How to sort IP addresses stored in dictionary in Python?

You can use a custom key function to return a sortable representation of your strings:

def split_ip(ip):    """Split a IP address given as string into a 4-tuple of integers."""    return tuple(int(part) for part in ip.split('.'))def my_key(item):    return split_ip(item[0])items = sorted(ipCount.items(), key=my_key)

The split_ip() function takes an IP address string like '192.168.102.105' and turns it into a tuple of integers (192, 168, 102, 105). Python has built-in support to sort tuples lexicographically.

UPDATE: This can actually be done even easier using the inet_aton() function in the socket module:

import socketitems = sorted(ipCount.items(), key=lambda item: socket.inet_aton(item[0]))

Use the key parameter of sorted to convert your ip to an integer, for example:

list_of_ips = ['192.168.204.111', '192.168.99.11', '192.168.102.105']sorted(list_of_ips, key=lambda ip: long(''.join(["%02X" % long(i) for i in ip.split('.')]), 16))

EDIT:

Gryphius proposes a solution with the socket module, and so why not use it to make the conversion from ip to long as it is cleaner:

from socket import inet_atonimport structlist_of_ips = ['192.168.204.111', '192.168.99.11', '192.168.102.105']sorted(list_of_ips, key=lambda ip: struct.unpack("!L", inet_aton(ip))[0])

A clean way of handling the right order is using Pythons ipaddress module. You can transform the Strings into IPv4Address representations and sort them afterwards. Here's a working example with list objects (Tested with Python3):

import ipaddressunsorted_list = [  '192.168.102.105',  '192.168.204.111',  '192.168.99.11']new_list = []for element in unsorted_list:  new_list.append(ipaddress.ip_address(element))new_list.sort()# [IPv4Address('192.168.99.11'), IPv4Address('192.168.102.105'), IPv4Address('192.168.204.111')]print(new_list)