Get class that defined method inside the method - python

How can I get the class that defined a method in Python?
For example
class A(object):
def meth(self):
return **get_current_class()**
class B(A):
def meth(self):
do_something
return super(B,self).meth()
>>> b=B()
>>> b.meth() ##that return the class A
Since b.__class__ is always the actual class of b(that is B),and what I want is the class which actual defined the method(that should be A),so self.__class__ is useless.

This is a bit of an odd thing to want to do, and you can't do it without getting extremely hacky. One possible solution is to realize that a method is a member of __dict__ on the class that defines it, so you can ascend the MRO to find the first one that has the method name in that dict:
def get_class_for_method(self, method_name):
for cls in self.__class__.mro():
if method_name in cls.__dict__:
return cls

To return the class that defined the method, the simplest way would be to just do it directly:
class A(object):
def meth(self):
return A
If meth is defined outside the class -- and presumably attached to more than one class -- it may not be feasible to hardcode the class in the body of meth. In that case, perhaps the easiest (only?) way for meth to know the class on which it is defined is to store that information in meth (as an attribute) at the time it is attached to the class.
Instead of attaching the function meth to the class A with
A.meth = meth
I propose using a function setmeth which performs both A.meth = meth and meth.thisclass = A:
def setmeth(cls, name, meth):
setattr(cls, name, meth)
setattr(meth, 'thisclass', cls)
def meth(self):
print(meth.thisclass)
class A(object): pass
setmeth(A, 'meth', meth)
class B(A):
def meth(self):
return super(B, self).meth()
b=B()
b.meth() ##that return the class A
prints
<class '__main__.A'>

Related

How to implement __getattr__() for class members (not instance members) [duplicate]

I have a class like:
class MyClass:
Foo = 1
Bar = 2
Whenever MyClass.Foo or MyClass.Bar is invoked, I need a custom method to be invoked before the value is returned. Is it possible in Python? I know it is possible if I create an instance of the class and I can define my own __getattr__ method. But my scnenario involves using this class as such without creating any instance of it.
Also I need a custom __str__ method to be invoked when str(MyClass.Foo) is invoked. Does Python provide such an option?
__getattr__() and __str__() for an object are found on its class, so if you want to customize those things for a class, you need the class-of-a-class. A metaclass.
class FooType(type):
def _foo_func(cls):
return 'foo!'
def _bar_func(cls):
return 'bar!'
def __getattr__(cls, key):
if key == 'Foo':
return cls._foo_func()
elif key == 'Bar':
return cls._bar_func()
raise AttributeError(key)
def __str__(cls):
return 'custom str for %s' % (cls.__name__,)
class MyClass:
__metaclass__ = FooType
# # in python 3:
# class MyClass(metaclass=FooType):
# pass
print(MyClass.Foo)
print(MyClass.Bar)
print(str(MyClass))
printing:
foo!
bar!
custom str for MyClass
And no, an object can't intercept a request for a stringifying one of its attributes. The object returned for the attribute must define its own __str__() behavior.
(I know this is an old question, but since all the other answers use a metaclass...)
You can use the following simple classproperty descriptor:
class classproperty(object):
""" #classmethod+#property """
def __init__(self, f):
self.f = classmethod(f)
def __get__(self, *a):
return self.f.__get__(*a)()
Use it like:
class MyClass(object):
#classproperty
def Foo(cls):
do_something()
return 1
#classproperty
def Bar(cls):
do_something_else()
return 2
For the first, you'll need to create a metaclass, and define __getattr__() on that.
class MyMetaclass(type):
def __getattr__(self, name):
return '%s result' % name
class MyClass(object):
__metaclass__ = MyMetaclass
print MyClass.Foo
For the second, no. Calling str(MyClass.Foo) invokes MyClass.Foo.__str__(), so you'll need to return an appropriate type for MyClass.Foo.
Surprised no one pointed this one out:
class FooType(type):
#property
def Foo(cls):
return "foo!"
#property
def Bar(cls):
return "bar!"
class MyClass(metaclass=FooType):
pass
Works:
>>> MyClass.Foo
'foo!'
>>> MyClass.Bar
'bar!'
(for Python 2.x, change definition of MyClass to:
class MyClass(object):
__metaclass__ = FooType
)
What the other answers say about str holds true for this solution: It must be implemented on the type actually returned.
Depending on the case I use this pattern
class _TheRealClass:
def __getattr__(self, attr):
pass
LooksLikeAClass = _TheRealClass()
Then you import and use it.
from foo import LooksLikeAClass
LooksLikeAClass.some_attribute
This avoid use of metaclass, and handle some use cases.

