On 05-Nov-2010, at 4:45 PM, Garry Willgoose wrote: > For a while now I have been using the command below in a python file to > determine the name of the python file that I use to launch an application > (i.e.IF I go 'python junk.py' I want to get 'junk.py'). The command I have > use that I came across somewhere some time ago was > > sys.modules['__main__'].__dict__['__file__'] > > Now this works fine for the standard python interpretter but when I use > ipython (I'm using the enthought distribution 6.1) this command returns > > /Library/Frameworks/Python.framework/Versions/6.1/lib/python2.6/site-packages/IPython/FakeModule.py > > I discovered this when I was tracking down inconsistencies/errors between > the matplotlib in the enthought distribution depending on whether you use > python or ipython. My question is this: Is the way I'm getting the file name > the recommended way or did I pick up a bit of flaky advice? > > I'm puzzled by the matplot lib errors (the y axes on imgshow() show problems > in python but appear to 1st order to be OK in ipython) ... but I'll pursue > that at the Matplotlib, enthought and ipython forums once I've been able to > run a broader cross-section of my codes. >
Try the following code: === #!/usr/bin/env python import sys print sys.argv === _______________________________________________ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
