python - 按 ip 地址对元组列表进行排序

Question

我有一个以 IP 地址作为第一个元素的元组列表

tuple_list = [
    ('192.168.0.15', 33, 60), 
    ('192.168.0.22', 34, 60), 
    ('192.168.0.1', 34, 60), 
    ('192.168.0.2', 34, 60), 
    ('192.168.0.8', 34, 60), 
    ('192.168.0.11', 34, 60)
] 

我想按 IP 地址对其进行排序，但它打印如下:

print(sorted(tuple_list))

[
    ('192.168.0.1', 34, 60), 
    ('192.168.0.11', 34, 60), 
    ('192.168.0.15', 33, 60), 
    ('192.168.0.2', 34, 60), 
    ('192.168.0.22', 34, 60), 
    ('192.168.0.8', 34, 60)
]

当我想要这样的时候:

[
    ('192.168.0.1', 34, 60), 
    ('192.168.0.2', 34, 60), 
    ('192.168.0.8', 34, 60), 
    ('192.168.0.11', 34, 60), 
    ('192.168.0.15', 33, 60), 
    ('192.168.0.22', 34, 60)
]

Answer 1

您可以使用ipaddress模块来存储IP地址而不是字符串。这样它们就可以相互比较，然后使用 operator.itemgetter 进行排序。在我看来，这是最有效的，并且不需要 lambda 表达式来排序。

from ipaddress import IPv4Address
from operator  import itemgetter

tuple_list = [(IPv4Address('192.168.0.15'), 33, 60), (IPv4Address('192.168.0.22'), 34, 60), (IPv4Address('192.168.0.1'), 34, 60), (IPv4Address('192.168.0.2'),34, 60), (IPv4Address('192.168.0.8'), 34, 60), (IPv4Address('192.168.0.11'), 34, 60)]

print(sorted(tuple_list, key=itemgetter(0)))
#[(IPv4Address('192.168.0.1'), 34, 60), (IPv4Address('192.168.0.2'), 34, 60), (IPv4Address('192.168.0.8'), 34, 60), (IPv4Address('192.168.0.11'), 34, 60), (IPv4Address('192.168.0.15'), 33, 60), (IPv4Address('192.168.0.22'), 34, 60)]

我只是使用这个简单的理解将字符串转换为 IPv4Address。

[(IPv4Address(x), y, z) for x, y, z in tuple_list]