这是另一个:

stats = {'a':1000, 'b':3000, 'c': 100}
max(stats.iterkeys(), key=lambda k: stats[k])

函数键只是返回应该用于排序的值，max()立即返回所需的元素。

d = {'A': 4,'B':10}

min_v = min(zip(d.values(), d.keys()))
# min_v is (4,'A')

max_v = max(zip(d.values(), d.keys()))
# max_v is (10,'B')

只是添加一个你想要选择某些键而不是所有键的情况:

stats = {'a':1000, 'b':3000, 'c': 100, 'd':3000, 'e':3000}
keys_to_search = ["a", "b", "c"]

max([k for k in keys_to_search], key=lambda x: stats[x])```

试试这个:

sorted(dict_name, key=dict_name.__getitem__, reverse=True)[0]

我测试了接受的答案和@thewolf的最快的解决方案，对一个非常基本的循环，循环比两者都快:

import time
import operator


d = {"a"+str(i): i for i in range(1000000)}

def t1(dct):
    mx = float("-inf")
    key = None
    for k,v in dct.items():
        if v > mx:
            mx = v
            key = k
    return key

def t2(dct):
    v=list(dct.values())
    k=list(dct.keys())
    return k[v.index(max(v))]

def t3(dct):
    return max(dct.items(),key=operator.itemgetter(1))[0]

start = time.time()
for i in range(25):
    m = t1(d)
end = time.time()
print ("Iterating: "+str(end-start))

start = time.time()
for i in range(25):
    m = t2(d)
end = time.time()
print ("List creating: "+str(end-start))

start = time.time()
for i in range(25):
    m = t3(d)
end = time.time()
print ("Accepted answer: "+str(end-start))

结果:

Iterating: 3.8201940059661865
List creating: 6.928712844848633
Accepted answer: 5.464320182800293

我对这些答案都不满意。Max总是选择第一个具有Max值的键。字典可以有多个具有该值的键。

def keys_with_top_values(my_dict):
    return [key  for (key, value) in my_dict.items() if value == max(my_dict.values())]

把这个答案贴出来，希望能帮助到别人。 请看下面的SO帖子

在平局的情况下，Python会选择哪个最大值?