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

Paul Hankin paul.hankin at gmail.com
Sat Feb 23 18:18:35 CET 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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