determine from which file a function is defined in python

Question

I am programmatically printing out a list of function in python. I can get the name from name

for ifunc,func in enumerate(list_of_functions):
    print(str(ifunc)+func.__name__)

How to get the source filename where the function is defined as well?

and in case the function it is attribute of a object, how to get the type of parent object?

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portability python2/3 is a must

Answer 1

func.__module__

Will return the module in witch it is defined

func.__globals__['__file__']

will return the whole path of the file where it is defined.Only for user defined functions

Answer 2

How to get the source filename where the function is defined ... ?

import inspect
inspect.getfile(func)

and in case the function it is attribute of a object, how to get the type of parent object?

To get the class of a method (i.e. when the function is an attribute of a object):

if inspect.ismethod(func):
    the_class = func.__self__.__class__

ref: https://docs.python.org/3/library/inspect.html

Answer 3

For getting the filename just use -

print(os.path.basename(__file__))

if you want the full file path you can just use

print __file__

Answer 4

As stated here https://docs.python.org/3/library/inspect.html func.__globals__ returns the global namespace where the function was defined.

Answer 5

You can use __file__ variable.

print(__file__)