Simple - looking for a way to do an element exists check..
Paul Hankin
paul.hankin at gmail.com
Sat Feb 23 12:18:35 EST 2008
On Feb 22, 7:01 pm, Paul McGuire <pt... at austin.rr.com> wrote:
> On Feb 22, 12:54 pm, Paul Rubin <http://phr...@NOSPAM.invalid> wrote:
>
> > Paul Rubin <http://phr...@NOSPAM.invalid> writes:
> > > if any(x==element[0] for x in a):
> > > a.append(element)
>
> > Should say:
>
> > if any(x[0]==element[0] for x in a):
> > a.append(element)
>
> I think you have this backwards. Should be:
>
> if not any(x[0]==element[0] for x in a):
> a.append(element)
IMO Jason's solution of testing containment in a generator is better
(more readable).
if element[0] not in (x[0] for x in a):
a.append(element)
--
Paul Hankin
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