How to properly determine current script directory?

I would like to see what is best way to determine current script directory in python?

I discovered that two to the many ways of calling python code, it is hard to find a good solution.

Here are some problems:

__file__ is not defined if the script is executed with exec , execfile

__module__ is defined only in modules

Use cases:

./myfile.py

python myfile.py

./somedir/myfile.py

python somedir/myfile.py

execfile('myfile.py') (from another script, that can be located in another directory and that can have another current directory.

I know that there is no perfect solution, because in some cases but I'm looking for the best approach that solved most of the cases.

The most used approach is os.path.dirname(os.path.abspath(__file__)) but this really doesn't work if you execute the script from another one with exec() .

Warning

Any solution that uses current directory will fail, this can be different based on the way the script is called or it can be changed inside the running script.

---

os.path.dirname(os.path.abspath(__file__))

is indeed the best you're going to get.

It's unusual to be executing a script with exec / execfile ; normally you should be using the module infrastructure to load scripts. If you must use these methods, I suggest setting __file__ in the globals you pass to the script so it can read that filename.

There's no other way to get the filename in execed code: as you note, the CWD may be in a completely different place.

---

If you really want to cover the case that a script is called via execfile(...) , you can use the inspect module to deduce the filename (including the path). As far as I am aware, this will work for all cases you listed:

filename = inspect.getframeinfo(inspect.currentframe()).filename
path = os.path.dirname(os.path.abspath(filename))

---

#!/usr/bin/env python
import inspect
import os
import sys

def get_script_dir(follow_symlinks=True):
    if getattr(sys, 'frozen', False): # py2exe, PyInstaller, cx_Freeze
        path = os.path.abspath(sys.executable)
    else:
        path = inspect.getabsfile(get_script_dir)
    if follow_symlinks:
        path = os.path.realpath(path)
    return os.path.dirname(path)

print(get_script_dir())

It works on CPython, Jython, Pypy. It works if the script is executed using execfile() ( sys.argv[0] and __file__ -based solutions would fail here). It works if the script is inside an executable zip file (/an egg). It works if the script is "imported" ( PYTHONPATH=/path/to/library.zip python -mscript_to_run ) from a zip file; it returns the archive path in this case. It works if the script is compiled into a standalone executable ( sys.frozen ). It works for symlinks ( realpath eliminates symbolic links). It works in an interactive interpreter; it returns the current working directory in this case.

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