CODEWITHSUNDEEP
pythonunit-testingpython-3.xpytest
Make42
Make42

Reputation: 13108

How to assert that exactly the expected elements are in a list?

I want to unit-test a function that takes a list and returns (a very different) list. I know what elements I expect, but I do not know in what order they will me in the returned list. How can I assert that exactly those elements are in the list an no other? I would like to do this using py.test. Also: I expect to have multiple duplicates.

Upvotes: 1

Views: 3510

Answers (4)

scf
scf

Reputation: 406

If I understand you correctly, you want to test if list A and list B have the exact same elements, including duplicates but disregarding order.
This calls for use of Counters:

from collections import Counter
wanted_elements = [1, 1, 2, 3, 4, 4, 4, 4]
input_list = [1, 2, 1, 4, 4, 4, 4, 3]
c1 = Counter(wanted_elements)
c2 = Counter(input_list)
assert c1 == c2

Upvotes: 1

TigerhawkT3
TigerhawkT3

Reputation: 49330

Assuming the elements are hashable and unique, just use sets:

set(l) == set(f(l))

If those conditions don't apply, sort them:

sorted(l) == sorted(f(l))

And there's collections.Counter, if the elements are hashable and non-unique:

import collections

collections.Counter(l) == collections.Counter(f(l))

Upvotes: 4

scf
scf

Reputation: 406

You might want to use the set operation difference:

list_to_test = [1,2,3,4,5,5,4,3,2,1,2]
wanted_elements = [1,2,3]
d = set(list_to_test).difference(set(wanted_elements))
print d
set([4, 5])

and so eventually you would assert that len(d) == 0.

Upvotes: 0

Patrick Haugh
Patrick Haugh

Reputation: 61042

If order doesn't matter, sort both lists and check their equality

assert sorted(exampleOutput) == sorted(func(inputList))

Upvotes: 1

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