For example, given the list
['one', 'two', 'one'], the algorithm should return
True, whereas given
['one', 'two', 'three'] it should return
set() to remove duplicates if all values are hashable:
>>> your_list = ['one', 'two', 'one'] >>> len(your_list) != len(set(your_list)) True
Recommended for short lists only:
any(thelist.count(x) > 1 for x in thelist)
Do not use on a long list -- it can take time proportional to the square of the number of items in the list!
For longer lists with hashable items (strings, numbers, &c):
def anydup(thelist): seen = set() for x in thelist: if x in seen: return True seen.add(x) return False
If your items are not hashable (sublists, dicts, etc) it gets hairier, though it may still be possible to get O(N logN) if they're at least comparable. But you need to know or test the characteristics of the items (hashable or not, comparable or not) to get the best performance you can -- O(N) for hashables, O(N log N) for non-hashable comparables, otherwise it's down to O(N squared) and there's nothing one can do about it:-(.