Get class that defined method

user_88696 0
Viewed 70526

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)
python-datamodel python-2.6 python
6 Answers
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
user_958100

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__.

user_26810570

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
user_21323120

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

user_37303970
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