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。

短而甜

{"key1"， "key2"} <= {*dict_name}

my_dict = {
    'name': 'Askavy',
    'country': 'India',
    'age': 30
}

if set(('name', 'country','age')).issubset(my_dict.keys()):
     print("All keys are present in the dictionary") 
else: 
    print("All keys are not present in  the dictionary") 

使用集:

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

另外:

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

>>> ok
{'five': '5', 'two': '2', 'one': '1'}

>>> if ('two' and 'one' and 'five') in ok:
...   print "cool"
... 
cool

这似乎有用

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