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

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): ...

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

Jason，()在Python中不是必需的。

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

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

使用集:

if set(("foo", "bar")).issubset(foo):
    #do stuff

另外:

if set(("foo", "bar")) <= set(foo):
    #do stuff