根据排名重新映射数据

Python dicts不遵守秩序。如果你想你需要一个 OrderedDict .

Counter 获得军衔。然后把它变成一个元组列表或者一个 订购信息 .

from collections import Counter, OrderedDict

d = {"a":3,"b":2,"c":3,"d":2,"e":2,"f":3,"g":4, "h":6}
c = Counter(d)

# if you want a list of tuples
ranked_list = [(pair[0],rank+1) for rank,pair in enumerate(c.most_common())]
# [('h', 1),('g', 2),('f', 3),('a', 4),('c', 5),('b', 6),('d', 7), ('e', 8)]

# if you want a dict:
ranked_dict = OrderedDict(ranked_list)
# OrderedDict([('h', 1),('g', 2),('f', 3),('a', 4),('c', 5),('b', 6),('d', 7), ('e', 8)])

您可以使用:

d = {"a":3,"b":2,"c":3,"d":2,"e":2,"f":3,"g":4, "h":6}

# sort the dictionary items by -value, throw away old value and use the
# enumerate position starting at 1 instead - no backreferencing in the old
# dict needed here

k = {k:idx for idx,(k,_) in enumerate(sorted(d.items(), key = lambda x:-x[1]),1)}

print(k)

输出:

{'h': 1, 'g': 2, 'a': 3, 'c': 4, 'f': 5, 'b': 6, 'd': 7, 'e': 8}

def ranker(d):
    ranks = sorted(set(d.values()),reverse=True)
    ranks = {r:i+1 for i,r in enumerate(ranks)}
    return {k: ranks[v] for k,v in d.items()}