Elegant Format for a MAC Address in Python 3.2

Question

I am looking for a more elegant solution to formatting a MAC address with colons. I am using Python 3.2. A fancy list comprehension perhaps?

s=\"\"
h=\"0023         


        

Answer 1

Well, I might start with something pretty specific, since you know that it is an MAC address you know the exact size and format.

print "%s:%s:%s:%s:%s:%s" % (h[0:2], h[2:4], h[4:6], h[6:8], h[8:10], h[10:12])

But we can make this better if we create a class and then tell it how to format it's string.

class Mac():
    def __init__(self, mac):
        self.mac = mac
    def __str__(self):
        return "%s:%s:%s:%s:%s:%s" % (
            self.mac[0:2],
            self.mac[2:4],
            self.mac[4:6],
            self.mac[6:8],
            self.mac[8:10],
            self.mac[10:12])

m = Mac("123456789012")
print m

Answer 2

from netaddr import EUI, mac_unix_expanded
print(EUI('00:01:02:03:04:05', dialect=mac_unix_expanded))
print(EUI('1-2-3-4-5-6', dialect=mac_unix_expanded))

Answer 3

Not sure how pretty this is, but it will do what you ask:

':'.join([h[i:i+2] for i,j in enumerate(h) if not (i%2)])

gives:

'00:23:3a:99:0c:21'

Answer 4

Your code is easily converted to a comprehension form:

':'.join(h[i:i+2] for i in range(0,12,2))

Answer 5

This is not the shortest solution, but it accepts all common types of mac formats as inputs. It also does some validation checks.

import re

def format_mac(mac: str) -> str:
    mac = re.sub('[.:-]', '', mac).lower()  # remove delimiters and convert to lower case
    mac = ''.join(mac.split())  # remove whitespaces
    assert len(mac) == 12  # length should be now exactly 12 (eg. 008041aefd7e)
    assert mac.isalnum()  # should only contain letters and numbers
    # convert mac in canonical form (eg. 00:80:41:ae:fd:7e)
    mac = ":".join(["%s" % (mac[i:i+2]) for i in range(0, 12, 2)])
    return mac

Here is a list of mac address strings and whether they will be considered as valid or invalid:

'008041aefd7e',  # valid
'00:80:41:ae:fd:7e',  # valid
'00:80:41:AE:FD:7E',  # valid
'00:80:41:aE:Fd:7E',  # valid
'00-80-41-ae-fd-7e',  # valid
'0080.41ae.fd7e',  # valid
'00 : 80 : 41 : ae : fd : 7e',  # valid
'  00:80:41:ae:fd:7e  ',  # valid
'00:80:41:ae:fd:7e\n\t',  # valid

'aa:00:80:41:ae:fd:7e',  # invalid
'0:80:41:ae:fd:7e',  # invalid
'ae:fd:7e',  # invalid
'$$:80:41:ae:fd:7e',  # invalid

All valid ones will be returned in the canonical form as:

'00:80:41:ae:fd:7e'

Answer 6

The easiest solution here is just to use str.join()

>>> ":".join(str_grouper(2, "00233a990c21"))
'00:23:3a:99:0c:21'

Here using a modified version of the grouper() recipe from the itertools docs:

def str_grouper(n, iterable):
     args = [iter(iterable)] * n
     for part in zip(*args): #itertools.izip in 2.x for efficiency.
         yield "".join(part)