Get location of the .py source file

Question

Say I have a python file in directory e like this:

/a/b/c/d/e/file.py

Under directory e I have a few folders I wan

Answer 1

# in /a/b/c/d/e/file.py
import os
os.path.dirname(os.path.abspath(__file__)) # /a/b/c/d/e

Answer 2

In Python +3.4 use of pathlib is more handy:

from pathlib import Path

source_path = Path(__file__).resolve()
source_dir = source_path.parent

Answer 3

Here is my solution which (a) gets the .py file rather than the .pyc file, and (b) sorts out symlinks.

Working on Linux, the .py files are often symlinked to another place, and the .pyc files are generated in the directory next to the symlinked py files. To find the real path of the source file, here's part of a script that I use to find the source path.

try:
    modpath = module.__file__
except AttributeError:
    sys.exit('Module does not have __file__ defined.')
    # It's a script for me, you probably won't want to wrap it in try..except

# Turn pyc files into py files if we can
if modpath.endswith('.pyc') and os.path.exists(modpath[:-1]):
    modpath = modpath[:-1]

# Sort out symlinks
modpath = os.path.realpath(modpath)