On Sat, 26 Oct 2013 19:23:43 -0700, Peter Cacioppi wrote: > 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...
I'm afraid I don't find it useful. I won't say that I *never* need to check the location of a module, but it's certainly very rare. I'm curious -- what are you doing that you copy this function into every script? Also, as given, the above function fails if the source code is not available (say, you provide only the .pyc file without any corresponding source). A better way is to inspect the module object's __file__ attribute: py> import math py> math.__file__ '/usr/local/lib/python2.7/lib-dynload/math.so' (Built-in modules which don't exist as a separate file, like sys, have no __file__ attribute, so be prepared to catch the exception.) Inside the module itself, the easiest way to get your own path is to check the global variable __file__. -- Steven -- https://mail.python.org/mailman/listinfo/python-list
