你可以这样做:

>>> if all(k in foo for k in ("foo","bar")):
...     print "They're there!"
...
They're there!

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

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

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

Jason，()在Python中不是必需的。

短而甜

{"key1"， "key2"} <= {*dict_name}

那么使用呢?

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