Algorithm help per favore

[Jeremy Hylton]

On Wed, 2003-06-18 at 14:56, David Eppstein wrote:
> In article <3EF0AAFB.DDA5F555 at engcorp.com>,
>  Peter Hansen <peter at engcorp.com> wrote:
> 
> > > >>> L = [6,3,3,3,5,7,6,3,4,4,3]
> > > >>> [x for x, y in zip(L, [L]+L) if x != y]
> > > [6, 3, 5, 7, 6, 3, 4, 3]
> > 
> > But this won't work if L contains a reference to L!  :-) :-)
> 
> Ok, what's the quickest way to build a new object that's guaranteed 
> unequal to any previous object?  E.g.
> [x for x, y in zip(L, [lambda x:x]+L) if x != y]
> 
> There must be a better way to do this than with a lambda.

How about "object()"?

Jeremy