views:

883

answers:

4

What's the best way to turn a string in this form into an IP address: "0200A8C0". The "octets" present in the string are in reverse order, i.e. the given example string should generate 192.168.0.2.

+2  A: 

You could do something like this:

>>> s = '0200A8C0'
>>> octets = [s[i:i+2] for i in range(0, len(s), 2)]
>>> ip = [int(i, 16) for i in reversed(octets)]
>>> ip_formatted = '.'.join(str(i) for i in ip)
>>> print ip_formatted
192.168.0.2

The octet splitting could probably be done more elegantly, but I can't think of a simpler way off the top of my head.

EDIT: Or on one line:

>>> s = '0200A8C0'
>>> print '.'.join(str(int(i, 16)) for i in reversed([s[i:i+2] for i in range(0, len(s), 2)]))
192.168.0.2
Max Shawabkeh
yeah teh grouping in particular gets to me. i can't find a nice way to "chunk" the octets out, and iterate _those_ in reverse order, i'm hoping someone knows a way
Matt Joiner
@Max, Roger's answer contains a very intriguing way to do this.
Matt Joiner
+2  A: 
>>> s = "0200A8C0"
>>> bytes = ["".join(x) for x in zip(*[iter(s)]*2)]
>>> bytes
['02', '00', 'A8', 'C0']
>>> bytes = [int(x, 16) for x in bytes]
>>> bytes
[2, 0, 168, 192]
>>> print ".".join(str(x) for x in reversed(bytes))
192.168.0.2

It is short and clear; wrap it up in a function with error checking to suit your needs.


Handy grouping functions:

def group(iterable, n=2, missing=None, longest=True):
  """Group from a single iterable into groups of n.

  Derived from http://bugs.python.org/issue1643
  """
  if n < 1:
    raise ValueError("invalid n")
  args = (iter(iterable),) * n
  if longest:
    return itertools.izip_longest(*args, fillvalue=missing)
  else:
    return itertools.izip(*args)

def group_some(iterable, n=2):
  """Group from a single iterable into groups of at most n."""
  if n < 1:
    raise ValueError("invalid n")
  iterable = iter(iterable)
  while True:
    L = list(itertools.islice(iterable, n))
    if L:
      yield L
    else:
      break
Roger Pate
@Roger, how does that `zip(*[iter(s)]*2)` work? I'm very interested in that.
Matt Joiner
This is quite the hack! `iter(s)` returns an iterator over the string. Multiplying the list of this iterator by 2 will create two references **to the same iterator**. `zip()` returns a list of tuples containing an element from each of its arguments. Since both arguments are the same iterator, it will take from it twice for each tuple, returning a tuple of each 2 adjacent characters. I would never have thought of trying anything like this. :D
Max Shawabkeh
@Matt: It's the core of most "grouping" recipes I've seen, and Max explained it well. Will update to include two short functions in that vein.
Roger Pate
@Roger, thanks very much, I was hoping a solution like yours would come up.
Matt Joiner
I've created a new question, which directly asks what you've solved here: http://stackoverflow.com/questions/2202461/yield-multiple-objects-at-a-time-from-an-iterable-object
Matt Joiner
+9  A: 

Network address manipulation is provided by the socket module.

socket.inet_ntoa(packed_ip)

Convert a 32-bit packed IPv4 address (a string four characters in length) to its standard dotted-quad string representation (for example, ‘123.45.67.89’). This is useful when conversing with a program that uses the standard C library and needs objects of type struct in_addr, which is the C type for the 32-bit packed binary data this function takes as an argument.

You can translate your hex string to packed ip using struct.pack() and the little endian, unsigned long format.

>>> import socket
>>> import struct
>>> addr_long = int("0200A8C0",16)
>>> hex(addr_long)
'0x200a8c0'
>>> struct.pack("<L", addr_long)
'\xc0\xa8\x00\x02'

>>> socket.inet_ntoa(struct.pack("<L", addr_long))
'192.168.0.2'
>>> 
gimel
The most useful answer, IMO.
Alok
You want "<I", not "L" as the first argument to `pack`, though "<L" will work too. Otherwise you're leaving yourself open to whatever the platform endianess is.
Omnifarious
@Omnifarious, how exactly does it's "long/int"ness effect it here?
Matt Joiner
@Matt Joiner, it doesn't really, at least if you do "<L". Otherwise, it's platform specific exactly how big "L" or "I" is. I have a tendency to use the conceptually smallest type that works.
Omnifarious
Edited to "<L". Thanks, @Omnifarious .
gimel
A: 

My try:

a = '0200A8C0'
indices = range(0, 8, 2)
data = [str(int(a[x:x+2], 16)) for x in indices]
'.'.join(reversed(data))
Alok