open() can't find file given path relative to PYTHONPATH

Question

I did export PYTHONPATH=$PYTHONPATH:/home/User/folder/test. Then I ran python when I was in /home/User/ and checked sys.path - it was

Answer 1

Open is relative to the current directory and does not use PYTHONPATH. The current directory defaults to whatever it was when python was started on the command line.

You can change the current directory with os.chdir

Answer 2

If I'm reading your question correctly, you want your data to reside in a location relative to the module. If that's the case, you can use:

full_path = os.path.join(os.path.split(__file__)[:-1]+['pics','text','text.txt'])

__file__ is the path to the module (including modulename.py). So I split that path, pull off modulename.py ([:-1]) and add the rest of the relative path via os.path.join

Answer 3

Whenever I want to import a script, relative to current and don't use packages, I usually use

sys.path = [os.path.dirname(__file__) + "/../another_dir"] + sys.path