re: normalize of mac address

[ogogon]

Colleagues, tell me, please, the answer to my question.

I have a variable that contains the mac address.
Here is a regular expression describing it -

r'^\w{1,2}:\w{1,2}:\w{1,2}:\w{1,2}:\w{1,2}:\w{1,2}$'

Now I want to one-sign bytes of address be supplemented by zero.

For this, I wrote this code:

mac = re.sub(r'^(\w):',r'0\1:',mac)
mac = re.sub(r':(\w):',r':0\1:',mac)
mac = re.sub(r':(\w)$',r':0\1',mac)

Unfortunately, the second line of code does not replace all the substrings matching the regular expression.

If I write to a variable mac value

'1:2:3:4:5:6'

, then after code execution, its value

01:02:3:04:5:06

Please tell me what am I doing wrong?

Ogogon.

[snippsat]

Can try this.

>>> import re
>>> 
>>> mac = '1:2:3:4:5:6'
>>> re.sub(r'(\d):?', r'0\1:', mac).strip(':')
'01:02:03:04:05:06'

[ogogon]

Can try this.

I think this is not a good idea.
Firstly, all bytes in the mac address are specified in hexadecimal encoding, and therefore in the regular expression need to use the '\w' macro, but not '\d'.
Secondly, your option, unfortunately, does not take into account the fact that some bytes can already be two-digit.

If I enter this mac-address

01:2:03:04:ff:6

I will get an answer

00:01:02:00:03:00:04:0f:0f:06

.
Instead

01:02:03:04:ff:06

Ogogon.

[micseydel]

How about this?

def normalize_mac(mac):
    parts = [int(x, 16) for x in mac.split(':')]
    return ':'.join(f"{part:02x}" for part in parts)

I would guess there's a more direct way than converting to an int first, but I think this works for your requirements. I'll leave optimizing it as an exercise to the reader :)