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Refer to a superclass from the class body

I've got some code where I need to refer to a superclass when defining stuff in a derived class:
class Base:
def foo(self):
print('foo')
def bar(self):
print('bar')
class Derived_A(Base):
meth = Base.foo
class Derived_B(Base):
meth = Base.bar
Derived_A().meth()
Derived_B().meth()
This works, but I don't like verbatim references to Base in derived classes. Is there a way to use super or alike for this?

You can't do that.
class keyword in Python is used to create classes which are instances of type type. In it's simplified version, it does the following:
Python creates a namespace and executes the body of the class in that namespace so that it will be populated with all methods and attributes and so on...
Then calls the three-arguments form of type(). The result of this call is your class which is then assign to a symbol which is the name of your class.
The point is when the body of the class is being executed. It doesn't know about the "bases". Those bases are passed to the type() after that.
I also explained the reasons why you can't use super() here.

Does this work for you?
class Base:
def foo(self):
print('foo')
def bar(self):
print('bar')
class Derived_A(Base):
def __init__(self):
self.meth = super().foo
class Derived_B(Base):
def __init__(self):
self.meth = super().bar
a = Derived_A().meth()
b = Derived_B().meth()

You'll need to lookup the method on the base class after the new type is created. In the body of the class definition, the type and base classes are not accessible.
Something like:
class Derived_A(Base):
def meth(self):
return super().foo()
Now, it is possible to do some magic behind the scenes to expose Base to the scope of the class definition as its being executed, but that's much dirtier, and would mean that you'd need to supply a metaclass in your class definition.

Since you want "magic", there is still one sane option we can take before diving into metaclasses. Requires Python 3.9+
def alias(name):
def inner(cls):
return getattr(cls, name).__get__(cls)
return classmethod(property(inner))
class Base:
def foo(self):
...
class Derived_A(Base):
meth = alias("foo")
Derived_A().meth() # works
Derived_A.meth() # also works
Yes, this does require passing the method name as a string, which destroys your IDE and typechecker's ability to reason about it. But there isn't a good way to get what you are wanting without some compromises like that.
Really, a bit of redundancy for readability is probably worth it here.

correct way of extending a class in python

I am given a designated factory of A-type objects. I would like to make a new version of A-type objects that also have the methods in a Mixin class. For reasons that are too long to explain here, I can't use class A(Mixin), I have to use the A_factory. Below I try to give a bare bones example.
I thought naively that it would be sufficient to inherit from Mixin to endow A-type objects with the mixin methods, but the attempts below don't work:
class A: pass
class A_factory:
def __new__(self):
return A()
class Mixin:
def method(self):
print('aha!')
class A_v2(Mixin): # attempt 1
def __new__(cls):
return A_factory()
class A_v3(Mixin): # attempt 2
def __new__(cls):
self = A_factory()
super().__init__(self)
return self
In fact A_v2().method() and A_v3().method() raises AttributeError: 'A' object has no attribute 'method'.
What is the correct way of using A_factory within class A_vn(Mixin) so that A-type objects created by the factory inherit the mixin methods?

There's no obvious reason why you should need __new__ for what you're showing here. There's a nice discussion here on the subject: Why is __init__() always called after __new__()?
If you try the below it should work:
class Mixin:
def method(self):
print('aha!')
class A(Mixin):
def __init__(self):
super().__init__()
test = A()
test.method()
If you need to use a factory method, it should be a function rather than a class. There's a very good discussion of how to use factory methods here: https://realpython.com/factory-method-python/

How to override a method used by a 3rd party library

This would be the layout
some_function.py
def some_function():
print("some_function")
some_library.py
from some_function import some_function
class A:
def xxx(self):
some_function()
main.py
from some_library import A
from some_function import some_function
def new_some_function():
print("new_some_function")
if __name__ == '__main__':
some_function = new_some_function
a = A()
a.xxx()
In the class A, the method xxx, calls some_function, so is it possible to override it with something else, without re-implementing the entire class?

