How to assert two list contain the same elements in Python?
Slightly faster version of the implementation (If you know that most couples lists will have different lengths):
def checkEqual(L1, L2):
return len(L1) == len(L2) and sorted(L1) == sorted(L2)
Comparing:
>>> timeit(lambda: sorting([1,2,3], [3,2,1]))
2.42745304107666
>>> timeit(lambda: lensorting([1,2,3], [3,2,1]))
2.5644469261169434 # speed down not much (for large lists the difference tends to 0)
>>> timeit(lambda: sorting([1,2,3], [3,2,1,0]))
2.4570400714874268
>>> timeit(lambda: lensorting([1,2,3], [3,2,1,0]))
0.9596951007843018 # speed up
Given
l1 = [a,b]
l2 = [b,a]
In Python >= 3.0
assertCountEqual(l1, l2) # True
In Python >= 2.7, the above function was named:
assertItemsEqual(l1, l2) # True
In Python < 2.7
import unittest2
assertItemsEqual(l1, l2) # True
Via six
module (Any Python version)
import unittest
import six
class MyTest(unittest.TestCase):
def test(self):
six.assertCountEqual(self, self.l1, self.l2) # True
Converting your lists to sets will tell you that they contain the same elements. But this method cannot confirm that they contain the same number of all elements. For example, your method will fail in this case:
L1 = [1,2,2,3]
L2 = [1,2,3,3]
You are likely better off sorting the two lists and comparing them:
def checkEqual(L1, L2):
if sorted(L1) == sorted(L2):
print "the two lists are the same"
return True
else:
print "the two lists are not the same"
return False
Note that this does not alter the structure/contents of the two lists. Rather, the sorting creates two new lists
As of Python 3.2 unittest.TestCase.assertItemsEqual
(doc) has been replaced by unittest.TestCase.assertCountEqual
(doc) which does exactly what you are looking for, as you can read from the python standard library documentation. The method is somewhat misleadingly named but it does exactly what you are looking for.
a and b have the same elements in the same number, regardless of their order
Here a simple example which compares two lists having the same elements but in a different order.
- using
assertCountEqual
the test will succeed - using
assertListEqual
the test will fail due to the order difference of the two lists
Here a little example script.
import unittest
class TestListElements(unittest.TestCase):
def setUp(self):
self.expected = ['foo', 'bar', 'baz']
self.result = ['baz', 'foo', 'bar']
def test_count_eq(self):
"""Will succeed"""
self.assertCountEqual(self.result, self.expected)
def test_list_eq(self):
"""Will fail"""
self.assertListEqual(self.result, self.expected)
if __name__ == "__main__":
unittest.main()
Side Note : Please make sure that the elements in the lists you are comparing are sortable.