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.


matomo