How do you properly determine the current script directory?

I would like to see what is the best way to determine the current script directory in Python.

I discovered that, due 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, but I'm looking for the best approach that solves 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.

Solution 1:

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.

Solution 2:

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))

Solution 3:

#!/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.