Function for the path of the script?
Peter Cacioppi
peter.cacioppi at gmail.com
Sat Oct 26 22:23:43 EDT 2013
Am I the only one who finds this function super useful?
def _code_file() :
return os.path.abspath(inspect.getsourcefile(_code_file))
I've got one in every script. It's the only one I have to copy around. For my workflow ... so handy.
I've got os.path.dirname aliased to dn, so its dn(_code_file()) that I find myself reaching for fairly often...
More information about the Python-list
mailing list