is there an equivalent for__getattr__ to intercept accesses to a python class's static attributes? [duplicate]

I have a class like:
class MyClass:
Foo = 1
Bar = 2
Whenever MyClass.Foo or MyClass.Bar is invoked, I need a custom method to be invoked before the value is returned. Is it possible in Python? I know it is possible if I create an instance of the class and I can define my own __getattr__ method. But my scnenario involves using this class as such without creating any instance of it.
Also I need a custom __str__ method to be invoked when str(MyClass.Foo) is invoked. Does Python provide such an option?
__getattr__() and __str__() for an object are found on its class, so if you want to customize those things for a class, you need the class-of-a-class. A metaclass.
class FooType(type):
def _foo_func(cls):
return 'foo!'
def _bar_func(cls):
return 'bar!'
def __getattr__(cls, key):
if key == 'Foo':
return cls._foo_func()
elif key == 'Bar':
return cls._bar_func()
raise AttributeError(key)
def __str__(cls):
return 'custom str for %s' % (cls.__name__,)
class MyClass(metaclass=FooType):
pass
# # in python 2:
# class MyClass:
# __metaclass__ = FooType
print(MyClass.Foo)
print(MyClass.Bar)
print(str(MyClass))
printing:
foo!
bar!
custom str for MyClass
And no, an object can't intercept a request for a stringifying one of its attributes. The object returned for the attribute must define its own __str__() behavior.
Updated 2023-02-20 for Python 3.x default implementation (python 2 as a comment).
(I know this is an old question, but since all the other answers use a metaclass...)
You can use the following simple classproperty descriptor:
class classproperty(object):
""" #classmethod+#property """
def __init__(self, f):
self.f = classmethod(f)
def __get__(self, *a):
return self.f.__get__(*a)()
Use it like:
class MyClass(object):
#classproperty
def Foo(cls):
do_something()
return 1
#classproperty
def Bar(cls):
do_something_else()
return 2
For the first, you'll need to create a metaclass, and define __getattr__() on that.
class MyMetaclass(type):
def __getattr__(self, name):
return '%s result' % name
class MyClass(object):
__metaclass__ = MyMetaclass
print MyClass.Foo
For the second, no. Calling str(MyClass.Foo) invokes MyClass.Foo.__str__(), so you'll need to return an appropriate type for MyClass.Foo.
Surprised no one pointed this one out:
class FooType(type):
#property
def Foo(cls):
return "foo!"
#property
def Bar(cls):
return "bar!"
class MyClass(metaclass=FooType):
pass
Works:
>>> MyClass.Foo
'foo!'
>>> MyClass.Bar
'bar!'
(for Python 2.x, change definition of MyClass to:
class MyClass(object):
__metaclass__ = FooType
)
What the other answers say about str holds true for this solution: It must be implemented on the type actually returned.
Depending on the case I use this pattern
class _TheRealClass:
def __getattr__(self, attr):
pass
LooksLikeAClass = _TheRealClass()
Then you import and use it.
from foo import LooksLikeAClass
LooksLikeAClass.some_attribute
This avoid use of metaclass, and handle some use cases.

