Python: Generating a "set of tuples" from a "list of tuples" that doesn't take order into consideration

Question

If I have the list of tuples as the following:

[('a', 'b'), ('c', 'd'), ('a', 'b'), ('b', 'a')]

I would like to remove duplicate tuples (duplicate in terms of both content and order of items inside) so that the output would be:

[('a', 'b'), ('c', 'd')]

Or

[('b', 'a'), ('c', 'd')]

I tried converting it to set then to list but the output would maintain both ('b', 'a') and ('a', 'b') in the resulting set!

Answer 1

Try this :

a = [('a', 'b'), ('c', 'd'), ('a', 'b'), ('b', 'a')]
b = list(set([ tuple(sorted(t)) for t in a ]))
[('a', 'b'), ('c', 'd')]

Let's break this down :

If you sort a tuple, it becomes a sorted list.

>>> t = ('b', 'a')
>>> sorted(t)
['a', 'b']

For each tuple t in a, sort it and convert it back to a tuple.

>>> b = [ tuple(sorted(t)) for t in a ]
>>> b
[('a', 'b'), ('c', 'd'), ('a', 'b'), ('a', 'b')]

Convert the resulting list b to a set : values are now unique. Convert it back to a list.

>>> list(set(b))
[('a', 'b'), ('c', 'd')]

Et voilà !

Note that you can skip the creation of the intermediate list b by using a generator instead of a list comprehension.

>>> list(set(tuple(sorted(t)) for t in a))
[('a', 'b'), ('c', 'd')]

Answer 2

If you did not mind using a frozenset with a set:

l = [('a', 'b'), ('c', 'd'), ('a', 'b'), ('b', 'a')]

print(set(map(frozenset,l)))
{frozenset({'a', 'b'}), frozenset({'c', 'd'})}

You can convert back to tuple if preferable:

l = [('a', 'b'), ('c', 'd'), ('a', 'b'), ('b', 'a')]

print(list(map(tuple,set(map(frozenset ,l)))))
[('a', 'b'), ('d', 'c')]

Or using a set and reversing the order of the tuples:

l = [('a', 'b'), ('c', 'd'), ('a', 'b'), ('b', 'a')]

seen, pairs = set(), []
for a,b in l:
    if (a,b) not in seen and (b,a) not in seen:
        pairs.append((a,b))
    seen.add((a,b))