Check if two unordered lists are equal [duplicate] Check if two unordered lists are equal [duplicate] python python

# 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)``

Documentation on `set`

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

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