Get __name__ of calling function’s module in Python

Suppose myapp/foo.py contains:

def info(msg):
    caller_name = ????
    print '[%s] %s' % (caller_name, msg)

And myapp/bar.py contains:

import foo
foo.info('Hello') # => [myapp.bar] Hello

I want caller_name to be set to the __name__ attribute of the calling functions’ module (which is ‘myapp.foo’) in this case. How can this be done?

Answers:

Thank you for visiting the Q&A section on Magenaut. Please note that all the answers may not help you solve the issue immediately. So please treat them as advisements. If you found the post helpful (or not), leave a comment & I’ll get back to you as soon as possible.

Method 1

Check out the inspect module:

inspect.stack() will return the stack information.

Inside a function, inspect.stack()[1] will return your caller’s stack. From there, you can get more information about the caller’s function name, module, etc.

See the docs for details:

http://docs.python.org/library/inspect.html

Also, Doug Hellmann has a nice writeup of the inspect module in his PyMOTW series:

http://pymotw.com/2/inspect/index.html#module-inspect

EDIT: Here’s some code which does what you want, I think:

import inspect 

def info(msg):
    frm = inspect.stack()[1]
    mod = inspect.getmodule(frm[0])
    print '[%s] %s' % (mod.__name__, msg)

Method 2

Confronted with a similar problem, I have found that sys._current_frames() from the sys module contains interesting information that can help you, without the need to import inspect, at least in specific use cases.

>>> sys._current_frames()
{4052: <frame object at 0x03200C98>}

You can then “move up” using f_back :

>>> f = sys._current_frames().values()[0]
>>> # for python3: f = list(sys._current_frames().values())[0]

>>> print f.f_back.f_globals['__file__']
'/base/data/home/apps/apricot/1.6456165165151/caller.py'

>>> print f.f_back.f_globals['__name__']
'__main__'

For the filename you can also use f.f_back.f_code.co_filename, as suggested by Mark Roddy above. I am not sure of the limits and caveats of this method (multiple threads will most likely be a problem) but I intend to use it in my case.

Method 3

I don’t recommend do this, but you can accomplish your goal with the following method:

def caller_name():
    frame=inspect.currentframe()
    frame=frame.f_back.f_back
    code=frame.f_code
    return code.co_filename

Then update your existing method as follows:

def info(msg):
    caller = caller_name()
    print '[%s] %s' % (caller, msg)


All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0

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