Develop Reference

How do I get the class which the function belongs to? [duplicate]

How can I get the class that defined a method in Python?
I'd want the following example to print "__main__.FooClass":
class FooClass:
def foo_method(self):
print "foo"
class BarClass(FooClass):
pass
bar = BarClass()
print get_class_that_defined_method(bar.foo_method)

import inspect
def get_class_that_defined_method(meth):
for cls in inspect.getmro(meth.im_class):
if meth.__name__ in cls.__dict__:
return cls
return None

I don't know why no one has ever brought this up or why the top answer has 50 upvotes when it is slow as hell, but you can also do the following:
def get_class_that_defined_method(meth):
return meth.im_class.__name__
For python 3 I believe this changed and you'll need to look into .__qualname__.

In Python 3, if you need the actual class object you can do:
import sys
f = Foo.my_function
vars(sys.modules[f.__module__])[f.__qualname__.split('.')[0]] # Gets Foo object
If the function could belong to a nested class you would need to iterate as follows:
f = Foo.Bar.my_function
vals = vars(sys.modules[f.__module__])
for attr in f.__qualname__.split('.')[:-1]:
vals = vals[attr]
# vals is now the class Foo.Bar

Thanks Sr2222 for pointing out I was missing the point...
Here's the corrected approach which is just like Alex's but does not require to import anything. I don't think it's an improvement though, unless there's a huge hierarchy of inherited classes as this approach stops as soon as the defining class is found, instead of returning the whole inheritance as getmro does. As said, this is a very unlikely scenario.
def get_class_that_defined_method(method):
method_name = method.__name__
if method.__self__:
classes = [method.__self__.__class__]
else:
#unbound method
classes = [method.im_class]
while classes:
c = classes.pop()
if method_name in c.__dict__:
return c
else:
classes = list(c.__bases__) + classes
return None
And the Example:
>>> class A(object):
... def test(self): pass
>>> class B(A): pass
>>> class C(B): pass
>>> class D(A):
... def test(self): print 1
>>> class E(D,C): pass
>>> get_class_that_defined_method(A().test)
<class '__main__.A'>
>>> get_class_that_defined_method(A.test)
<class '__main__.A'>
>>> get_class_that_defined_method(B.test)
<class '__main__.A'>
>>> get_class_that_defined_method(C.test)
<class '__main__.A'>
>>> get_class_that_defined_method(D.test)
<class '__main__.D'>
>>> get_class_that_defined_method(E().test)
<class '__main__.D'>
>>> get_class_that_defined_method(E.test)
<class '__main__.D'>
>>> E().test()
1
Alex solution returns the same results. As long as Alex approach can be used, I would use it instead of this one.

Python 3
Solved it in a very simple way:
str(bar.foo_method).split(" ", 3)[-2]
This gives
'FooClass.foo_method'
Split on the dot to get the class and the function name separately

I found __qualname__ is useful in Python3.
I test it like that:
class Cls(object):
def func(self):
print('1')
c = Cls()
print(c.func.__qualname__)
# output is: 'Cls.func'
def single_func():
print(2)
print(single_func.__module__)
# output: '__main__'
print(single_func.__qualname__)
# output: 'single_func'
After my test, I found another answer here.

I started doing something somewhat similar, basically the idea was checking whenever a method in a base class had been implemented or not in a sub class. Turned out the way I originally did it I could not detect when an intermediate class was actually implementing the method.
My workaround for it was quite simple actually; setting a method attribute and testing its presence later. Here's an simplification of the whole thing:
class A():
def method(self):
pass
method._orig = None # This attribute will be gone once the method is implemented
def run_method(self, *args, **kwargs):
if hasattr(self.method, '_orig'):
raise Exception('method not implemented')
self.method(*args, **kwargs)
class B(A):
pass
class C(B):
def method(self):
pass
class D(C):
pass
B().run_method() # ==> Raises Exception: method not implemented
C().run_method() # OK
D().run_method() # OK
UPDATE: Actually call method() from run_method() (isn't that the spirit?) and have it pass all arguments unmodified to the method.
P.S.: This answer does not directly answer the question. IMHO there are two reasons one would want to know which class defined a method; first is to point fingers at a class in debug code (such as in exception handling), and the second is to determine if the method has been re-implemented (where method is a stub meant to be implemented by the programmer). This answer solves that second case in a different way.

