# Check if two unordered lists are equal [duplicate]

Python has a built-in datatype for an unordered collection of (hashable) things, called a `set`

. If you convert both lists to sets, the comparison will be unordered.

`set(x) == set(y)`

EDIT: @mdwhatcott points out that you want to check for duplicates. `set`

ignores these, so you need a similar data structure that also keeps track of the number of items in each list. This is called a multiset; the best approximation in the standard library is a `collections.Counter`

:

`>>> import collections>>> compare = lambda x, y: collections.Counter(x) == collections.Counter(y)>>> >>> compare([1,2,3], [1,2,3,3])False>>> compare([1,2,3], [1,2,3])True>>> compare([1,2,3,3], [1,2,2,3])False>>> `

If elements are always nearly sorted as in your example then builtin `.sort()`

(timsort) should be fast:

`>>> a = [1,1,2]>>> b = [1,2,2]>>> a.sort()>>> b.sort()>>> a == bFalse`

If you don't want to sort inplace you could use `sorted()`

.

In practice it might always be faster then `collections.Counter()`

(despite asymptotically `O(n)`

time being better then `O(n*log(n))`

for `.sort()`

). Measure it; If it is important.

`sorted(x) == sorted(y)`

Copying from here: Check if two unordered lists are equal

I think this is the best answer for this question because

- It is better than using counter as pointed in this answer
- x.sort() sorts x, which is a side effect. sorted(x) returns a new list.