Subsets() with same element multiple times?

edit

Python's itertools module has lots of useful functions, including:

sage: import itertools
sage: ListA = [1,2,3]
sage: S = itertools.combinations_with_replacement(ListA, 2)
sage: list(S)
[(1, 1), (1, 2), (1, 3), (2, 2), (2, 3), (3, 3)]

Edit: If you want to use Sage-specific code, you can use Combinations. This will return duplicates only if they appear in the original list, so you can use ListA * n to get n copies of your list, and then choose combinations from that:

sage: S = Combinations(ListA * 2, 2)
sage: list(S)
[[1, 1], [1, 2], [1, 3], [2, 2], [2, 3], [3, 3]]

The itertools versions seem to be faster, in my limited testing.

Well... Let's try to be lazy, and reuse work already done :

sage: subsets?
Docstring:     
   Iterator over the *list* of all subsets of the iterable X, in no
   particular order. Each list appears exactly once, up to order.

   INPUT:

   * "X" - an iterable

   OUTPUT: iterator of lists

So, what's wrong with :

sage: [u for u in subsets(ListA) if len(u) == 2]
[[1, 2], [1, 3], [2, 3]]

Of course, the subsets operator will generate all subsets of ListA. It might be interesting to write a specialized generator...

OTOH, someone already did the work (and probably better than I can do) :

sage: Subsets?
Docstring:     
   Return the combinatorial class of the subsets of the finite set
   "s". The set can be given as a list, Set or any iterable
   convertible to a set. Alternatively, a non-negative integer n can
   be provided in place of "s"; in this case, the result is the
   combinatorial class of the subsets of the set \{1,2,...,n\} (i.e.
   of the Sage "range(1,n+1)").

   A second optional parameter "k" can be given. In this case,
   "Subsets" returns the combinatorial class of subsets of "s" of size
   "k".

Therefore :

sage: list(Subsets(ListA, 2))
[{1, 2}, {1, 3}, {2, 3}]

Beware : the returned elements are Sets (and not sets). If you insist on lists :

sage: list(map(list, Subsets(ListA, 2)))
[[1, 2], [1, 3], [2, 3]]

HTH,