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
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")
其他回答
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", "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}
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
并不是说这不是你没有想过的事情,但我发现最简单的事情通常是最好的:
if ("foo" in foo) and ("bar" in foo):
# do stuff