Efficiently remove duplicates, order-agnostic, from list of lists

Question

The following list has some duplicated sublists, with elements in different order:

l1 = [
    [\'The\', \'quick\', \'brown\', \'fox\'],
    [\'hi\', \'there\'         


        

Answer 1

This one is a little tricky. You want to key a dict off of frozen counters, but counters are not hashable in Python. For a small degradation in the asymptotic complexity, you could use sorted tuples as a substitute for frozen counters:

seen = set()
result = []
for x in l1:
    key = tuple(sorted(x))
    if key not in seen:
        result.append(x)
        seen.add(key)

The same idea in a one-liner would look like this:

[*{tuple(sorted(k)): k for k in reversed(l1)}.values()][::-1]