Pythonic way of removing reversed duplicates in list

Question

I have a list of pairs:

[0, 1], [0, 4], [1, 0], [1, 4], [4, 0], [4, 1]

and I want to remove any duplicates where

[a,b] == [         


        

Answer 1

If the order of pairs and pair-items matters, creating a new list by testing for membership might be the way to go here.

pairs = [0, 1], [0, 4], [1, 0], [1, 4], [4, 0], [4, 1]
no_dups = []
for pair in pairs:
    if not any( all( i in p for i in pair ) for p in no_dups ):
        no_dups.append(pair)

Otherwise, I'd go with Styvane's answer.

Incidentally, the above solution will not work for cases in which you have matching pairs. For example, [0,0] would not be added to the list. For that, you'd need to add an additional check:

for pair in pairs:
    if not any( all( i in p for i in pair ) for p in no_dups ) or ( len(set(pair)) == 1 and not pair in no_dups ):
        no_dups.append(pair)

However, that solution will not pick up empty "pairs" (eg, []). For that, you'll need one more adjustment:

    if not any( all( i in p for i in pair ) for p in no_dups ) or ( len(set(pair)) in (0,1) and not pair in no_dups ):
        no_dups.append(pair)

The and not pair in no_dups bit is required to prevent adding the [0,0] or [] to no_dups twice.