Python: __file__ of the caller

Question

I want to define a utility function f() in my module m such that

1. if m.f() is invoked in /foo/a.py, it returns /foo/a.py
2. if m.f() invoked in /bar/baz/b.py, it returns /bar/baz/b.py

Intuitively I'd explain f() as "it returns __file__ of the caller" but how can I really implement that (if possible at all?)

Note def f(): return __file__ returns /path/to/m.py wherever I call m.f() and that's not what I want.

Answer 1

some.py:

m.f()

m.py:

import inspect
import os

def f():
    return os.path.relpath(
            inspect.stack()[1][1],
            basePath)
    # returns path to caller file relative to some basePath

Answer 2

__file__ is an attribute of the module loaded. So m.__file__ will always give the path of the file from which m was loaded. To make it work for any module, you should call the attribute for that particular module.

import module

def f(m): return m.__file__

print f(module)

Answer 3

Take a look at the inspect module. It has a handy stack() method that returns the entire call stack, each element of which includes the filename. You just need to extract the last one.