1、使用 collections.defaultdict实现

文档：collections.defaultdict

from collections import defaultdict
pN ={'dave': 10, 'jacinta': 10, 'james': 8, 'john': 6, 'jack': 3, 'sam': 2}
#交换key和value
dct = defaultdict(list)

for k, v in pN.items():
  dct[v].append(k)
#排序
for k, v in sorted(dct.items(), reverse=True):
  print(k, ', '.join(v))
#返回前n个元素
def top_n(d, n):
  dct = defaultdict(list) 
  for k, v in d.items():
    dct[v].append(k)      
  return sorted(dct.items())[-n:][::-1]

top_n(pN, 3)

输出结果：

[(10, ['dave', 'jacinta']), (8, ['james']), (6, ['john'])]

2、使用max来实现

pN ={'dave': 10, 'jacinta': 10, 'james': 8, 'john': 6, 'jack': 3, 'sam': 2}
def top_n_scores(n, score_dict):
  ''' returns the n scores from a name:score dict'''
  lot = [(k,v) for k, v in pN.items()] #make list of tuple from scores dict
  nl = []
  while len(lot)> 0:
      nl.append(max(lot, key=lambda x: x[1]))
      lot.remove(nl[-1])
  return nl[0:n]   
top_n_scores(4, pN)  

输出结果：

[('dave', 10), ('jacinta', 10), ('james', 8), ('john', 6)]