if you get this error:
'function' object has no attribute 'im_class'
try this:
import inspect
def get_class_that_defined_method(meth):
class_func_defided = meth.__globals__[meth.__qualname__.split('.')[0]]
#full_func_name = "%s.%s.%s"%(class_func_defided.__module__,class_func_defided.__name__,meth.__name__)
if inspect.isfunction(class_func_defided):
print("%s is not part of a class."%meth.__name__)
return None
return class_func_defided
sample test:
class ExampleClass:
#staticmethod
def ex_static_method():
print("hello from static method")
def ex_instance_method(self):
print("hello from instance method")
def ex_funct(self):
print("hello from simple function")
if __name__ == "__main__":
static_method_class = get_class_that_defined_method(ExampleClass.ex_static_method)
static_method_class.ex_static_method()
instance_method_class = get_class_that_defined_method(ExampleClass.ex_instance_method)
instance_method_class().ex_instance_method()
function_class = get_class_that_defined_method(ex_funct)

Python generate a list of classes from folders and subfolders only of a certain type [duplicate]

I need a working approach of getting all classes that are inherited from a base class in Python.

New-style classes (i.e. subclassed from object, which is the default in Python 3) have a __subclasses__ method which returns the subclasses:
class Foo(object): pass
class Bar(Foo): pass
class Baz(Foo): pass
class Bing(Bar): pass
Here are the names of the subclasses:
print([cls.__name__ for cls in Foo.__subclasses__()])
# ['Bar', 'Baz']
Here are the subclasses themselves:
print(Foo.__subclasses__())
# [<class '__main__.Bar'>, <class '__main__.Baz'>]
Confirmation that the subclasses do indeed list Foo as their base:
for cls in Foo.__subclasses__():
print(cls.__base__)
# <class '__main__.Foo'>
# <class '__main__.Foo'>
Note if you want subsubclasses, you'll have to recurse:
def all_subclasses(cls):
return set(cls.__subclasses__()).union(
[s for c in cls.__subclasses__() for s in all_subclasses(c)])
print(all_subclasses(Foo))
# {<class '__main__.Bar'>, <class '__main__.Baz'>, <class '__main__.Bing'>}
Note that if the class definition of a subclass hasn't been executed yet - for example, if the subclass's module hasn't been imported yet - then that subclass doesn't exist yet, and __subclasses__ won't find it.
You mentioned "given its name". Since Python classes are first-class objects, you don't need to use a string with the class's name in place of the class or anything like that. You can just use the class directly, and you probably should.
If you do have a string representing the name of a class and you want to find that class's subclasses, then there are two steps: find the class given its name, and then find the subclasses with __subclasses__ as above.
How to find the class from the name depends on where you're expecting to find it. If you're expecting to find it in the same module as the code that's trying to locate the class, then
cls = globals()[name]
would do the job, or in the unlikely case that you're expecting to find it in locals,
cls = locals()[name]
If the class could be in any module, then your name string should contain the fully-qualified name - something like 'pkg.module.Foo' instead of just 'Foo'. Use importlib to load the class's module, then retrieve the corresponding attribute:
import importlib
modname, _, clsname = name.rpartition('.')
mod = importlib.import_module(modname)
cls = getattr(mod, clsname)
However you find the class, cls.__subclasses__() would then return a list of its subclasses.

If you just want direct subclasses then .__subclasses__() works fine. If you want all subclasses, subclasses of subclasses, and so on, you'll need a function to do that for you.
Here's a simple, readable function that recursively finds all subclasses of a given class:
def get_all_subclasses(cls):
all_subclasses = []
for subclass in cls.__subclasses__():
all_subclasses.append(subclass)
all_subclasses.extend(get_all_subclasses(subclass))
return all_subclasses

The simplest solution in general form:
def get_subclasses(cls):
for subclass in cls.__subclasses__():
yield from get_subclasses(subclass)
yield subclass
And a classmethod in case you have a single class where you inherit from:
#classmethod
def get_subclasses(cls):
for subclass in cls.__subclasses__():
yield from subclass.get_subclasses()
yield subclass

Python 3.6 - __init_subclass__
As other answer mentioned you can check the __subclasses__ attribute to get the list of subclasses, since python 3.6 you can modify this attribute creation by overriding the __init_subclass__ method.
class PluginBase:
subclasses = []
def __init_subclass__(cls, **kwargs):
super().__init_subclass__(**kwargs)
cls.subclasses.append(cls)
class Plugin1(PluginBase):
pass
class Plugin2(PluginBase):
pass
This way, if you know what you're doing, you can override the behavior of of __subclasses__ and omit/add subclasses from this list.

