Algorithm help per favore

[David Eppstein]

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.

-- 
David Eppstein                      http://www.ics.uci.edu/~eppstein/
Univ. of California, Irvine, School of Information & Computer Science