So I have this class that helps me override the update method of a queryset:
class QuerySetUpdateOverriden(QuerySet, object):
def update(self, *args, **kwargs):
super().update(*args, **kwargs)
if hasattr(self, 'method_from_object'):
self.method_from_object()
return
and here's my class where I use it:
class MyObject:
objects = QuerySetUpdateOverriden.as_manager()
def method_from_object(self):
print("called")
the print statement is never reached.
And I get why - the objects field doesn't inherit MyObject too.
So, the question is - how can I make it inherit MyObject so method_from_object will be called?
Thanks.
You test if self has a method called 'method_from_object', but your QuerySetUpdateOverriden has no method call like this. And MyObject does not inherit from QuerySetUpdateOverriden.
This code would be work i think:
class MyObjectManager(QuerySetUpdateOverriden.as_manager()):
def method_from_object(self):
print("called")
class MyObject(models.Model):
objects = QuerySetUpdateOverriden.as_manager()
Is it possible to skip classes in the method resolution order when calling for methods?
For example,
super().super()
I read the docs here: https://docs.python.org/3/library/functions.html#super
that lead me to this code
class A:
def show(self):
print("A")
class B(A):
def __init__(self):
print("B")
def s(self):
return super()
class C(B):
def __init__(self):
super().s().show()
c = C()
c
See that super returns a proxy object that seems not to have the super method (because i tried and the interpreter told me it doesn't). But you do have the others methods from the class, so this way I could get a proxy from its grandparent to use its methods
super() in the class definition for FooClass is shorthand for super(FooClass, self). Using this, we can do this:
class Grandparent:
def test(self):
print('grandparent gets called')
class Parent(Grandparent):
def test(self):
super().test()
print('parent gets skipped')
class Child(Parent):
def test(self):
super(Parent, self).test()
print('child gets called')
This "cheats" the MRO by checking the parent's superclass instead of the child's superclass.
Lets define simple class decorator function, which creates subclass and adds 'Dec' to original class name only:
def decorate_class(klass):
new_class = type(klass.__name__ + 'Dec', (klass,), {})
return new_class
Now apply it on a simple subclass definition:
class Base(object):
def __init__(self):
print 'Base init'
#decorate_class
class MyClass(Base):
def __init__(self):
print 'MyClass init'
super(MyClass, self).__init__()
Now, if you try instantiate decorated MyClass, it will end up in an infinite loop:
c = MyClass()
# ...
# File "test.py", line 40, in __init__
# super(MyClass, self).__init__()
# RuntimeError: maximum recursion depth exceeded while calling a Python object
It seems, super can't handle this case and does not skip current class from inheritance chain.
The question, how correctly use class decorator on classes using super ?
Bonus question, how get final class from proxy-object created by super ? Ie. get object class from super(Base, self).__init__ expression, as determined parent class defining called __init__.
If you just want to change the class's .__name__ attribute, make a decorator that does that.
from __future__ import print_function
def decorate_class(klass):
klass.__name__ += 'Dec'
return klass
class Base(object):
def __init__(self):
print('Base init')
#decorate_class
class MyClass(Base):
def __init__(self):
print('MyClass init')
super(MyClass, self).__init__()
c = MyClass()
cls = c.__class__
print(cls, cls.__name__)
Python 2 output
MyClass init
Base init
<class '__main__.MyClassDec'> MyClassDec
Python 3 output
MyClass init
Base init
<class '__main__.MyClass'> MyClassDec
Note the difference in the repr of cls. (I'm not sure why you'd want to change a class's name though, it sounds like a recipe for confusion, but I guess it's ok for this simple example).
As others have said, an #decorator isn't intended to create a subclass. You can do it in Python 3 by using the arg-less form of super (i.e., super().__init__()). And you can make it work in both Python 3 and Python 2 by explicitly supplying the parent class rather than using super.
from __future__ import print_function
def decorate_class(klass):
name = klass.__name__
return type(name + 'Dec', (klass,), {})
class Base(object):
def __init__(self):
print('Base init')
#decorate_class
class MyClass(Base):
def __init__(self):
print('MyClass init')
Base.__init__(self)
c = MyClass()
cls = c.__class__
print(cls, cls.__name__)
Python 2 & 3 output
MyClass init
Base init
<class '__main__.MyClassDec'> MyClassDec
Finally, if we just call decorate_class using normal function syntax rather than as an #decorator we can use super.
from __future__ import print_function
def decorate_class(klass):
name = klass.__name__
return type(name + 'Dec', (klass,), {})
class Base(object):
def __init__(self):
print('Base init')
class MyClass(Base):
def __init__(self):
print('MyClass init')
super(MyClass, self).__init__()
MyClassDec = decorate_class(MyClass)
c = MyClassDec()
cls = c.__class__
print(cls, cls.__name__)
The output is the same as in the last version.
Since your decorator returns an entirely new class with different name, for that class MyClass object doesn't even exist. This is not the case class decorators are intended for. They are intended to add additional functionality to an existing class, not outright replacing it with some other class.
Still if you are using Python3, solution is simple -
#decorate_class
class MyClass(Base):
def __init__(self):
print 'MyClass init'
super().__init__()
Otherwise, I doubt there is any straight-forward solution, you just need to change your implementation. When you are renaming the class, you need to rewrite overwrite __init__ as well with newer name.
The problem is that your decorator creates a subclass of the original one. That means that super(Myclass) now point to... the original class itself!
I cannot even explain how the 0 arg form of super manages to do the job in Python 3, I could not find anything explicit in the reference manual. I assume it must use the class in which it is used at the time of declaration. But I cannot imagine a way to get that result in Python2.
If you want to be able to use super in the decorated class in Python 2, you should not create a derived class, but directly modify the original class in place.
For example, here is a decorator that prints a line before and after calling any method:
def decorate_class(klass):
for name, method in klass.__dict__.iteritems(): # iterate the class attributes
if isinstance(method, types.FunctionType): # identify the methods
def meth(*args, **kwargs): # define a wrapper
print "Before", name
method(*args, **kwargs)
print "After", name
setattr(klass, name, meth) # tell the class to use the wrapper
return klass
With your example it gives as expected:
>>> c = MyClass()
Before __init__
MyClass init
Base init
After __init__
I have a python abstract base class as follows:
class Node(object):
"""
All concrete node classes should inherit from this
"""
__metaclass__ = ABCMeta
def __init__(self, name):
self.name = name
self.inputs = dict()
def add_input(self, key, value=None, d=None):
self.inputs[key] = (d, value)
def bind_input(self):
print "Binding inputs"
#abstractmethod
def run(self):
pass
Now, various derived classes will inherit from this node class and override the run method. It is always the case that bind_input() must be the first thing that should be called in the run method. Currently, for all derived classes the developer has to make sure to first call self.bind_input(). This is not a huge problem per se but out of curiosity is it possible to ensure this somehow from the base class itself that bind_input is called before executing the child object's run?
The usual object-oriented approach is this:
def run(self):
self.bind_input()
return self.do_run()
#abstractmethod
def do_run(self):
pass # override this method
Have your subclasses override the inner method, instead of the outer one.
If I make a python class which overrides another class, does __init__ get called from base class? What if I want to specify arguments for it?
Base class:
class bar(object):
def __init__(self, somearg = False):
# ...
New class
class foo(bar):
def __init__(self)
# ???
What do I want?
obj = foo() # somearg = True
No, the base class __init__ method is not called, since your derived class provides a new version of the method.
You'd call the parent __init__ method explicitly through the super() proxy and pass in the argument to set a different default:
class foo(bar):
def __init__(self)
super().__init__(somearg=True)