Note: I see that someone (not #unutbu) changed the referenced answer so that it no longer uses vars()['Foo'] — so the primary point of my post no longer applies.
FWIW, here's what I meant about #unutbu's answer only working with locally defined classes — and that using eval() instead of vars() would make it work with any accessible class, not only those defined in the current scope.
For those who dislike using eval(), a way is also shown to avoid it.
First here's a concrete example demonstrating the potential problem with using vars():
class Foo(object): pass
class Bar(Foo): pass
class Baz(Foo): pass
class Bing(Bar): pass
# unutbu's approach
def all_subclasses(cls):
return cls.__subclasses__() + [g for s in cls.__subclasses__()
for g in all_subclasses(s)]
print(all_subclasses(vars()['Foo'])) # Fine because Foo is in scope
# -> [<class '__main__.Bar'>, <class '__main__.Baz'>, <class '__main__.Bing'>]
def func(): # won't work because Foo class is not locally defined
print(all_subclasses(vars()['Foo']))
try:
func() # not OK because Foo is not local to func()
except Exception as e:
print('calling func() raised exception: {!r}'.format(e))
# -> calling func() raised exception: KeyError('Foo',)
print(all_subclasses(eval('Foo'))) # OK
# -> [<class '__main__.Bar'>, <class '__main__.Baz'>, <class '__main__.Bing'>]
# using eval('xxx') instead of vars()['xxx']
def func2():
print(all_subclasses(eval('Foo')))
func2() # Works
# -> [<class '__main__.Bar'>, <class '__main__.Baz'>, <class '__main__.Bing'>]
This could be improved by moving the eval('ClassName') down into the function defined, which makes using it easier without loss of the additional generality gained by using eval() which unlike vars() is not context-sensitive:
# easier to use version
def all_subclasses2(classname):
direct_subclasses = eval(classname).__subclasses__()
return direct_subclasses + [g for s in direct_subclasses
for g in all_subclasses2(s.__name__)]
# pass 'xxx' instead of eval('xxx')
def func_ez():
print(all_subclasses2('Foo')) # simpler
func_ez()
# -> [<class '__main__.Bar'>, <class '__main__.Baz'>, <class '__main__.Bing'>]
Lastly, it's possible, and perhaps even important in some cases, to avoid using eval() for security reasons, so here's a version without it:
def get_all_subclasses(cls):
""" Generator of all a class's subclasses. """
try:
for subclass in cls.__subclasses__():
yield subclass
for subclass in get_all_subclasses(subclass):
yield subclass
except TypeError:
return
def all_subclasses3(classname):
for cls in get_all_subclasses(object): # object is base of all new-style classes.
if cls.__name__.split('.')[-1] == classname:
break
else:
raise ValueError('class %s not found' % classname)
direct_subclasses = cls.__subclasses__()
return direct_subclasses + [g for s in direct_subclasses
for g in all_subclasses3(s.__name__)]
# no eval('xxx')
def func3():
print(all_subclasses3('Foo'))
func3() # Also works
# -> [<class '__main__.Bar'>, <class '__main__.Baz'>, <class '__main__.Bing'>]

Here is a simple but efficient version of code:
def get_all_subclasses(cls):
subclass_list = []
def recurse(klass):
for subclass in klass.__subclasses__():
subclass_list.append(subclass)
recurse(subclass)
recurse(cls)
return set(subclass_list)
Its time complexity is O(n) where n is the number of all subclasses if there's no multiple inheritance.
It's more efficient than the functions that recursively create lists or yield classes with generators, whose complexity could be (1) O(nlogn) when the class hierarchy is a balanced tree or (2) O(n^2) when the class hierarchy is a biased tree.

