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

那么使用呢?

 if reduce( (lambda x, y: x and foo.has_key(y) ), [ True, "foo", "bar"] ): # do stuff

这里有一个替代的解决方案，以防你想要得到不匹配的项目……

not_existing_keys = [item for item in ["foo","bar"] if item not in foo]
if not_existing_keys:
  log.error('These items are missing', not_existing_keys)

短而甜

{"key1"， "key2"} <= {*dict_name}

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

你不需要把左边包在一个集合里。你可以这样做:

if {'foo', 'bar'} <= set(some_dict):
    pass

这也比all(k in d…)解决方案性能更好。