All N Combinations of All Subsets

问题

Given a vector of elements, I would like to obtain a list of all possible n-length combinations of subsets of elements. For example, given the (simplest) sequence 1:2, I would like to obtain a list object of the form

{ {{1},{1}}, {{1},{2}}, {{2},{2}}, {{1},{1,2}}, {{2},{1,2}}, {{1,2},{1,2}} }

when n=2.

I was able to generate a list of all non-empty subsets using the following:

listOfAllSubsets <- function (s) {
  n <- length(s)
  unlist(lapply(1:n, function (n) {
    combn(s, n, simplify=FALSE)
  }), recursive=FALSE)
}

However, I'm not sure the best way to proceed from here. Essentially, I want a Cartesian product of this list with itself (for n=2).

Any suggestions? A non-iterative solution would be preferable (i.e., no for loops).

回答1:

This is what I would do, with, e.g., s=1:2:

1) Represent subsets with a 0/1 matrix for each element's membership.

subsets = as.matrix(do.call(expand.grid,replicate(length(s),0:1,simplify=FALSE)))

which gives

     Var1 Var2
[1,]    0    0
[2,]    1    0
[3,]    0    1
[4,]    1    1

Here, the first row is the empty subset; the second, {1}; the third, {2}; and the fourth, {1,2}. To get the subset itself, use mysubset = s[subsets[row,]], where row is the row of the subset you want.

2) Represent pairs of subsets as pairs of rows of the matrix:

pairs <- expand.grid(Row1=1:nrow(subsets),Row2=1:nrow(subsets))

which gives

   Row1 Row2
1     1    1
2     2    1
3     3    1
4     4    1
5     1    2
6     2    2
7     3    2
8     4    2
9     1    3
10    2    3
11    3    3
12    4    3
13    1    4
14    2    4
15    3    4
16    4    4

Here, the fourteenth row corresponds to the second and fourth rows of subsets, so {1} & {1,2}. This assumes the order of the pair matters (which is implicit in taking the Cartesian product). To recover the subsets, use mypairosubsets=lapply(pairs[p,],function(r) s[subsets[r,]]) where p is the row of the pair you want.

Expanding beyond pairs to the P(s)^n case (where P(s) is the power set of s) would look like

setsosets = as.matrix(do.call(expand.grid,replicate(n,1:nrow(subsets),simplify=FALSE)))

Here, each row will have a vector of numbers. Each number corresponds to a row in the subsets matrix.

Making copies of the elements of s is probably not necessary for whatever you are doing after this. However, you could do it from here by using lapply(1:nrow(pairs),function(p)lapply(pairs[p,],function(r) s[subsets[r,]])), which starts like...

[[1]]
[[1]]$Row1
integer(0)

[[1]]$Row2
integer(0)


[[2]]
[[2]]$Row1
[1] 1

[[2]]$Row2
integer(0)

回答2:

It is easier to start with a Cartesian product of the indices. Then duplication can be avoided by making sure the tuple of indices is sorted.

combosn <- function(items,n) {
  i <- seq_along(items)
  idx <-do.call(expand.grid,rep(list(i),n))
  idx <- idx[!apply(idx,1,is.unsorted),]
  apply(idx,1,function(x) items[x])
}

ss<-listOfAllSubsets(1:2)

str(combosn(ss,2))

List of 6
 $ :List of 2
  ..$ : int 1
  ..$ : int 1
 $ :List of 2
  ..$ : int 1
  ..$ : int 2
 $ :List of 2
  ..$ : int 2
  ..$ : int 2
 $ :List of 2
  ..$ : int 1
  ..$ : int [1:2] 1 2
 $ :List of 2
  ..$ : int 2
  ..$ : int [1:2] 1 2
 $ :List of 2
  ..$ : int [1:2] 1 2
  ..$ : int [1:2] 1 2

Or, for n=3,

str(combosn(ss,3))

List of 10
 $ :List of 3
  ..$ : int 1
  ..$ : int 1
  ..$ : int 1
 $ :List of 3
  ..$ : int 1
  ..$ : int 1
  ..$ : int 2
 $ :List of 3
  ..$ : int 1
  ..$ : int 2
  ..$ : int 2
 $ :List of 3
  ..$ : int 2
  ..$ : int 2
  ..$ : int 2
 $ :List of 3
  ..$ : int 1
  ..$ : int 1
  ..$ : int [1:2] 1 2
 $ :List of 3
  ..$ : int 1
  ..$ : int 2
  ..$ : int [1:2] 1 2
 $ :List of 3
  ..$ : int 2
  ..$ : int 2
  ..$ : int [1:2] 1 2
 $ :List of 3
  ..$ : int 1
  ..$ : int [1:2] 1 2
  ..$ : int [1:2] 1 2
 $ :List of 3
  ..$ : int 2
  ..$ : int [1:2] 1 2
  ..$ : int [1:2] 1 2
 $ :List of 3
  ..$ : int [1:2] 1 2
  ..$ : int [1:2] 1 2
  ..$ : int [1:2] 1 2

回答3:

allSubsets<-function(n,# size of initial set
                     m,# number of subsets
                    includeEmpty=FALSE)# should the empty set be consiered a subset?

{

    # m can't exceed the number of possible subsets
    if(includeEmpty)
        stopifnot(m <= 2^n)
    else
        stopifnot(m <= 2^n-1)

    # get the subsets of the initial set (of size n)
    if(includeEmpty){
        ll <- split(t(combn(2^n,m)),seq(choose(2^n,m)))
    }else
        ll <- split(t(combn(2^n-1,m)),seq(choose(2^n-1,m)))

    # get the subets 
    subsets <- apply(do.call(expand.grid,rep(list(c(F,T)),n)),
                     1,which)
    # remove the empty subset if desired
    if(!includeEmpty)
        subsets <- subsets[-1]

    # covert the subsets to vector
    subsets <- lapply(subsets,as.vector)

    # return the list of subsets
    apply(t(mapply('[',list(subsets),ll)),1,function(x)x)

}

# returns a list where each element is a list of length 2 with 
# subsets of the initial set of length 4
x = allSubsets(4,2,F)

来源：https://stackoverflow.com/questions/28353736/all-n-combinations-of-all-subsets

标签

combinatorics