a = [1, 2, 3, 1, 2, 3]
b = [3, 2, 1, 3, 2, 1]
a & b should be considered equal, because they have exactly the same elements, only in different
Let a,b lists
def ass_equal(a,b):
try:
map(lambda x: a.pop(a.index(x)), b) # try to remove all the elements of b from a, on fail, throw exception
if len(a) == 0: # if a is empty, means that b has removed them all
return True
except:
return False # b failed to remove some items from a
No need to make them hashable or sort them.