Develop Reference

Why are the 2 Magic Mocks instances different in ID?

I am trying to use patch() to inject mock into a call after reading this tutorial: https://kimsereylam.com/python/2021/03/19/how-to-use-patch-in-python-unittest.html, but the expect mock ID is different than actual.
Here is snippet of code to illustrate:
my_class.py
class MyClass:
def __init__(self):
self.value = 5
run.py
from my_class import MyClass
def call_first():
o = MyClass()
print(o)
call_second(o)
def call_second(myclass):
print("second is called")
print(myclass)
main.py
from unittest.mock import patch
from my_class import MyClass
from run import call_first, call_second
with patch("run.MyClass") as myclass_mock:
print(myclass_mock)
print(call_first())
with patch("run.call_second") as call_second_mock:
call_second_mock.assert_called_once_with(myclass_mock)
The output is:
<MagicMock name='MyClass' id='140711608367712'>
<MagicMock name='MyClass()' id='140711598743312'>
second is called
<MagicMock name='MyClass()' id='140711598743312'>
None
Traceback (most recent call last):
File "main.py", line 9, in <module>
call_second_mock.assert_called_once_with(myclass_mock)
File "/usr/lib/python3.8/unittest/mock.py", line 924, in assert_called_once_with
raise AssertionError(msg)
AssertionError: Expected 'call_second' to be called once. Called 0 times.
What did I miss? I am trying to use patch() so I can mock the MyClass() object that is instantiated inside call_first() and passed to call_second().

with patch("run.MyClass") as myclass_mock:
with patch("run.call_second") as call_second_mock:
#print(myclass_mock)
instance = myclass_mock.return_value
print(instance)
print(call_first())
call_second_mock.assert_called_once_with(instance)
I need to use .return_value to get the MagicMock for the class instance
I should mock the call_second first

How to define a custom function of an instance using setattr

I'm trying to add a method to a class dynamically, but I keep running into an error where self is not passed to a the new function. For instance:
class Dummy():
def say_hi(self):
print("hi")
def new_method(self):
print("bye")
dummy = Dummy()
setattr(dummy, "say_bye", new_method)
dummy.say_bye()
results in the following error:
Traceback (most recent call last):
File "main.py", line 13, in <module>
dummy.say_bye()
TypeError: new_method() missing 1 required positional argument: 'self'
What am I doing wrong?

Use types.MethodType feature:
from types import MethodType
class Dummy():
def say_hi(self):
print("hi")
def new_method(self):
print("bye")
dummy = Dummy()
dummy.say_bye = MethodType(new_method, dummy)
dummy.say_bye() # bye

You are setting the function new_method as an attribute of the dummy object.
If you do print(dummy.__dict__) you'll see something like this:
{'say_bye': <function new_method at 0x7fcd949af668>}
This means that your dummy object has the function new_method as an attribute, so when you do dummy.say_bye(), you're calling the function you have as an attribute without any argument.
It is not a function of the Dummy class, it is just a function that your dummy object has as an attribute.
You can achieve the functionality you are looking for using RomanPerekhrest's answer.
Hope it helps.
Cheers!

super() in a decorated subclass in Python 2

I'm trying to use super in a subclass which is wrapped in another class using a class decorator:
def class_decorator(cls):
class WrapperClass(object):
def make_instance(self):
return cls()
return WrapperClass
class MyClass(object):
def say(self, x):
print(x)
#class_decorator
class MySubclass(MyClass):
def say(self, x):
super(MySubclass, self).say(x.upper())
However, the call to super fails:
>>> MySubclass().make_instance().say('hello')
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "<stdin>", line 4, in say
TypeError: super(type, obj): obj must be an instance or subtype of type
The problem is that, when say is called, MySubclass doesn't refer to the original class anymore, but to the return value of the decorator.
One possible solution would be to store the value of MySubclass before decorating it:
class MySubclass(MyClass):
def say(self, x):
super(_MySubclass, self).say(x.upper())
_MySubclass = MySubclass
MySubclass = class_decorator(MySubclass)
This works, but isn't intuitive and would need to be repeated for each decorated subclass. I'm looking for a way that doesn't need additional boilerplate for each decorated subclass -- adding more code in one place (say, the decorator) would be OK.
Update: In Python 3 this isn't a problem, since you can use __class__ (or the super variant without arguments), so the following works:
#class_decorator
class MySubclass(MyClass):
def say(self, x):
super().say(x.upper())
Unfortunately, I'm stuck with Python 2.7 for this project.

