对我来说，简单和容易与pydash裁判中间的无键

import pydash as _
_.get(d, 'key1.key2.key3.whatevermaybeNone.inthemiddle', default=None) )

if {"foo", "bar"} <= myDict.keys(): ...

如果你还在使用python2，你可以这样做

if {"foo", "bar"} <= myDict.viewkeys(): ...

如果你仍然使用非常老的Python <= 2.6，你可以在字典上调用set，但它会遍历整个字典来构建集合，这是很慢的:

if set(("foo", "bar")) <= set(myDict): ...

短而甜

{"key1"， "key2"} <= {*dict_name}

检测是否所有键都在字典中的另一个选项:

dict_to_test = { ... }  # dict
keys_sought = { "key_sought_1", "key_sought_2", "key_sought_3" }  # set

if keys_sought & dict_to_test.keys() == keys_sought: 
    # True -- dict_to_test contains all keys in keys_sought
    # code_here
    pass

并不是说这不是你没有想过的事情，但我发现最简单的事情通常是最好的:

if ("foo" in foo) and ("bar" in foo):
    # do stuff

简单的基准测试钻机3的替代品。

输入D和Q的值

>>> from timeit import Timer
>>> setup='''from random import randint as R;d=dict((str(R(0,1000000)),R(0,1000000)) for i in range(D));q=dict((str(R(0,1000000)),R(0,1000000)) for i in range(Q));print("looking for %s items in %s"%(len(q),len(d)))'''

>>> Timer('set(q) <= set(d)','D=1000000;Q=100;'+setup).timeit(1)
looking for 100 items in 632499
0.28672504425048828

#This one only works for Python3
>>> Timer('set(q) <= d.keys()','D=1000000;Q=100;'+setup).timeit(1)
looking for 100 items in 632084
2.5987625122070312e-05

>>> Timer('all(k in d for k in q)','D=1000000;Q=100;'+setup).timeit(1)
looking for 100 items in 632219
1.1920928955078125e-05