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?

[Raymond Hettinger]

On Apr 22, 10:49 am, John Nagle <na... at animats.com> wrote:
> Chris Rebert wrote:
> > 2010/4/22 Jo Chan <csj... at 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