sorting on IP addresses

[Max Møller Rasmussen]

From: Sam Wun [mailto:swun at esec.com.au]

>Does anyone know what is the quickly way to sort a list of IP addresses?

>ie. 203.21.254.89 should be larger than 203.21.254.9

it probably doesn't get any quikclier than this:

-----------------------------------

from string import split, join

ips = """
192.168.1.1
192.168.1.2
192.168.1.3
192.169.1.5
127.1.2.3
192.167.1.6
191.168.1.4
193.168.1.2
127.1.1.1
194.168.1.8
195.168.1.2
"""

ipList = split(ips, '\n')

ipTupleList = []
for ip in ipList:
    if ip != '': # no empty lines
        n1, n2, n3, n4 = tuple(split(ip, '.'))
        ipTupleList.append((int(n1), int(n2), int(n3), int(n4)))
ipTupleList.sort()

for ipTuple in ipTupleList:
    print "%s.%s.%s.%s" % ipTuple

-----------------------------------

Regards Max M