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How to set return value of method of mock.patch result?

I have a test function which is getting a mocked class via a #mock.patch decorator. Within the function, I want to force a return value of one of the method (which is needed for the test to succeed) as well as make assertions on another of the methods; these methods are exercised by calling a free function in the class's module.
Example module bla.py:
class Tool:
def __init__(self):
self.items = list()
def m1(self, val):
self.items.append(val)
def act(self):
return len(self.items)
def exercise_tool():
toolio = Tool()
for i in range(2):
toolio.m1(i)
return toolio.act() <= 2
and its test:
from unittest import mock
import bla
#mock.patch("bla.Tool")
def test_exercise_tool(mtool):
mtool.act.return_value = 3
assert not bla.exercise_tool() # verify the mocked return value
mtool.m1.assert_was_called()
The test fails with TypeError: '<=' not supported between instances of 'MagicMock' and 'int'.
So how do I get this test to pass?

Python How to create method of class in runtime

I am curious how to create a custom method for a class at runtime...
I mean for example with name of method, name of parameters, body of method read from database and assign this method to a class or to an instance.
I have a found possibility to add method that is already written:
class A:
def __init__(self):
pass
def method(self):
return True
A.method = method
a = A()
print(a.method())
but I am interested in completely assembling a new method from scratch:
name = "method"
params = ["self"] # Params in list should be a strings
body = "return True"
# To create method from pieces
Is it possible using __dict__ ? Or how else this be done?

Methods are another attribute on the object that is the class. They can be added like other attributes:
Code:
class A:
def __init__(self):
pass
def method(self):
return True
def another_method(self):
return False
setattr(A, 'another_method', another_method)
Test Code:
a = A()
print(a.another_method())
Results:
False
Methods from a string:
Add if you really need to get your methods from a database or such you can use exec like:
method_string = """
def yet_another_method(self):
return type(self).__name__
"""
exec(method_string)
setattr(A, 'yet_another_method', yet_another_method)
a = A()
print(a.yet_another_method())
Results:
A

This answer has to be treated with care, using exec or eval can run arbitary code and may compromise your system. So if you rely on user-input to create the function you mustn't use this!!!
The warning aside you can simply create anything using exec:
exec("""
def method():
return True
""")
>>> method()
True
So what you basically need is just a way to get your requirements in there:
functionname = 'funfunc'
parameters = ['a', 'b']
body = 'return a + b'
exec("""
def {functionname}({parameters}):
{body}
""".format(
functionname=functionname,
parameters=', '.join(parameters),
body='\n'.join([' {line}'.format(line=line) for line in body.split('\n')])))
The body will be indented so that it's valid syntax and the parameter list will be joined using ,. And the test:
>>> funfunc(1, 2)
3

One of the best solutions that I have found is the following:
def import_code(code, name, add_to_sys_modules=0):
"""
Import dynamically generated code as a module. code is the
object containing the code (a string, a file handle or an
actual compiled code object, same types as accepted by an
exec statement). The name is the name to give to the module,
and the final argument says wheter to add it to sys.modules
or not. If it is added, a subsequent import statement using
name will return this module. If it is not added to sys.modules
import will try to load it in the normal fashion.
import foo
is equivalent to
foofile = open("/path/to/foo.py")
foo = importCode(foofile,"foo",1)
Returns a newly generated module.
"""
import sys,imp
module = imp.new_module(name)
exec(code,module.__dict__)
if add_to_sys_modules:
sys.modules[name] = module
return module
class A:
def __init__(self):
pass
name = "method"
params = ["self"] # Params in list should be a strings
body = "return True"
scratch = "def {0}({1}):\n\t{2}".format(name, ','.join(params), body)
new_module = import_code(scratch, "test")
A.method = new_module.method
a = A()
print(a.method())
Original function import_code by the following link http://code.activestate.com/recipes/82234-importing-a-dynamically-generated-module/
Using this solution I can dynamically create methods, load them in runtime and link to whatever I want object !!

