Function for the path of the script?
Ben Finney
ben+python at benfinney.id.au
Sun Oct 27 00:10:11 EDT 2013
Peter Cacioppi <peter.cacioppi at gmail.com> writes:
> Am I the only one who finds this function super useful?
>
> def _code_file() :
> return os.path.abspath(inspect.getsourcefile(_code_file))
I've used ‘os.path.dirname(os.path.abspath(__file__))’ to find the
directory containing the current file, in the past. But that was before
Python's ‘unittest’ module could discover where the test code lives.
> I've got one in every script. It's the only one I have to copy around.
> For my workflow ... so handy.
What workflow requires you to know the filename of the module, within
the module?
--
\ “Time wounds all heels.” —Groucho Marx |
`\ |
_o__) |
Ben Finney
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