finding items that occur more than once in a list

[Justin Bozonier]

On Mar 19, 2:48 pm, John Machin <sjmac... at lexicon.net> wrote:
> On Mar 19, 10:08 am, sturlamolden <sturlamol... at yahoo.no> wrote:
>
>
>
> > On 18 Mar, 23:45, Arnaud Delobelle <arno... at googlemail.com> wrote:
>
> > > > def nonunique(lst):
> > > >    slst = sorted(lst)
> > > >    dups = [s[0] for s in
> > > >         filter(lambda t : t[0] == t[1], zip(slst[:-1],slst[1:]))]
> > > >    return [dups[0]] + [s[1] for s in
> > > >         filter(lambda t : t[0] != t[1], zip(dups[:-1],dups[1:]))]
>
> > > Argh!  What's wrong with something like:
>
> > > def duplicates(l):
> > >     i = j = object()
> > >     for k in sorted(l):
> > >         if i != j == k: yield k
> > >         i, j = j, k
>
> > Nice, and more readable. But I'd use Paul Robin's solution. It is O(N)
> > as opposed to ours which are O(N log N).
>
> I'd use Raymond Hettinger's solution. It is as much O(N) as Paul's,
> and is IMHO more readable than Paul's.

It's not as much O(N)... Paul Robin's uses a sort first which is
definitely not O(N). Paul's could be prettied up a bit but the general
principle is sound.