The problem is that your decorator returns a different class than python (or anyone who uses your code) expects. super not working is just one of the many unfortunate consequences:
>>> isinstance(MySubclass().make_instance(), MySubclass)
False
>>> issubclass(MySubclass, MyClass)
False
>>> pickle.dumps(MySubclass().make_instance())
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
_pickle.PicklingError: Can't pickle <class '__main__.MySubclass'>: it's not the same object as __main__.MySubclass
This is why a class decorator should modify the class instead of returning a different one. The correct implementation would look like this:
def class_decorator(wrapped_cls):
#classmethod
def make_instance(cls):
return cls()
wrapped_cls.make_instance = make_instance
return wrapped_cls
Now super and everything else will work as expected:
>>> MySubclass().make_instance().say('hello')
HELLO

The problem occurs because at the time when MySubclass.say() is called, the global symbol MySubclass no longer refers to what's defined in your code as 'class MySubclass'. It is an instance of WrapperClass, which isn't in any way related to MySubclass.
If you are using Python3, you can get around this by NOT passing any arguments to 'super', like this:
super().say(x.upper())
I don't really know why you use the specific construct that you have, but it does look strange that a sub-class of MyClass that defines 'say()' - and has itself a 'say()' method in the source code would have to end up as something that does not have that method - which is the case in your code.
Note you could change the class WrapperClass line to make it read
class WrapperClass(cls):
this will make your wrapper a sub-class of the one you just decorated. This doesn't help with your super(SubClass, self) call - you still need to remove the args (which is OK only on Python3), but at least an instance created as x=MySubclass() would have a 'say' method, as one would expect at first glance.
EDIT: I've come up with a way around this, but it really looks odd and has the disadvantage of making the 'wrapped' class know that it is being wrapped (and it becomes reliant on that, making it unusable if you remove the decorator):
def class_decorator(cls):
class WrapperClass(object):
def make_instance(self):
i = cls()
i._wrapped = cls
return i
return WrapperClass
class MyClass(object):
def say(self, x):
print(x)
#class_decorator
class MySubclass(MyClass):
def say(self, x):
super(self._wrapped, self).say(x.upper())
# make_instance returns inst of the original class, non-decorated i = MySubclass().make_instance() i.say('hello')
In essence, _wrapped saves a class reference as it was at declaration time, consistent with using the regular super(this_class_name, self) builtin call.

Using a method from the parent class

I want to call a method from the parent class in a child class.
I use XX.__init__() in my child class and call the press function from the parent class. But it fails when I run the following code:
Func.py
class PC:
def __init__(self):
PCKeyDis = {}
self.PCKeyDis = PCKeyDis
def Press(self,key):
KeyDis = self.PCKeyDis
if len(key)==1 and key.islower():
key = key.upper()
win32api.keybd_event(KeyDis[key],0,0,0)
time.sleep(0.1)
win32api.keybd_event(KeyDis[key],0,win32con.KEYEVENTF_KEYUP,0)
class PCFunc(PC):
def __init__(self):
pass
def Sentence(self,string):
PC.__init__()
strlist = list(string)
for i in xrange(len(strlist)):
if strlist[i] == ' ':
strlist[i] = 'Space'
PC.Press(strlist[i]) #use this function
action.py
import Func
import win32gui
PC = Func.PC()
PCFunc = Func.PCFunc ()
win32gui.SetForegroundWindow(win32gui.FindWindow(winclass,winnm))
PCFunc.Sentence(path)
I get:
unbound method Sentence() must be called with PCFunc instance as first argument (got str instance instead)

If you want to call the constructor of the base class, then you do it on instantiation in the __init__() method, not in the Sentence() method:
def __init__(self):
super(self.__class__, self).__init__()
Since Sentence() is an instance method, you need to call it via an instance of the class (like the error tells you):
pc_func = PCFunc()
pc_func.Sentence(var)
Here you are calling the method with an undefined variable:
PCFunc.Sentence(path)
Instead you need to give a string as parameter, so either write Sentence('path'), or define the variable first:
path = 'my path'
pc_func.Sentence(path)
Do not use the same name as the class name for an instance of the class:
PCFunc = Func.PCFunc ()
Otherwise the variable name storing the instance overwrites the class name.
Apart from that, it is unclear what your code is actually supposed to do. Have a look at the Python code conventions for a first step to making your code more readible. Then do some research about classes and inheritance.

