# Check if two unordered lists are equal

### Question

I'm looking for an easy (and quick) way to determine if two unordered lists contain the same elements:

For example:

``````['one', 'two', 'three'] == ['one', 'two', 'three'] :  true
['one', 'two', 'three'] == ['one', 'three', 'two'] :  true
['one', 'two', 'three'] == ['one', 'two', 'three', 'three'] :  false
['one', 'two', 'three'] == ['one', 'two', 'three', 'four'] :  false
['one', 'two', 'three'] == ['one', 'two', 'four'] :  false
['one', 'two', 'three'] == ['one'] :  false
``````

I'm hoping to do this without using a map.

1
224
1/25/2019 1:07:20 PM

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
>>>
``````
395
3/8/2012 7:04:24 PM

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 == b
False
``````

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.