打印字典中最前面的n个词条

Question

这是我当前的代码

pN ={'dave': 10, 'jacinta': 10, 'james': 8, 'john': 6, 'jack': 3, 'sam': 2}
highestCount = max(pN.values())

for k, v in pN.items():
  if v == highestCount:
    print(v,k)

但是，这只打印顶级用户，如果该位置是共享的，则再次打印

^{pr2}$

我需要能够打印任意数量的顶级用户（n）并将其格式化，例如n = 5：

10 john, jacinta, 
8 james
6 john
3 jack
2 sam

Answer 1

您可以使用sorted和字典理解：

from typing import Dict, List
def ranking(d, n:'n_users', top = True) -> Dict[int, List[str]]:
  _ranks = sorted(d.values())
  _ranks = _ranks[-n:] if top else _ranks[:n]
  return {i:[a for a, b in d.items() if b == i] for i in _ranks}

pN ={'dave': 10, 'jacinta': 10, 'james': 8, 'john': 6, 'jack': 3, 'sam': 2}
for a, b in sorted(ranking(pN, 10).items(), key=lambda x:x[0], reverse=True):
  print('{} {}'.format(a, ', '.join(b)))

输出：

^{pr2}$

编辑：对于任意数量的顶级用户，将值传递给函数：

_r = ranking(pN, 5) #for the top 5 users

Answer 2

使用^{}，交换你的keys和values

from collections import defaultdict
dct = defaultdict(list)

for k, v in pN.items():
  dct[v].append(k)

# defaultdict(<class 'list'>, {10: ['dave', 'jacinta'], 8: ['james'], 6: ['john'], 3: ['jack'], 2: ['sam']})

使用sorted输出：

^{pr2}$

function返回topn用户（将ties视为一个条目）：

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)

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

使用defaultdict既简单又快速，下面是一些时间来证明它：

将定时的函数

def chris_z(d, n):
  dct = defaultdict(list) 
  for k, v in d.items():
    dct[v].append(k)      
  return sorted(dct.items())[-n:][::-1]

def tim_lombard(score_dict, n):
  lot = [(k,v) for k, v in score_dict.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])

def ajax(d, n:'n_users', top = True):
  _ranks = sorted(d.values())
  _ranks = _ranks[-n:] if top else _ranks[:n]
  return {i:[a for a, b in d.items() if b == i] for i in _ranks}

结果

x = [''.join(i) for i in itertools.permutations('chrisz', 6)]    
y = [random.randint(0, 100) for _ in range(720)]  
z = dict(zip(x, y))

In [40]: %timeit chris_z(z, 500)
110 µs ± 259 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)

In [42]: %timeit tim_lombard(z, 500)
26.2 ms ± 60 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

In [43]: %timeit ajax(z, 500)
15.3 ms ± 227 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

Answer 3

这个对你有用吗？在

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]   

要获得前4名：

^{pr2}$