Filter top N elements of dict, looking for a more pythonic | performant | cleaner way to do this

dict_top_n.py

from operator import itemgetter

def dict_top_n_items(d, n):
  """Take dictionary of form {key: count} and return a list
  with only the top n entries (based on count)

  If there is a tie for n-th place, include all keys that
  meet the tie (thus, return list could have more than n items)
  """
  if n == 0:
    return []

  items = sorted(d.items(), key=itemgetter(1), reverse=True)

  if len(items) <= n:
    return items

  last = items[n-1][1]
  for i, item in enumerate(items[n:]):
    if item[1] != last:
      return items[:n+i]
  return items
      
if __name__ == '__main__':
  d = {'a':5, 'b':4, 'c':3, 'd':3, 'e':2, 'f':2, 'g':1, 'h':0}
  for x in xrange(10):
    print "top", x, ":", dict_top_n_items(d, x)

output

top 0 : []
top 1 : [('a', 5)]
top 2 : [('a', 5), ('b', 4)]
top 3 : [('a', 5), ('b', 4), ('c', 3), ('d', 3)]
top 4 : [('a', 5), ('b', 4), ('c', 3), ('d', 3)]
top 5 : [('a', 5), ('b', 4), ('c', 3), ('d', 3), ('e', 2), ('f', 2)]
top 6 : [('a', 5), ('b', 4), ('c', 3), ('d', 3), ('e', 2), ('f', 2)]
top 7 : [('a', 5), ('b', 4), ('c', 3), ('d', 3), ('e', 2), ('f', 2), ('g', 1)]
top 8 : [('a', 5), ('b', 4), ('c', 3), ('d', 3), ('e', 2), ('f', 2), ('g', 1), ('h', 0)]
top 9 : [('a', 5), ('b', 4), ('c', 3), ('d', 3), ('e', 2), ('f', 2), ('g', 1), ('h', 0)]