How do you properly determine the current script directory in Python?

Question

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 ha

Answer 1

In Python 3.4+ you can use the simpler pathlib module:

from inspect import currentframe, getframeinfo
from pathlib import Path

filename = getframeinfo(currentframe()).filename
parent = Path(filename).resolve().parent

You can also use __file__ to avoid the inspect module altogether:

from pathlib import Path
parent = Path(__file__).resolve().parent

Answer 2

For .py scripts as well as interactive usage:

I frequently use the directory of my scripts (for accessing files stored along side them), but I also frequently run these scripts in an interactive shell for debugging purposes. I define __dirpath__ as:

• When running or importing a .py file, the file's base directory. This is always the correct path.
• When running an .ipyn notebook, the current working directory. This is always the correct path, since Jupyter sets the working directory as the .ipynb base directory.
• When running in a REPL, the current working directory. Hmm, what is the actual "correct path" when the code is detached from a file? Rather, make it your responsibility to change into the "correct path" before invoking the REPL.

Python 3.4 (and above):

from pathlib import Path
__dirpath__ = Path(globals().get("__file__", "./_")).absolute().parent

Python 2 (and above):

import os
__dirpath__ = os.path.dirname(os.path.abspath(globals().get("__file__", "./_")))

Explanation:

• globals() returns all the global variables as a dictionary.
• .get("__file__", "./_") returns the value from the key "__file__" if it exists in globals(), otherwise it returns the provided default value "./_".
• The rest of the code just expands __file__ (or "./_") into an absolute filepath, and then returns the filepath's base directory.

Answer 3

Here is a partial solution, still better than all published ones so far.

import sys, os, os.path, inspect

#os.chdir("..")

if '__file__' not in locals():
    __file__ = inspect.getframeinfo(inspect.currentframe())[0]

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

Now this works will all calls but if someone use chdir() to change the current directory, this will also fail.

Notes:

• sys.argv[0] is not going to work, will return -c if you execute the script with python -c "execfile('path-tester.py')"
• I published a complete test at https://gist.github.com/1385555 and you are welcome to improve it.

Answer 4

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))

Answer 5

Hopefully this helps:- If you run a script/module from anywhere you'll be able to access the __file__ variable which is a module variable representing the location of the script.

On the other hand, if you're using the interpreter you don't have access to that variable, where you'll get a name NameError and os.getcwd() will give you the incorrect directory if you're running the file from somewhere else.

This solution should give you what you're looking for in all cases:

from inspect import getsourcefile
from os.path import abspath
abspath(getsourcefile(lambda:0))

I haven't thoroughly tested it but it solved my problem.

Answer 6

#!/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.