Check if two unordered lists are equal


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.

1/25/2019 1:07:20 PM

Accepted Answer

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])
>>> compare([1,2,3], [1,2,3])
>>> compare([1,2,3,3], [1,2,2,3])
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

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.

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