在确定是否只有一些键匹配的情况下，这是有效的:

any_keys_i_seek = ["key1", "key2", "key3"]

if set(my_dict).intersection(any_keys_i_seek):
    # code_here
    pass

还有另一个选项，如果只有一些键匹配:

any_keys_i_seek = ["key1", "key2", "key3"]

if any_keys_i_seek & my_dict.keys():
    # code_here
    pass

检查字典中是否存在所有键:

{'key_1', 'key_2', 'key_3'} <= set(my_dict)

检查字典中是否存在一个或多个键:

{'key_1', 'key_2', 'key_3'} & set(my_dict)

>>> if 'foo' in foo and 'bar' in foo:
...     print 'yes'
... 
yes

Jason，()在Python中不是必需的。

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

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

如果你想:

还可以获取键的值 多查字典

然后:

from operator import itemgetter
foo = {'foo':1,'zip':2,'zam':3,'bar':4}
keys = ("foo","bar") 
getter = itemgetter(*keys) # returns all values
try:
    values = getter(foo)
except KeyError:
    # not both keys exist
    pass