How to get the file path of a module from a function executed but not declared in it?

Question

If I want the path of the current module, I\'ll use __file__.

Now let\'s say I want a function to return that. I can\'t do:

def get_path         


        

Answer 1

I found a way to do it with the inspect module. I'm ok with this solution, but if somebody find a way to do it without dumping the whole stacktrace, it would be cleaner and I would accept his answer gratefully:

def get_path():
    frame, filename, line_number, function_name, lines, index =\
        inspect.getouterframes(inspect.currentframe())[1]
    return filename

Answer 2

Get it from the globals dict in that case:

def get_path():
    return globals()['__file__']

Edit in response to the comment: given the following files:

# a.py
def get_path():
    return 'Path from a.py: ' + globals()['__file__']

# b.py
import a

def get_path():
    return 'Path from b.py: ' + globals()['__file__']

print get_path()
print a.get_path()

Running this will give me the following output:

C:\workspace>python b.py
Path from b.py: b.py
Path from a.py: C:\workspace\a.py

Next to the absolute/relative paths being different (for brevity, lets leave that out), it looks good to me.

Answer 3

This is how I would do it:

import sys

def get_path():
    namespace = sys._getframe(1).f_globals  # caller's globals
    return namespace.get('__file__')