How do I get the path of the current executed file in Python

Questions

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:

http://www.py2exe.org/index.cgi/WhereAmI http://www.py2exe.org/index.cgi/WhereAmI
• 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:

http://stackoverflow.com/questions/1296501/python-find-path-to-file-being-run http://stackoverflow.com/questions/1296501/python-find-path-to-file-being-run http://stackoverflow.com/questions/1483827/python-path-to-current-file-depends-on-how-i-execute-the-program http://stackoverflow.com/questions/1483827/python-path-to-current-file-depends-on-how-i-execute-the-program http://stackoverflow.com/questions/2259503/how-to-know-the-path-of-the-running-script-in-python http://stackoverflow.com/questions/2259503/how-to-know-the-path-of-the-running-script-in-python http://stackoverflow.com/questions/509742/python-chdir-to-dir-the-py-script-is-in http://stackoverflow.com/questions/509742/python-chdir-to-dir-the-py-script-is-in

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

Answers

You can t directly determine the location of the main script being executed. After all, sometimes the script didn t come from a file at all. For example, it could come from the interactive interpreter or dynamically generated code stored only in memory.

However, you can reliably determine the location of a module, since modules are always loaded from a file. If you create a module with the following code and put it in the same directory as your main script, then the main script can import the module and use that to locate itself.

some_path/module_locator.py:

def we_are_frozen():
    # All of the modules are built-in to the interpreter, e.g., by py2exe
    return hasattr(sys, "frozen")

def module_path():
    encoding = sys.getfilesystemencoding()
    if we_are_frozen():
        return os.path.dirname(unicode(sys.executable, encoding))
    return os.path.dirname(unicode(__file__, encoding))

some_path/main.py:

import module_locator
my_path = module_locator.module_path()

If you have several main scripts in different directories, you may need more than one copy of module_locator.

Of course, if your main script is loaded by some other tool that doesn t let you import modules that are co-located with your script, then you re out of luck. In cases like that, the information you re after simply doesn t exist anywhere in your program. Your best bet would be to file a bug with the authors of the tool.

Source

License : cc by-sa 3.0

http://stackoverflow.com/questions/2632199/how-do-i-get-the-path-of-the-current-executed-file-in-python

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