A much shorter version for getting a list of all subclasses:
from itertools import chain
def subclasses(cls):
return list(
chain.from_iterable(
[list(chain.from_iterable([[x], subclasses(x)])) for x in cls.__subclasses__()]
)
)

Here's a version without recursion:
def get_subclasses_gen(cls):
def _subclasses(classes, seen):
while True:
subclasses = sum((x.__subclasses__() for x in classes), [])
yield from classes
yield from seen
found = []
if not subclasses:
return
classes = subclasses
seen = found
return _subclasses([cls], [])
This differs from other implementations in that it returns the original class.
This is because it makes the code simpler and:
class Ham(object):
pass
assert(issubclass(Ham, Ham)) # True
If get_subclasses_gen looks a bit weird that's because it was created by converting a tail-recursive implementation into a looping generator:
def get_subclasses(cls):
def _subclasses(classes, seen):
subclasses = sum(*(frozenset(x.__subclasses__()) for x in classes))
found = classes + seen
if not subclasses:
return found
return _subclasses(subclasses, found)
return _subclasses([cls], [])

How can I find all subclasses of a class given its name?
We can certainly easily do this given access to the object itself, yes.
Simply given its name is a poor idea, as there can be multiple classes of the same name, even defined in the same module.
I created an implementation for another answer, and since it answers this question and it's a little more elegant than the other solutions here, here it is:
def get_subclasses(cls):
"""returns all subclasses of argument, cls"""
if issubclass(cls, type):
subclasses = cls.__subclasses__(cls)
else:
subclasses = cls.__subclasses__()
for subclass in subclasses:
subclasses.extend(get_subclasses(subclass))
return subclasses
Usage:
>>> import pprint
>>> list_of_classes = get_subclasses(int)
>>> pprint.pprint(list_of_classes)
[<class 'bool'>,
<enum 'IntEnum'>,
<enum 'IntFlag'>,
<class 'sre_constants._NamedIntConstant'>,
<class 'subprocess.Handle'>,
<enum '_ParameterKind'>,
<enum 'Signals'>,
<enum 'Handlers'>,
<enum 'RegexFlag'>]

This isn't as good an answer as using the special built-in __subclasses__() class method which #unutbu mentions, so I present it merely as an exercise. The subclasses() function defined returns a dictionary which maps all the subclass names to the subclasses themselves.
def traced_subclass(baseclass):
class _SubclassTracer(type):
def __new__(cls, classname, bases, classdict):
obj = type(classname, bases, classdict)
if baseclass in bases: # sanity check
attrname = '_%s__derived' % baseclass.__name__
derived = getattr(baseclass, attrname, {})
derived.update( {classname:obj} )
setattr(baseclass, attrname, derived)
return obj
return _SubclassTracer
def subclasses(baseclass):
attrname = '_%s__derived' % baseclass.__name__
return getattr(baseclass, attrname, None)
class BaseClass(object):
pass
class SubclassA(BaseClass):
__metaclass__ = traced_subclass(BaseClass)
class SubclassB(BaseClass):
__metaclass__ = traced_subclass(BaseClass)
print subclasses(BaseClass)
Output:
{'SubclassB': <class '__main__.SubclassB'>,
'SubclassA': <class '__main__.SubclassA'>}

While I'm very partial to the __init_subclass__ approach, this will preserve definition order, and avoid combinatorial order of growth if you have a very dense hierarchy with multiple inheritance everywhere:
def descendents(cls):
'''Does not return the class itself'''
R = {}
def visit(cls):
for subCls in cls.__subclasses__():
R[subCls] = True
visit(subCls)
visit(cls)
return list(R.keys())
This works because dictionaries remember the insertion order of their keys. A list approach would also work.

how to get the class name of a statics method belongs to it [duplicate]

How can I get the class that defined a method in Python?
I'd want the following example to print "__main__.FooClass":
class FooClass:
def foo_method(self):
print "foo"
class BarClass(FooClass):
pass
bar = BarClass()
print get_class_that_defined_method(bar.foo_method)

import inspect
def get_class_that_defined_method(meth):
for cls in inspect.getmro(meth.im_class):
if meth.__name__ in cls.__dict__:
return cls
return None

I don't know why no one has ever brought this up or why the top answer has 50 upvotes when it is slow as hell, but you can also do the following:
def get_class_that_defined_method(meth):
return meth.im_class.__name__
For python 3 I believe this changed and you'll need to look into .__qualname__.