__getattr__ on Exception subclass [duplicate]

I have a class like:
class MyClass:
Foo = 1
Bar = 2
Whenever MyClass.Foo or MyClass.Bar is invoked, I need a custom method to be invoked before the value is returned. Is it possible in Python? I know it is possible if I create an instance of the class and I can define my own __getattr__ method. But my scnenario involves using this class as such without creating any instance of it.
Also I need a custom __str__ method to be invoked when str(MyClass.Foo) is invoked. Does Python provide such an option?
__getattr__() and __str__() for an object are found on its class, so if you want to customize those things for a class, you need the class-of-a-class. A metaclass.
class FooType(type):
def _foo_func(cls):
return 'foo!'
def _bar_func(cls):
return 'bar!'
def __getattr__(cls, key):
if key == 'Foo':
return cls._foo_func()
elif key == 'Bar':
return cls._bar_func()
raise AttributeError(key)
def __str__(cls):
return 'custom str for %s' % (cls.__name__,)
class MyClass(metaclass=FooType):
pass
# # in python 2:
# class MyClass:
# __metaclass__ = FooType
print(MyClass.Foo)
print(MyClass.Bar)
print(str(MyClass))
printing:
foo!
bar!
custom str for MyClass
And no, an object can't intercept a request for a stringifying one of its attributes. The object returned for the attribute must define its own __str__() behavior.
Updated 2023-02-20 for Python 3.x default implementation (python 2 as a comment).
(I know this is an old question, but since all the other answers use a metaclass...)
You can use the following simple classproperty descriptor:
class classproperty(object):
""" #classmethod+#property """
def __init__(self, f):
self.f = classmethod(f)
def __get__(self, *a):
return self.f.__get__(*a)()
Use it like:
class MyClass(object):
#classproperty
def Foo(cls):
do_something()
return 1
#classproperty
def Bar(cls):
do_something_else()
return 2
For the first, you'll need to create a metaclass, and define __getattr__() on that.
class MyMetaclass(type):
def __getattr__(self, name):
return '%s result' % name
class MyClass(object):
__metaclass__ = MyMetaclass
print MyClass.Foo
For the second, no. Calling str(MyClass.Foo) invokes MyClass.Foo.__str__(), so you'll need to return an appropriate type for MyClass.Foo.
Surprised no one pointed this one out:
class FooType(type):
#property
def Foo(cls):
return "foo!"
#property
def Bar(cls):
return "bar!"
class MyClass(metaclass=FooType):
pass
Works:
>>> MyClass.Foo
'foo!'
>>> MyClass.Bar
'bar!'
(for Python 2.x, change definition of MyClass to:
class MyClass(object):
__metaclass__ = FooType
)
What the other answers say about str holds true for this solution: It must be implemented on the type actually returned.
Depending on the case I use this pattern
class _TheRealClass:
def __getattr__(self, attr):
pass
LooksLikeAClass = _TheRealClass()
Then you import and use it.
from foo import LooksLikeAClass
LooksLikeAClass.some_attribute
This avoid use of metaclass, and handle some use cases.

Python: How to register all child classes with the father class upon creation

I have python class trees, each made up of an abstract base class and many deriving concrete classes. I want all concrete classes to be accessible through a base-class method, and I do not want to specify anything during child-class creation.
This is what my imagined solution looks like:
class BaseClassA(object):
# <some magic code around here>
#classmethod
def getConcreteClasses(cls):
# <some magic related code here>
class ConcreteClassA1(BaseClassA):
# no magic-related code here
class ConcreteClassA2(BaseClassA):
# no magic-related code here
As much as possible, I'd prefer to write the "magic" once as a sort of design pattern. I want to be able to apply it to different class trees in different scenarios (i.e. add a similar tree with "BaseClassB" and its concrete classes).
Thanks Internet!
you can use meta classes for that:
class AutoRegister(type):
def __new__(mcs, name, bases, classdict):
new_cls = type.__new__(mcs, name, bases, classdict)
#print mcs, name, bases, classdict
for b in bases:
if hasattr(b, 'register_subclass'):
b.register_subclass(new_cls)
return new_cls
class AbstractClassA(object):
__metaclass__ = AutoRegister
_subclasses = []
#classmethod
def register_subclass(klass, cls):
klass._subclasses.append(cls)
#classmethod
def get_concrete_classes(klass):
return klass._subclasses
class ConcreteClassA1(AbstractClassA):
pass
class ConcreteClassA2(AbstractClassA):
pass
class ConcreteClassA3(ConcreteClassA2):
pass
print AbstractClassA.get_concrete_classes()
I'm personnaly very wary of this kind of magic. Don't put too much of this in your code.
Here is a simple solution using modern python's (3.6+) __init__subclass__ defined in PEP 487. It allows you to avoid using a meta-class.
class BaseClassA(object):
_subclasses = []
#classmethod
def get_concrete_classes(cls):
return list(cls._subclasses)
def __init_subclass__(cls):
BaseClassA._subclasses.append(cls)
class ConcreteClassA1(BaseClassA):
pass # no magic-related code here
class ConcreteClassA2(BaseClassA):
pass # no magic-related code here
print(BaseClassA.get_concrete_classes())
You should know that part of the answer you're looking for is built-in. New-style classes automatically keep a weak reference to all of their child classes which can be accessed with the __subclasses__ method:
#classmethod
def getConcreteClasses(cls):
return cls.__subclasses__()
This won't return sub-sub-classes. If you need those, you can create a recursive generator to get them all:
#classmethod
def getConcreteClasses(cls):
for c in cls.__subclasses__():
yield c
for c2 in c.getConcreteClasses():
yield c2
Another way to do this, with a decorator, if your subclasses are either not defining __init__ or are calling their parent's __init__:
def lister(cls):
cls.classes = list()
cls._init = cls.__init__
def init(self, *args, **kwargs):
cls = self.__class__
if cls not in cls.classes:
cls.classes.append(cls)
cls._init(self, *args, **kwargs)
cls.__init__ = init
#classmethod
def getclasses(cls):
return cls.classes
cls.getclasses = getclasses
return cls
#lister
class A(object): pass
class B(A): pass
class C(A):
def __init__(self):
super(C, self).__init__()
b = B()
c = C()
c2 = C()
print 'Classes:', c.getclasses()
It will work whether or not the base class defines __init__.