I think you are looking for monkey patching (means changing classes/modules dynamically while running). This way you don't need to overwrite the class A and use inheritance as suggested by other comments - you said you don't want that, so try this solution:
import some_class # import like this or will not work (cos of namespaces)
def new_some_function():
print("new_some_function")
if __name__ == '__main__':
# Import and use like this, other ways of import will not work.
# Because other way imports method to your namespace and then change it in your namespace,
# but you need to change it in the original namespace
some_class.some_function = new_some_function
That way replace the original method and even other classes will use it then. Be careful, if the original method is a class/instance method, you need to create new function with proper params, like this:
def new_some_function(self):
# for instance methods, you may add other args, but 'self' is important
def new_some_function(cls):
# for class methods, you may add other args, but 'cls' is important

You provide very little information about your use case here. As one of the comments points out, this might be a case for inheritance. If you are in a testing context, you may not want to use inheritance though, but you might rather want to use a mock-object.
Here is the inheritance version:
from some_library import A
def new_some_function():
print("new_some_function")
class B(A):
def xxx(self):
new_some_function()
if __name__ == '__main__':
a = B()
a.xxx()
Note, how class B derives from class A through the class B(A) statement. This way, class B inherits all functionality from A and the definition of class B only consists of the parts where B differs from A. In your example, that is the fact that the xxx method should call new_some_function instead of some_function.
Here is the mock version:
from unittest import mock
from some_library import A
def new_some_function():
print("new_some_function")
if __name__ == '__main__':
with mock.patch('some_library.some_function') as mock_some_function:
mock_some_function.side_effect = new_some_function
a = A()
a.xxx()
As mentioned above this approach is mostly useful if you are in a testing context and if some_function does something costly and/or unpredictable. In order to test code that involves a call to some_function, you may temporarily want to replace some_function by something else, that is cheap to call and behaves in a predictable way. In fact, for this scenario, replacing some_function by new_some_function might even be more than what is actually needed. Maybe, you just want an empty hull that can be called and that always returns the same value (instead of the side_effect line, you can specify a constant .return_value in the above code example). One of the key functionalities of mock objects is that you can later check if that function has been called. If testing is your use case, I would very much recommend looking at the documentation of the python mock module.
Note that the example uses the mock.patch context manager. This means that within the managed context (i.e. the block inside the with-statement) some_library.some_function is replaced by a mock object, but once you leave the managed context, the original functionality is put back in place.

You may just create another class and override the method you need.
Just as an example:
class myInt(int):
def __pow__(self, x):
return 0
a = myInt(10)
a+10 # 20
a**2 # 0
In this case a is an int and has access to all the method of the int class, but will use the __pow__ method I've defined.

What you need is inheritance, You can subclass a class and with super method you can inherit all the parent class functions. If you want to override parent class functions, you just need to provide a different implementation by the same name in child class.
from some_library import A
from some_function import some_function
def new_some_function():
print("new_some_function")
class B(A):
def __init__(*args, **kwargs):
super().__init__(self)
pass
def xxx(self):
new_some_function()
if __name__ == '__main__':
a = B()
a.xxx()
Output:
new_some_function
you syntax may differ depending upon the python version.
In python3
class B(A):
def __init__(self):
super().__init__()
In Python 2,
class B(A):
def __init__(self):
super(ChildB, self).__init__()

How can I refer to the currently being defined class? [duplicate]

For a recursive function we can do:
def f(i):
if i<0: return
print i
f(i-1)
f(10)
However is there a way to do the following thing?
class A:
# do something
some_func(A)
# ...

If I understand your question correctly, you should be able to reference class A within class A by putting the type annotation in quotes. This is called forward reference.
class A:
# do something
def some_func(self, a: 'A')
# ...
See ref below
https://github.com/python/mypy/issues/3661
https://www.youtube.com/watch?v=AJsrxBkV3kc

In Python you cannot reference the class in the class body, although in languages like Ruby you can do it.
In Python instead you can use a class decorator but that will be called once the class has initialized. Another way could be to use metaclass but it depends on what you are trying to achieve.

