Determine if 2 lists have the same elements, regardless of order?
Inferring from your example:
x = ['a', 'b']
y = ['b', 'a']
that the elements of the lists won't be repeated (they are unique) as well as hashable (which strings and other certain immutable python objects are), the most direct and computationally efficient answer uses Python's builtin sets, (which are semantically like mathematical sets you may have learned about in school).
set(x) == set(y) # prefer this if elements are hashable
In the case that the elements are hashable, but non-unique, the collections.Counter also works semantically as a multiset, but it is far slower:
from collections import Counter
Counter(x) == Counter(y)
Prefer to use sorted:
sorted(x) == sorted(y)
if the elements are orderable. This would account for non-unique or non-hashable circumstances, but this could be much slower than using sets.
An empirical experiment concludes that one should prefer set, then sorted. Only opt for Counter if you need other things like counts or further usage as a multiset.
First setup:
import timeit
import random
from collections import Counter
data = [str(random.randint(0, 100000)) for i in xrange(100)]
data2 = data[:] # copy the list into a new one
def sets_equal():
return set(data) == set(data2)
def counters_equal():
return Counter(data) == Counter(data2)
def sorted_lists_equal():
return sorted(data) == sorted(data2)
And testing:
>>> min(timeit.repeat(sets_equal))
13.976069927215576
>>> min(timeit.repeat(counters_equal))
73.17287588119507
>>> min(timeit.repeat(sorted_lists_equal))
36.177085876464844
So we see that comparing sets is the fastest solution, and comparing sorted lists is second fastest.