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