Patch a method outside python class

I am interested in patching a method which is called by another method in one file. Example - original.py file contains -
def A():
a = 10
b = 5
return a*b;
def B():
c = A()
return c* 10
I want to write unit test for this file , say call it test.py
import mock
import unittest
class TestOriginal(unitest.TestCase):
def test_Original_method(self):
with patch(''):
How can I use patch and mock modules to test original.py. I want A() to always return MagicMock() object instead of an integer.

You simply patch out the A global in the module under test. I'd use the #patch decorator syntax here:
import mock
import unittest
import module_under_test
class TestOriginal(unitest.TestCase):
#patch('module_under_test.A')
def test_Original_method(self, mocked_A):
mocked_A.return_value = 42
result = module_under_test.B()
mocked_A.assert_called_with()
self.assertEqual(result, 420)
This passes in the MagicMock mock object for A() as an extra argument to the test method.
Note that we explicitly named the module here. You could also use patch.object(), just naming the attribute on the module (which are your module globals):
class TestOriginal(unitest.TestCase):
#patch.object(module_under_test, 'A')
def test_Original_method(self, mocked_A):
mocked_A.return_value = 42
result = module_under_test.B()
mocked_A.assert_called_with()
self.assertEqual(result, 420)
You can still use a with statement too, of course:
class TestOriginal(unitest.TestCase):
def test_Original_method(self):
with patch('module_under_test.A') as mocked_A:
mocked_A.return_value = 42
result = module_under_test.B()
mocked_A.assert_called_with()
self.assertEqual(result, 420)

Class method as a decorator

I have a class where I have multiple methods. I want to use one of the methods as a decorator for other methods. For this I am using following syntax:
#self.action
def execute(self,req):
where action is other method in my class. But it doesn't work and throws exception as
name 'self' is not defined

You cannot use a method of the class while defining it; there is no self within the class nor is the class 'baked' yet to even access any class.
You can treat methods as functions to use as a decorator:
class SomeClass():
def action(func):
# decorate
return wrapper
#action
def execute(self, req):
# something
If action is defined on a base class, then you'd have to refer to the name via the base class:
class Base():
#staticmethod
def action(func):
# decorate
return wrapper
class Derived(Base):
#Base.action
def execute(self, req):
# something
For Python 2, you'd have to make action a static method here, as otherwise you get an unbound method that'll complain you cannot call it without an instance as the first argument. In Python 3, you can leave off the #staticmethod decorator there, at least for the purposes of the decorator.
But note that action cannot then be used as a method directly; perhaps it should not be part of the class at all at that point. It is not part of the end-user API here, presumably the decorator is not used by consumers of the instances of these classes.

Just beware that both the decorator and the decorated function are unbound methods, so you can only access the self (or cls for classmethods) in the inner scope of the decorator, and must manually bind the decorated method to the instance bound in the inner decorator.
class A:
x = 5
y = 6
def decorate(unbound):
def _decorator(self):
bound = unbound.__get__(self)
return bound() * self.x
return _decorator
#decorate
def func(self):
return self.y
A().func() # 30!!
Still trying to wrap my head around how decorators could be inherited and overridden.
Beware that for the decorator to work it can't be bound to an instance. That is: there is no way to make this work
a = A()
#a.decorate
def func(*args):
return 1
Despite this pattern is much more common than the asked here.
At this point the question raises: is it a method at all or just code that you happen to hide in a class?
The only way to prevent the decorator being wrongfully bound is to declare it as a staticmethod, but then it must be in a previous super class because to be used it must be bound to the static class reference which would not be yet defined, just as the self.
class A:
x = 1
#staticmethod
def decorate(unbound):
def _decorator(self):
bound = unbound.__get__(self)
return bound() * self.x
return _decorator
class B(A):
#A.decorate
def func(self):
return 1
class C():
x = 2
#B.decorate
def func(self):
return 1
a = A()
class D():
x = 3
#a.decorate
def func(self):
return 1
B().func() # 1
C().func() # 2
D().func() # 3
But as you can see, there is no way for the decorator to use the state of its own class. class A from this last example just happens to be a mixin with a default x variable and an "unrelated" static decorator.
So, again, is it a method?
To overcome all of this, you can bind the staticmethod in your same class to an arbitrary type. Namely, the builtin type will do.
class A:
x = 1
#staticmethod
def decorate(unbound):
def _decorator(self):
bound = unbound.__get__(self)
return bound() * self.x
return _decorator
#decorate.__get__(type)
def func(self):
return 1
class B:
x = 2
#A.decorate
def func(self):
return 1
class C:
x = 3
#(A().decorate) # Only for Python 3.9+, see PEP-614
def func(self):
return 1
A().func() # 1
B().func() # 2
C().func() # 3
But this features too much magic for my taste. And still not a method for my gut.