You can't with the specific syntax you're describing due to the time at which they are evaluated. The reason the example function given works is that the call to f(i-1) within the function body is because the name resolution of f is not performed until the function is actually called. At this point f exists within the scope of execution since the function has already been evaluated. In the case of the class example, the reference to the class name is looked up during while the class definition is still being evaluated. As such, it does not yet exist in the local scope.
Alternatively, the desired behavior can be accomplished using a metaclass like such:
class MetaA(type):
def __init__(cls):
some_func(cls)
class A(object):
__metaclass__=MetaA
# do something
# ...
Using this approach you can perform arbitrary operations on the class object at the time that the class is evaluated.

Maybe you could try calling __class__.
Right now I'm writing a code that calls a class method from within the same class.
It is working well so far.
I'm creating the class methods using something like:
#classmethod
def my_class_method(cls):
return None
And calling then by using:
x = __class__.my_class_method()

It seems most of the answers here are outdated. From python3.7:
from __future__ import annotations
Example:
$ cat rec.py
from __future__ import annotations
class MyList:
def __init__(self,e):
self.data = [e]
def add(self, e):
self.data.append(e)
return self
def score(self, other:MyList):
return len([e
for e in self.data
if e in other.data])
print(MyList(8).add(3).add(4).score(MyList(4).add(9).add(3)))
$ python3.7 rec.py
2

Nope. It works in a function because the function contents are executed at call-time. But the class contents are executed at define-time, at which point the class doesn't exist yet.
It's not normally a problem because you can hack further members into the class after defining it, so you can split up a class definition into multiple parts:
class A(object):
spam= 1
some_func(A)
A.eggs= 2
def _A_scramble(self):
self.spam=self.eggs= 0
A.scramble= _A_scramble
It is, however, pretty unusual to want to call a function on the class in the middle of its own definition. It's not clear what you're trying to do, but chances are you'd be better off with decorators (or the relatively new class decorators).

There isn't a way to do that within the class scope, not unless A was defined to be something else first (and then some_func(A) will do something entirely different from what you expect)
Unless you're doing some sort of stack inspection to add bits to the class, it seems odd why you'd want to do that. Why not just:
class A:
# do something
pass
some_func(A)
That is, run some_func on A after it's been made. Alternately, you could use a class decorator (syntax for it was added in 2.6) or metaclass if you wanted to modify class A somehow. Could you clarify your use case?

If you want to do just a little hacky thing do
class A(object):
...
some_func(A)
If you want to do something more sophisticated you can use a metaclass. A metaclass is responsible for manipulating the class object before it gets fully created. A template would be:
class AType(type):
def __new__(meta, name, bases, dct):
cls = super(AType, meta).__new__(meta, name, bases, dct)
some_func(cls)
return cls
class A(object):
__metaclass__ = AType
...
type is the default metaclass. Instances of metaclasses are classes so __new__ returns a modified instance of (in this case) A.
For more on metaclasses, see http://docs.python.org/reference/datamodel.html#customizing-class-creation.

If the goal is to call a function some_func with the class as an argument, one answer is to declare some_func as a class decorator. Note that the class decorator is called after the class is initialized. It will be passed the class that is being decorated as an argument.
def some_func(cls):
# Do something
print(f"The answer is {cls.x}")
return cls # Don't forget to return the class
#some_func
class A:
x = 1
If you want to pass additional arguments to some_func you have to return a function from the decorator:
def some_other_func(prefix, suffix):
def inner(cls):
print(f"{prefix} {cls.__name__} {suffix}")
return cls
return inner
#some_other_func("Hello", " and goodbye!")
class B:
x = 2
Class decorators can be composed, which results in them being called in the reverse order they are declared:
#some_func
#some_other_func("Hello", "and goodbye!")
class C:
x = 42
The result of which is:
# Hello C and goodbye!
# The answer is 42

