sorting on IP addresses

[Jacob Kaplan-Moss]

In article <3A7F26CB.BE85C96F at esec.com.au>, Sam Wun <swun at esec.com.au> 
wrote:
> Hi,
> 
> 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
> 
> Thanks
> Sam
> 

Sam --

Not sure about quick, but this is the way I've been doing it.  If any of 
the gurus out there would like to point out a better method, I would be 
happy to read it.

iplist = ["203.21.254.89",
        "203.21.254.9",
        "1.2.3.4",
        "1.2.3.5",
        "211.1.50.124",
        "207.1.23.4",
        "192.168.1.101",
        "192.168.1.1"]

def ip_compare( a, b ):
   """a and b are valid IP addresses of the form w.x.y.z."""
   
   la = [int(n) for n in a.split(".")]
   lb = [int(n) for n in b.split(".")]
   
   for pair in zip(la, lb):
      if pair[0] > pair[1]:
         return -1
      elif pair[0] < pair[1]:
         return 1
   
   return 0
   
iplist.sort(ip_compare)
print iplist

Hope that helps, 

Jacob