Adding a __filename__ predefined attribute to 2.5?
fredrik at pythonware.com
Wed Oct 12 11:31:46 CEST 2005
Rune Strand wrote:
> I know it's several ways to isolate the filename. I just want to avoid
> the overhead of importing sys or os to achieve it.
those modules are already imported when Python gets to your code, so
the only "overhead" you're saving is a little typing.
> Currently I have this in my scripts:
> __filename__ = __file__.replace('\\', '/').rsplit('/', 1)[-1]
wow. that's one lousy optimization...
here's a *shorter* piece of code, which is also readable and portable, and
a lot easier to type on most keyboards:
__filename__ = os.path.basename(__file__)
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