filter() and booleanness

[Michael Hudson]

Peter Hansen <peter at engcorp.com> writes:

> rihad wrote:
> > 
> > Hi there, the following two calls will both print ('hi', 'there') due
> > to None being substituted for `identity function' (as the docs say):
> > 
> > print filter(None, ('hi', None, 0, 'there'))
> > print filter(lambda x: x, ('hi', None, 0, 'there'))
> > 
> > Then I thought I'd clamp the x to [0..1] but things like x != 0 missed
> > None and x != None missed 0. :) Now I have lambda x: not not x, is it
> > bulletproof? (it seems to work).
> 
> I believe I understand what you're asking :-), and yes, 
> it _is_ bulletproof.  That's an idiom for mapping false
> to 0 and true to 1, regardless of the original values.

There's also operator.truth().

It's all a little pointless, though.

Cheers,
M.

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