Hello python devs.

One thing which appears to be pretty common in many scripts is figuring out module's parent directory path.

I often find myself writing:

HERE = os.path.abspath(os.path.dirname(__file__))

...at the top of a module.

Also, it is not rare to do further concatenations in order to declare where static files are located:

CONF_FILE = os.path.abspath(os.path.join(HERE, 'app.conf'))

CERT_FILE = os.path.abspath(os.path.join(HERE, 'cert.pem'))

THIS_FILE = os.path.abspath(os.path.join(HERE, __file__))

A quick search within Python source code shows this is indeed a very common pattern:

$ find . -name \*.py | xargs grep "os\.path" | grep __file__

Just some examples (there are *many*):

I think there's space for improvements here so my proposal is to add a simple os.path.here() function working like this:

>>> # assuming /home/giampaolo/app/run.py is the current module:

>>> os.path.here()

/home/giampaolo/app/

>>> os.path.here(__file__)

/home/giampaolo/app/run.py

>>> os.path.here('..')

/home/giampaolo/

The implementation is pretty straightforward:

def here(concat=None):

"""Return the absolute path of the parent directory where the

script is defined.

"""

here = os.path.abspath(os.path.dirname(__file__))

if concat is not None:

here = os.path.abspath(os.path.join(here, concat))

return here

Thoughts?