Mocking out methods on any instance of a python class

Needing to mock out methods when testing is very common and there are lots of tools to help you with it in Python. The danger with "monkey patching" classes like this is that if you don't undo it afterwards then the class has been modified for all other uses throughout your tests.

My library mock, which is one of the most popular Python mocking libraries, includes a helper called "patch" that helps you to safely patch methods or attributes on objects and classes during your tests.

The mock module is available from:

http://pypi.python.org/pypi/mock

The patch decorator can be used as a context manager or as a test decorator. You can either use it to patch out with functions yourself, or use it to automatically patch with Mock objects that are very configurable.

from a import A
from b import B

from mock import patch

def new_foo(self):
    return "New foo"

with patch.object(A, 'foo', new_foo):
    y = B()
    if y.bar() == "New foo":
        print "Success!"

This handles the unpatching for you automatically. You could get away without defining the mock function yourself:

from mock import patch

with patch.object(A, 'foo') as mock_foo:
    mock_foo.return_value = "New Foo"

    y = B()
    if y.bar() == "New foo":
        print "Success!"

Mock is the way to do it, alright. It can be a bit tricky to make sure you're patching the instance method on any instances created from the class.

# a.py
class A(object):
    def foo(self):
        return "A's foo"

# b.py
from a import A

class B(object):
    def bar(self):
        x = A()
        return x.foo()

# test.py
from a import A
from b import B
import mock

mocked_a_class = mock.Mock()
mocked_a_instance = mocked_a_class.return_value
mocked_a_instance.foo.return_value = 'New foo'

with mock.patch('b.A', mocked_a_class):  # Note b.A not a.A
    y = B()
    if y.bar() == "New foo":
        print "Success!"

Referenced in the docs, at the para starting "To configure return values on methods of instances on the patched class..."

Easiest way is probably to use a class method. You really should use an instance method, but it's a pain to create those, whereas there's a built-in function that creates a class method. With a class method, your stub will get a reference to the class (rather than the instance) as the first argument, but since it's a stub this probably doesn't matter. So:

Product.name = classmethod(lambda cls: "stubbed_name")

Note that the signature of the lambda must match the signature of the method you're replacing. Also, of course, since Python (like Ruby) is a dynamic language, there is no guarantee that someone won't switch out your stubbed method for something else before you get your hands on the instance, though I expect you will know pretty quickly if that happens.

Edit: On further investigation, you can leave out the classmethod():

Product.name = lambda self: "stubbed_name"

I was trying to preserve the original method's behavior as closely as possible, but it looks like it's not actually necessary (and doesn't preserve the behavior as I'd hoped, anyhow).