In Python 3, if you need the actual class object you can do:
import sys
f = Foo.my_function
vars(sys.modules[f.__module__])[f.__qualname__.split('.')[0]] # Gets Foo object
If the function could belong to a nested class you would need to iterate as follows:
f = Foo.Bar.my_function
vals = vars(sys.modules[f.__module__])
for attr in f.__qualname__.split('.')[:-1]:
vals = vals[attr]
# vals is now the class Foo.Bar

Thanks Sr2222 for pointing out I was missing the point...
Here's the corrected approach which is just like Alex's but does not require to import anything. I don't think it's an improvement though, unless there's a huge hierarchy of inherited classes as this approach stops as soon as the defining class is found, instead of returning the whole inheritance as getmro does. As said, this is a very unlikely scenario.
def get_class_that_defined_method(method):
method_name = method.__name__
if method.__self__:
classes = [method.__self__.__class__]
else:
#unbound method
classes = [method.im_class]
while classes:
c = classes.pop()
if method_name in c.__dict__:
return c
else:
classes = list(c.__bases__) + classes
return None
And the Example:
>>> class A(object):
... def test(self): pass
>>> class B(A): pass
>>> class C(B): pass
>>> class D(A):
... def test(self): print 1
>>> class E(D,C): pass
>>> get_class_that_defined_method(A().test)
<class '__main__.A'>
>>> get_class_that_defined_method(A.test)
<class '__main__.A'>
>>> get_class_that_defined_method(B.test)
<class '__main__.A'>
>>> get_class_that_defined_method(C.test)
<class '__main__.A'>
>>> get_class_that_defined_method(D.test)
<class '__main__.D'>
>>> get_class_that_defined_method(E().test)
<class '__main__.D'>
>>> get_class_that_defined_method(E.test)
<class '__main__.D'>
>>> E().test()
1
Alex solution returns the same results. As long as Alex approach can be used, I would use it instead of this one.

Python 3
Solved it in a very simple way:
str(bar.foo_method).split(" ", 3)[-2]
This gives
'FooClass.foo_method'
Split on the dot to get the class and the function name separately

I found __qualname__ is useful in Python3.
I test it like that:
class Cls(object):
def func(self):
print('1')
c = Cls()
print(c.func.__qualname__)
# output is: 'Cls.func'
def single_func():
print(2)
print(single_func.__module__)
# output: '__main__'
print(single_func.__qualname__)
# output: 'single_func'
After my test, I found another answer here.

I started doing something somewhat similar, basically the idea was checking whenever a method in a base class had been implemented or not in a sub class. Turned out the way I originally did it I could not detect when an intermediate class was actually implementing the method.
My workaround for it was quite simple actually; setting a method attribute and testing its presence later. Here's an simplification of the whole thing:
class A():
def method(self):
pass
method._orig = None # This attribute will be gone once the method is implemented
def run_method(self, *args, **kwargs):
if hasattr(self.method, '_orig'):
raise Exception('method not implemented')
self.method(*args, **kwargs)
class B(A):
pass
class C(B):
def method(self):
pass
class D(C):
pass
B().run_method() # ==> Raises Exception: method not implemented
C().run_method() # OK
D().run_method() # OK
UPDATE: Actually call method() from run_method() (isn't that the spirit?) and have it pass all arguments unmodified to the method.
P.S.: This answer does not directly answer the question. IMHO there are two reasons one would want to know which class defined a method; first is to point fingers at a class in debug code (such as in exception handling), and the second is to determine if the method has been re-implemented (where method is a stub meant to be implemented by the programmer). This answer solves that second case in a different way.

if you get this error:
'function' object has no attribute 'im_class'
try this:
import inspect
def get_class_that_defined_method(meth):
class_func_defided = meth.__globals__[meth.__qualname__.split('.')[0]]
#full_func_name = "%s.%s.%s"%(class_func_defided.__module__,class_func_defided.__name__,meth.__name__)
if inspect.isfunction(class_func_defided):
print("%s is not part of a class."%meth.__name__)
return None
return class_func_defided
sample test:
class ExampleClass:
#staticmethod
def ex_static_method():
print("hello from static method")
def ex_instance_method(self):
print("hello from instance method")
def ex_funct(self):
print("hello from simple function")
if __name__ == "__main__":
static_method_class = get_class_that_defined_method(ExampleClass.ex_static_method)
static_method_class.ex_static_method()
instance_method_class = get_class_that_defined_method(ExampleClass.ex_instance_method)
instance_method_class().ex_instance_method()
function_class = get_class_that_defined_method(ex_funct)

Python: check if method is static

assume following class definition:
class A:
def f(self):
return 'this is f'
#staticmethod
def g():
return 'this is g'
a = A()
So f is a normal method and g is a static method.
Now, how can I check if the funcion objects a.f and a.g are static or not? Is there a "isstatic" funcion in Python?
I have to know this because I have lists containing many different function (method) objects, and to call them I have to know if they are expecting "self" as a parameter or not.

