那么使用呢?

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

那么使用呢?

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

你可以这样做:

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

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

这应该可以工作:

if all(key in foo for key in ["foo","bar"]):
    # do stuff
    pass

提示:

在all()中使用方括号来创建一个列表推导式:

if all([key in foo for key in ["foo","bar"]]):

不仅是不必要的，而且是非常有害的，因为它们阻碍了all()的正常短路行为。

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