Check if two unordered lists are equal [duplicate]
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.
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 == 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.