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

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

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

这似乎有用

那么使用呢?

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

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") 

你也可以使用. is子集()

>>> {"key1", "key2"}.issubset({"key1":1, "key2":2, "key3": 3})
True
>>> {"key4", "key2"}.issubset({"key1":1, "key2":2, "key3": 3})
False
>>>

这里有一个替代的解决方案，以防你想要得到不匹配的项目……

not_existing_keys = [item for item in ["foo","bar"] if item not in foo]
if not_existing_keys:
  log.error('These items are missing', not_existing_keys)