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
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
>>>
其他回答
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
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): ...
你不需要把左边包在一个集合里。你可以这样做:
if {'foo', 'bar'} <= set(some_dict):
pass
这也比all(k in d…)解决方案性能更好。
这只是我的看法,在所有给定的选项中,有两个方法很容易理解。所以我的主要标准是代码可读性强,而不是特别快。为了保持代码的可理解性,我更喜欢给定的可能性:
var <=var2.keys() var.issubset(var2)
事实上,“var <= var2.keys()”在我下面的测试中执行得更快,我更喜欢这个。
import timeit
timeit.timeit('var <= var2.keys()', setup='var={"managed_ip", "hostname", "fqdn"}; var2= {"zone": "test-domain1.var23.com", "hostname": "bakje", "api_client_ip": "127.0.0.1", "request_data": "", "request_method": "GET", "request_url": "hvar2p://127.0.0.1:5000/test-domain1.var23.com/bakje", "utc_datetime": "04-Apr-2019 07:01:10", "fqdn": "bakje.test-domain1.var23.com"}; var={"managed_ip", "hostname", "fqdn"}')
0.1745898080000643
timeit.timeit('var.issubset(var2)', setup='var={"managed_ip", "hostname", "fqdn"}; var2= {"zone": "test-domain1.var23.com", "hostname": "bakje", "api_client_ip": "127.0.0.1", "request_data": "", "request_method": "GET", "request_url": "hvar2p://127.0.0.1:5000/test-domain1.var23.com/bakje", "utc_datetime": "04-Apr-2019 07:01:10", "fqdn": "bakje.test-domain1.var23.com"}; var={"managed_ip", "hostname", "fqdn"};')
0.2644960229999924
>>> if 'foo' in foo and 'bar' in foo:
... print 'yes'
...
yes
Jason,()在Python中不是必需的。