我想做的事情是:
foo = {
'foo': 1,
'zip': 2,
'zam': 3,
'bar': 4
}
if ("foo", "bar") in foo:
#do stuff
我如何检查是否foo和酒吧都在dict foo?
我想做的事情是:
foo = {
'foo': 1,
'zip': 2,
'zam': 3,
'bar': 4
}
if ("foo", "bar") in foo:
#do stuff
我如何检查是否foo和酒吧都在dict foo?
当前回答
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
其他回答
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
>>>
>>> if 'foo' in foo and 'bar' in foo:
... print 'yes'
...
yes
Jason,()在Python中不是必需的。
我认为这是最聪明和最精辟的。
{'key1','key2'} <= my_dict.keys()
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): ...
短而甜
{"key1", "key2"} <= {*dict_name}