Remove duplicated lists in list of lists in Python

Question

I've seen some questions here very related but their answer doesn't work for me. I have a list of lists where some sublists are repeated but their elements may be disordered. For example

g = [[1, 2, 3], [3, 2, 1], [1, 3, 2], [9, 0, 1], [4, 3, 2]]

The output should be, naturally according to my question:

g = [[1,2,3],[9,0,1],[4,3,2]]

I've tried with set but only removes those lists that are equal (I thought It should work because sets are by definition without order). Other questions i had visited only has examples with lists exactly duplicated or repeated like this: Python : How to remove duplicate lists in a list of list?. For now order of output (for list and sublists) is not a problem.

Answer 1

(ab)using side-effects version of a list comp:

seen = set()

[x for x in g if frozenset(x) not in seen and not seen.add(frozenset(x))]
Out[4]: [[1, 2, 3], [9, 0, 1], [4, 3, 2]]

For those (unlike myself) who don't like using side-effects in this manner:

res = []
seen = set()

for x in g:
    x_set = frozenset(x)
    if x_set not in seen:
        res.append(x)
        seen.add(x_set)

The reason that you add frozensets to the set is that you can only add hashable objects to a set, and vanilla sets are not hashable.

Answer 2

If you don't care about the order for lists and sublists (and all items in sublists are unique):

result = set(map(frozenset, g))

If a sublist may have duplicates e.g., [1, 2, 1, 3] then you could use tuple(sorted(sublist)) instead of frozenset(sublist) that removes duplicates from a sublist.

If you want to preserve the order of sublists:

def del_dups(seq, key=frozenset):
    seen = {}
    pos = 0
    for item in seq:
        if key(item) not in seen:
            seen[key(item)] = True
            seq[pos] = item
            pos += 1
    del seq[pos:]

Example:

del_dups(g, key=lambda x: tuple(sorted(x)))

See In Python, what is the fastest algorithm for removing duplicates from a list so that all elements are unique while preserving order?

Answer 3

I would convert each element in the list to a frozenset (which is hashable), then create a set out of it to remove duplicates:

>>> g = [[1, 2, 3], [3, 2, 1], [1, 3, 2], [9, 0, 1], [4, 3, 2]]
>>> set(map(frozenset, g))
set([frozenset([0, 9, 1]), frozenset([1, 2, 3]), frozenset([2, 3, 4])])

If you need to convert the elements back to lists:

>>> map(list, set(map(frozenset, g)))
[[0, 9, 1], [1, 2, 3], [2, 3, 4]]

Answer 4

What about using mentioned by roippi frozenset this way:

>>> g = [list(x) for x in set(frozenset(i) for i in [set(i) for i in g])]

[[0, 9, 1], [1, 2, 3], [2, 3, 4]]