On Feb 22, 7:01 pm, Paul McGuire <[EMAIL PROTECTED]> wrote: > On Feb 22, 12:54 pm, Paul Rubin <http://[EMAIL PROTECTED]> wrote: > > > Paul Rubin <http://[EMAIL PROTECTED]> 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 -- http://mail.python.org/mailman/listinfo/python-list