What do you want to achieve? It's possible to access a class to tweak its definition using a metaclass, but it's not recommended.
Your code sample can be written simply as:
class A(object):
pass
some_func(A)

If you want to refer to the same object, just use 'self':
class A:
def some_func(self):
another_func(self)
If you want to create a new object of the same class, just do it:
class A:
def some_func(self):
foo = A()
If you want to have access to the metaclass class object (most likely not what you want), again, just do it:
class A:
def some_func(self):
another_func(A) # note that it reads A, not A()

Do remember that in Python, type hinting is just for auto-code completion therefore it helps IDE to infer types and warn user before runtime. In runtime, type hints almost never used(except in some cases) so you can do something like this:
from typing import Any, Optional, NewType
LinkListType = NewType("LinkList", object)
class LinkList:
value: Any
_next: LinkListType
def set_next(self, ll: LinkListType):
self._next = ll
if __name__ == '__main__':
r = LinkList()
r.value = 1
r.set_next(ll=LinkList())
print(r.value)
And as you can see IDE successfully infers it's type as LinkList:
Note: Since the next can be None, hinting this in the type would be better, I just didn't want to confuse OP.
class LinkList:
value: Any
next: Optional[LinkListType]

It's ok to reference the name of the class inside its body (like inside method definitions) if it's actually in scope... Which it will be if it's defined at top level. (In other cases probably not, due to Python scoping quirks!).
For on illustration of the scoping gotcha, try to instantiate Foo:
class Foo(object):
class Bar(object):
def __init__(self):
self.baz = Bar.baz
baz = 15
def __init__(self):
self.bar = Foo.Bar()
(It's going to complain about the global name 'Bar' not being defined.)
Also, something tells me you may want to look into class methods: docs on the classmethod function (to be used as a decorator), a relevant SO question. Edit: Ok, so this suggestion may not be appropriate at all... It's just that the first thing I thought about when reading your question was stuff like alternative constructors etc. If something simpler suits your needs, steer clear of #classmethod weirdness. :-)

Most code in the class will be inside method definitions, in which case you can simply use the name A.

Applying a decorator to every method in a class?

I have decorator #login_testuser applied to method test_1():
class TestCase(object):
#login_testuser
def test_1(self):
print "test_1()"
Is there a way I can apply #login_testuser on every method of the class prefixed with "test_"?
In other words, the decorator would apply to test_1(), test_2() methods below, but not on setUp().
class TestCase(object):
def setUp(self):
pass
def test_1(self):
print "test_1()"
def test_2(self):
print "test_2()"

In Python 2.6, a class decorator is definitely the way to go. e.g., here's a pretty general one for these kind of tasks:
import inspect
def decallmethods(decorator, prefix='test_'):
def dectheclass(cls):
for name, m in inspect.getmembers(cls, inspect.isfunction):
if name.startswith(prefix):
setattr(cls, name, decorator(m))
return cls
return dectheclass
#decallmethods(login_testuser)
class TestCase(object):
def setUp(self):
pass
def test_1(self):
print("test_1()")
def test_2(self):
print("test_2()")
will get you what you desire. In Python 2.5 or worse, the #decallmethods syntax doesn't work for class decoration, but with otherwise exactly the same code you can replace it with the following statement right after the end of the class TestCase statement:
TestCase = decallmethods(login_testuser)(TestCase)

Sure. Iterate all attributes of the class. Check each one for being a method and if the name starts with "test_". Then replace it with the function returned from your decorator
Something like:
from inspect import ismethod, getmembers
for name, obj in getmembers(TestCase, ismethod):
if name.startswith("test_"):
setattr(TestCase, name, login_testuser(obj))

Are you sure you wouldn't be better off by putting login_testuser's code into setUp instead? That's what setUp is for: it's run before every test method.

Yes, you can loop over the class's dir/__dict__ or have a metaclass that does so, identifying if the attributes start with "test". However, this will create less straightforward, explicit code than writing the decorator explicitly.