How do I get the path of the current executed file

Question

This may seem like a newbie question, but it is not. Some common approaches don't work in all cases:

sys.argv[0]

This means using path = os.path.abspath(os.path.dirname(sys.argv[0])), but this does not work if you are running from another Python script in another directory, and this can happen in real life.

__file__

This means using path = os.path.abspath(os.path.dirname(__file__)), but I found that this doesn't work:

• py2exe doesn't have a __file__ attribute, but there is a workaround
• When you run from IDLE with execute() there is no __file__ attribute
• OS X 10.6 where I get NameError: global name '__file__' is not defined

Related questions with incomplete answers:

• Python - Find Path to File Being Run
• Path to current file depends on how I execute the program
• How to know the path of the running script in Python?
• Change directory to the directory of a Python script

I'm looking for a generic solution, one that would work in all above use cases.

Update

Here is the result of a testcase:

Output of python a.py (on Windows)

a.py: __file__= a.py
a.py: os.getcwd()= C:\zzz

b.py: sys.argv[0]= a.py
b.py: __file__= a.py
b.py: os.getcwd()= C:\zzz

a.py

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

print "a.py: sys.argv[0]=", sys.argv[0]
print "a.py: __file__=", __file__
print "a.py: os.getcwd()=", os.getcwd()
print

execfile("subdir/b.py")

subdir/b.py

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

print "b.py: sys.argv[0]=", sys.argv[0]
print "b.py: __file__=", __file__
print "b.py: os.getcwd()=", os.getcwd()
print

tree

C:.
|   a.py
\---subdir
        b.py

Answer 1

You can use Path from the pathlib module:

from pathlib import Path

# ...

Path(__file__)

You can use call to parent to go further in the path:

Path(__file__).parent

Answer 2

The short answer is that there is no guaranteed way to get the information you want, however there are heuristics that work almost always in practice. You might look at How do I find the location of the executable in C?. It discusses the problem from a C point of view, but the proposed solutions are easily transcribed into Python.

Answer 3

this solution is robust even in executables

import inspect, os.path

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

Answer 4

This should do the trick in a cross-platform way (so long as you're not using the interpreter or something):

import os, sys
non_symbolic=os.path.realpath(sys.argv[0])
program_filepath=os.path.join(sys.path[0], os.path.basename(non_symbolic))

sys.path[0] is the directory that your calling script is in (the first place it looks for modules to be used by that script). We can take the name of the file itself off the end of sys.argv[0] (which is what I did with os.path.basename). os.path.join just sticks them together in a cross-platform way. os.path.realpath just makes sure if we get any symbolic links with different names than the script itself that we still get the real name of the script.

I don't have a Mac; so, I haven't tested this on one. Please let me know if it works, as it seems it should. I tested this in Linux (Xubuntu) with Python 3.4. Note that many solutions for this problem don't work on Macs (since I've heard that __file__ is not present on Macs).

Note that if your script is a symbolic link, it will give you the path of the file it links to (and not the path of the symbolic link).

Answer 5

If the code is coming from a file, you can get its full name

sys._getframe().f_code.co_filename

You can also retrieve the function name as f_code.co_name

Answer 6

First, you need to import from inspect and os

from inspect import getsourcefile
from os.path import abspath

Next, wherever you want to find the source file from you just use

abspath(getsourcefile(lambda:0))