On Apr 22, 10:49 am, John Nagle <na...@animats.com> wrote: > Chris Rebert wrote: > > 2010/4/22 Jo Chan <csj...@gmail.com>: > >> Hi,friends. > >> I wanna ask if there is a function which is able to take a list as > >> argument > >> and then return its top-k maximums? > >> I only know about max which is poorly a top-1 maximum function, now I want > >> more yet I am lazy enough that don't want to write one by myself. > > >http://docs.python.org/library/heapq.html#heapq.nlargest > > > Cheers, > > Chris > > -- > >http://blog.rebertia.com > > Is "nlargest" smart enough to decide when it's cheaper to track the > N largest entries on a linear pass through the list than to sort?
nlargest() has gotten smarter over time. So, the answer depends on which version you're using. Here's a snippet from http://svn.python.org/view/python/branches/py3k/Lib/heapq.py?revision=70695&view=markup def nlargest(n, iterable, key=None): """Find the n largest elements in a dataset. Equivalent to: sorted(iterable, key=key, reverse=True)[:n] """ # Short-cut for n==1 is to use max() when len(iterable)>0 if n == 1: it = iter(iterable) head = list(islice(it, 1)) if not head: return [] if key is None: return [max(chain(head, it))] return [max(chain(head, it), key=key)] # When n>=size, it's faster to use sort() try: size = len(iterable) except (TypeError, AttributeError): pass else: if n >= size: return sorted(iterable, key=key, reverse=True)[:n] # When key is none, use simpler decoration if key is None: it = zip(iterable, count(0,-1)) # decorate result = _nlargest(n, it) return [r[0] for r in result] # undecorate # General case, slowest method in1, in2 = tee(iterable) it = zip(map(key, in1), count(0,-1), in2) # decorate result = _nlargest(n, it) return [r[2] for r in result] # undecorate -- http://mail.python.org/mailman/listinfo/python-list
