Re: Yet another unique() function...

"MonkeeSage" <MonkeeSage@xxxxxxxxx> writes:
In your case optimized version, in the second try clause using
itertools, it should be like this, shouldn't it?

return t(g.next()[1] for k,g in groupby(s, lambda (i,v): v))

I didn't think so but I can't conveniently test it for now. Maybe
tonight.
.