CODEWITHSUNDEEP
pythonsortingip
Anderson
Anderson

Reputation: 406

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 address first and then through port number. I am able to sort IP address just fine, but when combining with port numbers, it becomes difficult.

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.

I would like to get the output sorted by IP address first, and then for each IP to sort via port number. So for example, the first few results would be:

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

Upvotes: 1

Views: 705

Answers (3)

Chris
Chris

Reputation: 29742

Using 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)

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.

Upvotes: 4

Anwarvic
Anwarvic

Reputation: 12992

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

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

Upvotes: 0

Keerthana Prabhakaran
Keerthana Prabhakaran

Reputation: 3787

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)|']]

Upvotes: 0

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