name of the class that contains the method code

I'm trying to find the name of the class that contains method code.
In the example underneath I use self.__class__.__name__, but of course this returns the name of the class of which self is an instance and not class that contains the test() method code. b.test() will print 'B' while I would like to get 'A'.
I looked into the inspect module documentation but did not find anything directly useful.
class A:
def __init__(self):
pass
def test(self):
print self.__class__.__name__
class B(A):
def __init__(self):
A.__init__(self)
a = A()
b = B()
a.test()
b.test()
In Python 3.x, you can simply use __class__.__name__. The __class__ name is mildly magic, and not the same thing as the __class__ attribute of self.
In Python 2.x, there is no good way to get at that information. You can use stack inspection to get the code object, then walk the class hierarchy looking for the right method, but it's slow and tedious and will probably break when you don't want it to. You can also use a metaclass or a class decorator to post-process the class in some way, but both of those are rather intrusive approaches. And you can do something really ugly, like accessing self.__nonexistant_attribute, catching the AttributeError and extracting the class name from the mangled name. None of those approaches are really worth it if you just want to avoid typing the name twice; at least forgetting to update the name can be made a little more obvious by doing something like:
class C:
...
def report_name(self):
print C.__name__
inspect.getmro gives you a tuple of the classes where the method might come from, in order. As soon as you find one of them that has the method's name in its dict, you're done:
for c in inspect.getmro(self.__class__):
if 'test' in vars(c): break
return c.__name__
Use __dict__ of class object itself:
class A(object):
def foo(self):
pass
class B(A):
pass
def find_decl_class(cls, method):
if method in cls.__dict__:
return cls
for b in cls.__bases__:
decl = find_decl_class(b, method)
if decl:
return decl
print 'foo' in A.__dict__
print 'foo' in B.__dict__
print find_decl_class(B, 'foo').__name__
Will print True, False, A
You can use (abuse?) private name mangling to accomplish this effect. If you look up an attribute on self that starts with __ from inside a method, python changes the name from __attribute to _classThisMethodWasDefinedIn__attribute.
Just somehow stash the classname you want in mangled-form where the method can see it. As an example, we can define a __new__ method on the base class that does it:
def mangle(cls, attrname):
if not attrname.startswith('__'):
raise ValueError('attrname must start with __')
return '_%s%s' % (cls.__name__, attrname)
class A(object):
def __new__(cls, *args, **kwargs):
obj = object.__new__(cls)
for c in cls.mro():
setattr(obj, mangle(c, '__defn_classname'), c.__name__)
return obj
def __init__(self):
pass
def test(self):
print self.__defn_classname
class B(A):
def __init__(self):
A.__init__(self)
a = A()
b = B()
a.test()
b.test()
which prints:
A
A
You can do
>>> class A(object):
... def __init__(self):
... pass
... def test(self):
... for b in self.__class__.__bases__:
... if hasattr(b, 'test'):
... return b.__name__
... return self.__class__.__name__
...
>>> class B(A):
... def __init__(self):
... A.__init__(self)
...
>>> B().test()
'A'
>>> A().test()
'A'
>>>
Keep in mind that you could simplify it by using __class__.__base__, but if you use multiple inheritance, this version will work better.
It simply checks first on its baseclasses for test. It's not the prettiest, but it works.

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