Identifying the path to the currently executing Python script is a crucial aspect for various applications. However, finding a "universal" approach that works across different scenarios can be challenging. This article explores several methods and addresses limitations associated with them, ultimately presenting a comprehensive solution.
path = os.path.abspath(os.path.dirname(__file__)): This method may not work in cases where:
To obtain the path of the currently executing file regardless of the execution context, a combination of functions from the inspect and os modules can be employed:
from inspect import getsourcefile from os.path import abspath path = abspath(getsourcefile(lambda:0))
This solution provides a consistent approach that retrieves the source file path in various scenarios, including:
Considering the following directory structure:
C:. | a.py \---subdir b.py
And the code within a.py:
#! /usr/bin/env python import os, sys print "a.py: os.getcwd()=", os.getcwd() print execfile("subdir/b.py")
And the code within subdir/b.py:
#! /usr/bin/env python import os, sys print "b.py: os.getcwd()=", os.getcwd() print
The output of python a.py is:
a.py: os.getcwd()= C:\ b.py: os.getcwd()= C:\zzz
This demonstrates that os.getcwd() reflects the working directory of the execution context, which may differ from the location of the script being executed. By contrast, the proposed solution (abspath(getsourcefile(lambda:0))) consistently yields the script's source file path, regardless of the execution context.
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