On Mar 20, 9:57 am, Justin Bozonier <[EMAIL PROTECTED]> wrote: > On Mar 19, 2:48 pm, John Machin <[EMAIL PROTECTED]> wrote: > > > > > 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. > > 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.
""" from collections import defaultdict def nonunique(lst): d = defaultdict(int) for x in lst: d[x] += 1 return [x for x,n in d.iterkeys() if n > 1] """ I see no sort here. -- http://mail.python.org/mailman/listinfo/python-list