The code you posted does not produce the error you posted. Here is an example that will produce that error:
class Dog:
def do_stuff(self, string):
print string
d = Dog()
d.do_stuff('hello')
Dog.do_stuff(d, 'goodbye')
Dog.do_stuff('goodbye')
--output:--
hello
goodbye
Traceback (most recent call last):
File "1.py", line 9, in <module>
Dog.do_stuff('goodbye')
TypeError: unbound method do_stuff() must be called with Dog instance as first argument (got str instance instead)
An __init__() function can also produce that error:
class Dog:
def __init__(self):
pass
def do_stuff(self, string):
print(string)
Dog.__init__()
--output:--
Traceback (most recent call last):
File "1.py", line 7, in <module>
Dog.__init__()
TypeError: unbound method __init__() must be called with Dog instance as first argument (got nothing instead)
In the line:
d.do_stuff('hello')
the fragment d.do_stuff causes python to create and return a bound method object--which is then immediately executed by the function execution operator () in the fragment ('hello’). The bound method is bound to the instance d, hence the reason it is called a bound method. A bound method automatically passes the instance it contains to the method when the method is executed.
On the other hand, when you write:
Dog.do_stuff(....)
the fragment Dog.do_stuff causes python to create and return an unbound method. An unbound method does not contain an instance, so when an unbound method is executed by the function execution operator (), you must manually pass an instance. (In python3, things changed and you can pass anything as the first argument--an instance of the class isn't required.)

How to create a non-instantiable class?

For one of the project I am currently working I was thinking of creating a class that could not be instantiate by a client and only be supplied an instance of through a particular interface i.e. the client would not be able create further instance out of it by some hackery such as:
>>> try:
... raise WindowsError
... except:
... foo = sys.exc_info()
...
>>> foo
(<type 'exceptions.WindowsError'>, WindowsError(), <traceback object at 0x0000000005503A48>)
>>> type(foo[2])()
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: cannot create 'traceback' instances
once he has one.
I was successfully able to create a class that couldn't be instantiated. i.e.
>>> class Foo():
... def __init__(self):
... raise TypeError("cannot create 'Foo' instances")
...
>>> bar = Foo()
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "<stdin>", line 3, in __init__
TypeError: cannot create 'Foo' instances
>>> bar
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
NameError: name 'bar' is not defined
But how could I use this every same definition to create an instance of the class?
Of course I could do something like this:
>>> class Foo():
... def __init__(self, instantiate = False):
... if not instantiate:
... raise TypeError("cannot create 'Foo' instances")
but I don't find it elegant enough nor does it completely prevent the client from further instantiating it. And no I aint going down the road of building a C++ module for it.
Any suggestions on how to achieve such a thing? import abc?
A brief rational to answer Martijn's question and for completeness:
Actual you could consider the instance of the particular, and related, classes, in question, as nodes in a tree and that both the parent and the children to remain connected, dependent on and cognizant of each other and have a single unique root throughout any instance of python(insured by the use package). Any state changes in a particular node would cause others to update themselves and the database to which they are connect, accordingly. Apart from that I was being curious to know how such a thing could be put in place (the traceback class was teasing me).

What you're doing is a bad idea, you shouldn't do it.
I'm sure there's an other, better solution.
If you do decide to go with your way anyways (you shouldn't), here's how you can create an object without using __init__():
Objects in python are created with the __new__() method. The method __init__() only edits the object which was created by __new__(). For example, __init__() usually initializes some attributes for the object.
When declaring something like x = Foo() what happens is this:
x = object.__new__(Foo) gets called first and creates the object.
Foo.__init__(x) gets called second, it simply initializes some attributes etc. to the already existing object.
This means that you are not required to call Foo() (and as a result, call __init__() too). Instead, you can just call __new__() directly:
class Foo(object):
def __init__(self):
raise TypeError("Cannot create 'Foo' instances.")
>>> x = object.__new__(Foo)
>>> x
<__main__.Foo object at 0x02B074F0>
Our x is now an instance of Foo, without any attributes that is, and it can use any methods defined in Foo class.
If you want, you can create your own replacement function of __init__ for initializing attributes:
def init_foo(foo, name):
foo.name = name
>>> init_foo(x, "Mike")
>>> x.name
'Mike'
This could of course be Foo's instance method too:
class Foo(object):
def __init__(self):
raise TypeError("Cannot create 'Foo' instances.")
def init(self, name):
self.name = name
>>> x = object.__new__(Foo)
>>> x.init("Mike")
>>> x.name
'Mike'
Going even step further, you can even use a classmethod for creating your object with only one call:
class Foo(object):
def __init__(self):
raise TypeError("Cannot create 'Foo' instances.")
#classmethod
def new(cls, name):
obj = object.__new__(cls)
obj.name = name
return obj
>>> x = Foo.new("Mike")
>>> x.name
'Mike'