Lets experiment a bit:
>>> import types
>>> class A:
... def f(self):
... return 'this is f'
... #staticmethod
... def g():
... return 'this is g'
...
>>> a = A()
>>> a.f
<bound method A.f of <__main__.A instance at 0x800f21320>>
>>> a.g
<function g at 0x800eb28c0>
>>> isinstance(a.g, types.FunctionType)
True
>>> isinstance(a.f, types.FunctionType)
False
So it looks like you can use types.FunctionType to distinguish static methods.

Your approach seems a bit flawed to me, but you can check class attributes:
(in Python 2.7):
>>> type(A.f)
<type 'instancemethod'>
>>> type(A.g)
<type 'function'>
or instance attributes in Python 3.x
>>> a = A()
>>> type(a.f)
<type 'method'>
>>> type(a.g)
<type 'function'>

To supplement the answers here, in Python 3 the best way is like so:
import inspect
class Test:
#staticmethod
def test(): pass
isstatic = isinstance(inspect.getattr_static(Test, "test"), staticmethod)
We use getattr_static rather than getattr, since getattr will retrieve the bound method or function, not the staticmethod class object. You can do a similar check for classmethod types and property's (e.g. attributes defined using the #property decorator)
Note that even though it is a staticmethod, don't assume it was defined inside the class. The method source may have originated from another class. To get the true source, you can look at the underlying function's qualified name and module. For example:
class A:
#staticmethod:
def test(): pass
class B: pass
B.test = inspect.getattr_static(A, "test")
print("true source: ", B.test.__qualname__)
Technically, any method can be used as "static" methods, so long as they are called on the class itself, so just keep that in mind. For example, this will work perfectly fine:
class Test:
def test():
print("works!")
Test.test()
That example will not work with instances of Test, since the method will be bound to the instance and called as Test.test(self) instead.
Instance and class methods can be used as static methods as well in some cases, so long as the first arg is handled properly.
class Test:
def test(self):
print("works!")
Test.test(None)
Perhaps another rare case is a staticmethod that is also bound to a class or instance. For example:
class Test:
#classmethod
def test(cls): pass
Test.static_test = staticmethod(Test.test)
Though technically it is a staticmethod, it is really behaving like a classmethod. So in your introspection, you may consider checking the __self__ (recursively on __func__) to see if the method is bound to a class or instance.

I happens to have a module to solve this. And it's Python2/3 compatible solution. And it allows to test with method inherit from parent class.
Plus, this module can also test:
regular attribute
property style method
regular method
staticmethod
classmethod
For example:
class Base(object):
attribute = "attribute"
#property
def property_method(self):
return "property_method"
def regular_method(self):
return "regular_method"
#staticmethod
def static_method():
return "static_method"
#classmethod
def class_method(cls):
return "class_method"
class MyClass(Base):
pass
Here's the solution for staticmethod only. But I recommend to use the module posted here.
import inspect
def is_static_method(klass, attr, value=None):
"""Test if a value of a class is static method.
example::
class MyClass(object):
#staticmethod
def method():
...
:param klass: the class
:param attr: attribute name
:param value: attribute value
"""
if value is None:
value = getattr(klass, attr)
assert getattr(klass, attr) == value
for cls in inspect.getmro(klass):
if inspect.isroutine(value):
if attr in cls.__dict__:
bound_value = cls.__dict__[attr]
if isinstance(bound_value, staticmethod):
return True
return False

Why bother? You can just call g like you call f:
a = A()
a.f()
a.g()

Get class that defined method

How can I get the class that defined a method in Python?
I'd want the following example to print "__main__.FooClass":
class FooClass:
def foo_method(self):
print "foo"
class BarClass(FooClass):
pass
bar = BarClass()
print get_class_that_defined_method(bar.foo_method)

import inspect
def get_class_that_defined_method(meth):
for cls in inspect.getmro(meth.im_class):
if meth.__name__ in cls.__dict__:
return cls
return None

I don't know why no one has ever brought this up or why the top answer has 50 upvotes when it is slow as hell, but you can also do the following:
def get_class_that_defined_method(meth):
return meth.im_class.__name__
For python 3 I believe this changed and you'll need to look into .__qualname__.

