On Mar 19, 10:08 am, sturlamolden <[EMAIL PROTECTED]> wrote: > On 18 Mar, 23:45, Arnaud Delobelle <[EMAIL PROTECTED]> 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. -- http://mail.python.org/mailman/listinfo/python-list
