Generate permutations of list with repeated elements

This web page looks promising.

def next_permutation(seq, pred=cmp):
    """Like C++ std::next_permutation() but implemented as
    generator. Yields copies of seq."""
    def reverse(seq, start, end):
        # seq = seq[:start] + reversed(seq[start:end]) + \
        #       seq[end:]
        end -= 1
        if end <= start:
            return
        while True:
            seq[start], seq[end] = seq[end], seq[start]
            if start == end or start+1 == end:
                return
            start += 1
            end -= 1
    if not seq:
        raise StopIteration
    try:
        seq[0]
    except TypeError:
        raise TypeError("seq must allow random access.")
    first = 0
    last = len(seq)
    seq = seq[:]
    # Yield input sequence as the STL version is often
    # used inside do {} while.
    yield seq[:]
    if last == 1:
        raise StopIteration
    while True:
        next = last - 1
        while True:
            # Step 1.
            next1 = next
            next -= 1
            if pred(seq[next], seq[next1]) < 0:
                # Step 2.
                mid = last - 1
                while not (pred(seq[next], seq[mid]) < 0):
                    mid -= 1
                seq[next], seq[mid] = seq[mid], seq[next]
                # Step 3.
                reverse(seq, next1, last)
                # Change to yield references to get rid of
                # (at worst) |seq|! copy operations.
                yield seq[:]
                break
            if next == first:
                raise StopIteration
    raise StopIteration

>>> for p in next_permutation([int(c) for c in "111222"]):
...     print p
... 
[1, 1, 1, 2, 2, 2]
[1, 1, 2, 1, 2, 2]
[1, 1, 2, 2, 1, 2]
[1, 1, 2, 2, 2, 1]
[1, 2, 1, 1, 2, 2]
[1, 2, 1, 2, 1, 2]
[1, 2, 1, 2, 2, 1]
[1, 2, 2, 1, 1, 2]
[1, 2, 2, 1, 2, 1]
[1, 2, 2, 2, 1, 1]
[2, 1, 1, 1, 2, 2]
[2, 1, 1, 2, 1, 2]
[2, 1, 1, 2, 2, 1]
[2, 1, 2, 1, 1, 2]
[2, 1, 2, 1, 2, 1]
[2, 1, 2, 2, 1, 1]
[2, 2, 1, 1, 1, 2]
[2, 2, 1, 1, 2, 1]
[2, 2, 1, 2, 1, 1]
[2, 2, 2, 1, 1, 1]
>>> 

---

2017-08-12

Seven years later, here is a better algorithm (better for clarity):

from itertools import permutations

def unique_perms(series):
    return {"".join(p) for p in permutations(series)}

print(sorted(unique_perms('1122')))