Simple - looking for a way to do an element exists check..

[Paul McGuire]

On Feb 22, 3:38 pm, Paul Rubin <http://phr...@NOSPAM.invalid> wrote:
> Paul McGuire <pt... at austin.rr.com> writes:
> > I think you have this backwards.  Should be:
>
> >      if not any(x[0]==element[0] for x in a):
> >         a.append(element)
>
> I think you are right, it was too early for me to be reading code when
> I posted that ;-)

I'm still getting used to 'any' and 'all' as new Python built-ins -
but they'll do the short-circuiting as well as a for-loop-with-break.
But I think a set- or dict-based solution will still surpass a list-
based one for the OP.

-- Paul