Peter Cacioppi <peter.cacio...@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 -- https://mail.python.org/mailman/listinfo/python-list
