如何用端口号对IP地址排序？

Question

我正在创建自己的报告工具，并试图找出如何将IP地址与端口号排序。你知道吗

如何使用端口号对IP地址进行排序，以便先对IP地址进行排序，然后再通过端口号进行排序。我能够很好地对IP地址进行排序，但是当与端口号结合使用时，它就变得很困难了。你知道吗

a = ['192.168.0.3 (443/tcp)|', '192.168.0.176 (443/tcp)|', '192.168.0.40 (443/tcp)|', '192.168.0.15 (8443/tcp)|', '192.168.0.16 (8443/tcp)|', '192.168.0.12 (443/tcp)|', '192.168.0.9 (3389/tcp)|', '192.168.0.15 (443/tcp)|', '192.168.0.16 (443/tcp)|', '192.168.0.3 (3389/tcp)|', '192.168.0.14 (443/tcp)|']

print(a.sort(key=lambda s: map(int, s.split('.')))) #this works fine with just IP address not with the current format of (xxx/tcp). The pipe is for delimiters so please ignore.

我想得到的输出排序的IP地址第一，然后为每个IP排序通过端口号。例如，前几个结果是：

a= ['192.168.0.3 (443/tcp)|', '192.168.0.3 (3389/tcp)|', 192.168.0.9 (3389/tcp)|, ...']

Answer 1

您可以在一行中使用sorted这样做：

sorted(a, key=lambda x:x.split(' ')[0])

Answer 2

使用re.findall：

import re

def get_ip_port(x):
    *ips, port = map(int, re.findall('\d+', x))
    return ips, port

sorted(a, key=get_ip_port)

输出：

['192.168.0.3 (443/tcp)|',
 '192.168.0.3 (3389/tcp)|',
 '192.168.0.9 (3389/tcp)|',
 '192.168.0.12 (443/tcp)|',
 '192.168.0.14 (443/tcp)|',
 '192.168.0.15 (443/tcp)|',
 '192.168.0.15 (8443/tcp)|',
 '192.168.0.16 (443/tcp)|',
 '192.168.0.16 (8443/tcp)|',
 '192.168.0.40 (443/tcp)|',
 '192.168.0.176 (443/tcp)|']

说明：

• map(int, re.findall('\d+', x))：查找所有数字并使其成为int
• *ips, port：将上面的int解包并重新打包到除最后一个（*ips）和最后一个（port）之外的所有文件中
• sorted(a, key=get_ip_port)：因为get_ip_port返回两个键（ips，port），所以sorted首先按ips对a排序，然后按需要对port排序。你知道吗

Answer 3

您可以使用多个条件进行排序（map(int,e[0].split('.'))作为条件1，int(e[1].lstrip('(').split('/')[0])作为条件2），如下所示

>>> a
['192.168.0.3 (443/tcp)|',
 '192.168.0.176 (443/tcp)|',
 '192.168.0.40 (443/tcp)|',
 '192.168.0.15 (8443/tcp)|',
 '192.168.0.16 (8443/tcp)|',
 '192.168.0.12 (443/tcp)|',
 '192.168.0.9 (3389/tcp)|',
 '192.168.0.15 (443/tcp)|',
 '192.168.0.16 (443/tcp)|',
 '192.168.0.3 (3389/tcp)|',
 '192.168.0.14 (443/tcp)|']
>>> [i.split() for i in a]
[['192.168.0.3', '(443/tcp)|'],
 ['192.168.0.176', '(443/tcp)|'],
 ['192.168.0.40', '(443/tcp)|'],
 ['192.168.0.15', '(8443/tcp)|'],
 ['192.168.0.16', '(8443/tcp)|'],
 ['192.168.0.12', '(443/tcp)|'],
 ['192.168.0.9', '(3389/tcp)|'],
 ['192.168.0.15', '(443/tcp)|'],
 ['192.168.0.16', '(443/tcp)|'],
 ['192.168.0.3', '(3389/tcp)|'],
 ['192.168.0.14', '(443/tcp)|']]

>>> sorted([i.split() for i in a],key=lambda e: (map(int,e[0].split('.')),int(e[1].strip('(').split('/')[0])))
[['192.168.0.3', '(443/tcp)|'],
 ['192.168.0.3', '(3389/tcp)|'],
 ['192.168.0.9', '(3389/tcp)|'],
 ['192.168.0.12', '(443/tcp)|'],
 ['192.168.0.14', '(443/tcp)|'],
 ['192.168.0.15', '(443/tcp)|'],
 ['192.168.0.15', '(8443/tcp)|'],
 ['192.168.0.16', '(443/tcp)|'],
 ['192.168.0.16', '(8443/tcp)|'],
 ['192.168.0.40', '(443/tcp)|'],
 ['192.168.0.176', '(443/tcp)|']]