In Python 3, if you need the actual class object you can do:
import sys
f = Foo.my_function
vars(sys.modules[f.__module__])[f.__qualname__.split('.')[0]] # Gets Foo object
If the function could belong to a nested class you would need to iterate as follows:
f = Foo.Bar.my_function
vals = vars(sys.modules[f.__module__])
for attr in f.__qualname__.split('.')[:-1]:
vals = vals[attr]
# vals is now the class Foo.Bar

Thanks Sr2222 for pointing out I was missing the point...
Here's the corrected approach which is just like Alex's but does not require to import anything. I don't think it's an improvement though, unless there's a huge hierarchy of inherited classes as this approach stops as soon as the defining class is found, instead of returning the whole inheritance as getmro does. As said, this is a very unlikely scenario.
def get_class_that_defined_method(method):
method_name = method.__name__
if method.__self__:
classes = [method.__self__.__class__]
else:
#unbound method
classes = [method.im_class]
while classes:
c = classes.pop()
if method_name in c.__dict__:
return c
else:
classes = list(c.__bases__) + classes
return None
And the Example:
>>> class A(object):
... def test(self): pass
>>> class B(A): pass
>>> class C(B): pass
>>> class D(A):
... def test(self): print 1
>>> class E(D,C): pass
>>> get_class_that_defined_method(A().test)
<class '__main__.A'>
>>> get_class_that_defined_method(A.test)
<class '__main__.A'>
>>> get_class_that_defined_method(B.test)
<class '__main__.A'>
>>> get_class_that_defined_method(C.test)
<class '__main__.A'>
>>> get_class_that_defined_method(D.test)
<class '__main__.D'>
>>> get_class_that_defined_method(E().test)
<class '__main__.D'>
>>> get_class_that_defined_method(E.test)
<class '__main__.D'>
>>> E().test()
1
Alex solution returns the same results. As long as Alex approach can be used, I would use it instead of this one.

Python 3
Solved it in a very simple way:
str(bar.foo_method).split(" ", 3)[-2]
This gives
'FooClass.foo_method'
Split on the dot to get the class and the function name separately

I found __qualname__ is useful in Python3.
I test it like that:
class Cls(object):
def func(self):
print('1')
c = Cls()
print(c.func.__qualname__)
# output is: 'Cls.func'
def single_func():
print(2)
print(single_func.__module__)
# output: '__main__'
print(single_func.__qualname__)
# output: 'single_func'
After my test, I found another answer here.

I started doing something somewhat similar, basically the idea was checking whenever a method in a base class had been implemented or not in a sub class. Turned out the way I originally did it I could not detect when an intermediate class was actually implementing the method.
My workaround for it was quite simple actually; setting a method attribute and testing its presence later. Here's an simplification of the whole thing:
class A():
def method(self):
pass
method._orig = None # This attribute will be gone once the method is implemented
def run_method(self, *args, **kwargs):
if hasattr(self.method, '_orig'):
raise Exception('method not implemented')
self.method(*args, **kwargs)
class B(A):
pass
class C(B):
def method(self):
pass
class D(C):
pass
B().run_method() # ==> Raises Exception: method not implemented
C().run_method() # OK
D().run_method() # OK
UPDATE: Actually call method() from run_method() (isn't that the spirit?) and have it pass all arguments unmodified to the method.
P.S.: This answer does not directly answer the question. IMHO there are two reasons one would want to know which class defined a method; first is to point fingers at a class in debug code (such as in exception handling), and the second is to determine if the method has been re-implemented (where method is a stub meant to be implemented by the programmer). This answer solves that second case in a different way.

if you get this error:
'function' object has no attribute 'im_class'
try this:
import inspect
def get_class_that_defined_method(meth):
class_func_defided = meth.__globals__[meth.__qualname__.split('.')[0]]
#full_func_name = "%s.%s.%s"%(class_func_defided.__module__,class_func_defided.__name__,meth.__name__)
if inspect.isfunction(class_func_defided):
print("%s is not part of a class."%meth.__name__)
return None
return class_func_defided
sample test:
class ExampleClass:
#staticmethod
def ex_static_method():
print("hello from static method")
def ex_instance_method(self):
print("hello from instance method")
def ex_funct(self):
print("hello from simple function")
if __name__ == "__main__":
static_method_class = get_class_that_defined_method(ExampleClass.ex_static_method)
static_method_class.ex_static_method()
instance_method_class = get_class_that_defined_method(ExampleClass.ex_instance_method)
instance_method_class().ex_instance_method()
function_class = get_class_that_defined_method(ex_funct)