How To Sort Ip Addresses With Port Number?

I am creating my own reporting tool, and I am trying to figure out how to sort IP address with port number. How do I sort an IP address with port number such that I sort the IP ad

Solution 1:

Using re.findall:

import re

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

sorted(a, key=get_ip_port)

Output:

['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)|']

Explanation:

• map(int, re.findall('\d+', x)): finds all digits and make them int
• *ips, port: unpacks the above ints and repack into all but last one (*ips) and last one (port)
• sorted(a, key=get_ip_port): as get_ip_port returns two keys (ips,port), sorted sorts the a first by ips and then port, just as desired.

Solution 2:

You can sort with multiple criteria(with map(int,e[0].split('.'))as criteria 1 and int(e[1].lstrip('(').split('/')[0]) as criteria2) as shown below,

>>> 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)|']]

Solution 3:

You can do it in just one line using sorted like so:

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