sorting ip address

[Matheus Suffi]

Hello,

im trying to sort an IP adddres list, this is the only result i getting

192.168.1.1

192.168.1.10

192.168.1.11

i want to list in that order:

192.168.1.1

192.168.1.2

192.168.1.3

192.168.1.4

Any ideias ?

[Niphlod]

use the power, luke!
aside from various networking libraries out there, the problem is rather simple....
what you're getting back is the "alphabetical order"  .
what you need instead is to sort every triplet as an integer.
In reality, python strings for ip addresses are good enough, if taken separately, i.e.
1 is lower than 10 even if "1" and "10" are strings.

so, to order your ip addresses, you just split every ip address by "."

given a list of
 
myips = [

'192.168.4.1',

'192.168.1.10',

'192.168.1.11',

]

what_you_want = sorted(myips, key=lambda x: x.split('.'))


[Niphlod]

nope, wrong........... sorry, I reaally need some bed time ^_^

You NEED to convert triplets to integers before sorting.

no worries though...

what_you_want = sorted(myips, key=lambda x:tuple(map(int, x.split('.'))))

is the correct one.

[Derek]

Wrong again. You really should just go to sleep.

First, you need to convert your IP addresses to decimal representation, then sort them.

import socket, struct

def ip2long_1(ip):
    return struct.unpack("!L", socket.inet_aton(ip))[0]

[Niphlod]

maybe........what "list of ip addresses" gets sorted wrongly converting triplets to integers and right converting addresses to a decimal representation ?

[Mirek Zvolský]

:)))

it works very well. The second one. Maybe tuple is not necessary:

what_you_want = sorted(myips, key=lambda x:map(int, x.split('.')))

But there are a little non-suitable testing values:  .4. versus .1.     :))

Dne středa 1. dubna 2015 0:27:11 UTC+2 Matheus Suffi napsal(a):

[rharde...@gmail.com]

Try using the netaddr library...