<-apply(A,2,function(x,ref) ref[!ref%in%x],ref=X)
B<-do.call(cbind,B)

Whereas in your previous question x and y were not sets, provided that the columns of A are proper sets, the above code should work.