How can I access the current executing module or class name in Python?

Question

I would like to be able to dynamically retrieve the current executing module or class name from within an imported module. Here is some code:

foo.py:

Answer 1

Using __file__ alone gives you a relative path for the main module and an absolute path for imported modules. Being aware this we can get the module file constantly either way with a little help from our os.path tools.

For filename only use __file__.split(os.path.sep)[-1].

For complete path use os.path.abspath(__file__).

Demo:

/tmp $ cat f.py
from pprint import pprint
import os
import sys

pprint({
    'sys.modules[__name__]': sys.modules[__name__],
    '__file__': __file__,
    '__file__.split(os.path.sep)[-1]': __file__.split(os.path.sep)[-1],
    'os.path.abspath(__file__)': os.path.abspath(__file__),
})

/tmp $ cat i.py
import f

Results:

## on *Nix ##

/tmp $ python3 f.py
{'sys.modules[__name__]': <module '__main__' from 'f.py'>,
 '__file__': 'f.py',
 '__file__.split(os.path.sep)[-1]': 'f.py',
 'os.path.abspath(__file__)': '/tmp/f.py'}

/tmp $ python3 i.py
{'sys.modules[__name__]': <module 'f' from '/tmp/f.pyc'>,
 '__file__': '/tmp/f.pyc',
 '__file__.split(os.path.sep)[-1]': 'f.pyc',
 'os.path.abspath(__file__)': '/tmp/f.pyc'}

## on Windows ##

PS C:\tmp> python3.exe f.py
{'sys.modules[__name__]': <module '__main__' from 'f.py'>,
 '__file__': 'f.py',
 '__file__.split(os.path.sep)[-1]': 'f.py',
 'os.path.abspath(__file__)': 'C:\\tools\\cygwin\\tmp\\f.py'}

PS C:\tmp> python3.exe i.py
{'sys.modules[__name__]': <module 'f' from 'C:\\tools\\cygwin\\tmp\\f.py'>,
 '__file__': 'C:\\tools\\cygwin\\tmp\\f.py',
 '__file__.split(os.path.sep)[-1]': 'f.py',
 'os.path.abspath(__file__)': 'C:\\tools\\cygwin\\tmp\\f.py'}

If you want to strip the '.py' off the end, you can do that easily. (But don't forget that you may run a '.pyc' instead.)

Answer 2

If you want only the name of the file:

file_name = __file__.split("/")[len(__file__.split("/"))-1]

Answer 3

I don't believe that's possible since that's out of foo's scope. foo will only be aware of its internal scope since it may be being called by countless other modules and applications.

Answer 4

To get the current file module, containing folder, here is what worked for me:

 import os
 parts = os.path.splitext(__name__)
 module_name = parts[len(parts) - 2]

Answer 5

To obtain a reference to the "__main__" module when in another:

import sys
sys.modules['__main__']

To then obtain the module's file path, which includes its name:

sys.modules['__main__'].__file__

If within the "__main__" module, simply use: __file__

To obtain just the file name from the file path:

import os
os.path.basename(file_path)

To separate the file name from its extension:

file_name.split(".")[0]

To obtain the name of a class instance:

instance.__class__.__name__

To obtain the name of a class:

class.__name__