In python "self" is passed to instance methods as an argument (the first), "self" is just a convention is possible to call it "foobarbaz" (of course it would be silly)… the point is that, from the outside "self" is not defined (because its scope is the method)… you can't decorate class methods with other class methods, instead you have to write a separate class!

How do you mock patch a python class and get a new Mock object for each instantiation?

OK,
I know this is mentioned in the manual, and probably has to do with side_effect and/or return_value, but a simple, direct example will help me immensely.
I have:
class ClassToPatch():
def __init__(self, *args):
_do_some_init_stuff()
def some_func():
_do_stuff()
class UUT():
def __init__(self, *args)
resource_1 = ClassToPatch()
resource_2 = ClassToPatch()
Now, I want to unit test the UUT class, and mock the ClassToPatch. Knowing the UUT class will instantiate exactly two ClassToPatch objects, I want the Mock framework to return a new Mock object for each instantiation, so I can later assert calls on each separately.
How do I achieve this using the #patch decorator in a test case? Namely, how to fix the following code sample?
class TestCase1(unittest.TestCase):
#patch('classToPatch.ClassToPatch',autospec=True)
def test_1(self,mock1,mock2):
_assert_stuff()

Here's a quick'n'dirty example to get you going:
import mock
import unittest
class ClassToPatch():
def __init__(self, *args):
pass
def some_func(self):
return id(self)
class UUT():
def __init__(self, *args):
resource_1 = ClassToPatch()
resource_2 = ClassToPatch()
self.test_property = (resource_1.some_func(), resource_2.some_func())
class TestCase1(unittest.TestCase):
#mock.patch('__main__.ClassToPatch', autospec = True)
def test_1(self, mock1):
ctpMocks = [mock.Mock(), mock.Mock()]
ctpMocks[0].some_func.return_value = "funky"
ctpMocks[1].some_func.return_value = "monkey"
mock1.side_effect = ctpMocks
u = UUT()
self.assertEqual(u.test_property, ("funky", "monkey"))
if __name__ == '__main__':
unittest.main()
I've added test_property to UUT so that the unit test does something useful. Now, without the mock test_property should be a tuple containing the ids of the two ClassToPatch instances. But with the mock it should be the tuple: ("funky", "monkey").
I've used the side_effect property of the mock object so that a different instance of ClassToPatch is returned on each call in the UUT initialiser.
Hope this helps.
Edit: Oh, by the way, when I run the unit test I get:
.
----------------------------------------------------------------------
Ran 1 test in 0.004s
OK

Here is another version which is more generic to handle any number of instances created:
class TestUUT:
def test_init(self, mocker):
class MockedClassToPatchMeta(type):
static_instance = mocker.MagicMock(spec=ClassToPatch)
def __getattr__(cls, key):
return MockedClassToPatchMeta.static_instance.__getattr__(key)
class MockedClassToPatch(metaclass=MockedClassToPatchMeta):
original_cls = ClassToPatch
instances = []
def __new__(cls, *args, **kwargs):
MockedClassToPatch.instances.append(
mocker.MagicMock(spec=MockedClassToPatch.original_cls))
MockedClassToPatch.instances[-1].__class__ = MockedClassToPatch
return MockedClassToPatch.instances[-1]
mocker.patch(__name__ + '.ClassToPatch', new=MockedClassToPatch)
UUT()
# since your original code created two instances
assert 2 == len(MockedClassToPatch.instances)
If you need more thorough validation for each instance you can access MockedClassToPatch.instances[0] or MockedClassToPatch.instances[1].
I've also created a helper library to generate the meta class boilerplate for me. To generate the needed code for your example I wrote:
print(PytestMocker(mocked=ClassToPatch, name=__name__).mock_classes().mock_classes_static().generate())