How to perform this RegExp in Python?

Question

I would like to transform a phone number of this form +33.300000000 in 03.00.00.00.00

+33 is the indicatif it could be 2 or 3 digits length.

Digits after the . are the phone number. It could be 9 or 10 digits length.

I try like this :

p = re.compile( "\+[0-9]+\.([0-9]+)", re.VERBOSE)
number = "+33.300000000"
p.sub("0\1", number)

But this doesn't work.

However, the search seams to work :

>>> p.search(number).groups()
('300000000',)

And after how to modify 0300000000 in 03.00.00.00.00 in Python ?

Thank you for your help,

Natim

For nine digits you add a leading 0, what happens for 10 digits? — John La Rooy
– John La Rooy, Commented Oct 11, 2009 at 23:58

Alex Martelli · Accepted Answer · 2009-10-12 00:09:04Z

3

The simplest approach is a mix of RE and pure string manipulation, e.g.:

import re

def doitall(number):
  # get 9 or 10 digits, or None:
  mo = re.search(r'\d{9,10}', number)
  if mo is None: return None
  # add a leading 0 if they were just 9
  digits = ('0' + mo.group())[-10:]
  # now put a dot after each 2 digits
  # and discard the resulting trailing dot
  return re.sub(r'(\d\d)', r'\1.', digits)[:-1]

number = "+33.300000000"
print doitall(number)

emits 03.00.00.00.00, as required.

Yes, you can do it all in a RE, but it's not worth the headache -- the mix works fine;-).

answered Oct 12, 2009 at 0:09

Alex Martelli

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2 Comments

John La Rooy Over a year ago

you can also use digits = mo.group().zfill(10)

John La Rooy Over a year ago

and return ".".join(re.findall(r'(\d\d)',digits))

Natim · Accepted Answer · 2009-10-11 23:59:44Z

0

In the terminal it works like that :

p = re.compile( "\+[0-9]+\.([0-9]+)", re.VERBOSE)
number = "+33.300000000"
p.sub("0\\1", number)

And in the python script it works like that :

p = re.compile( "\+[0-9]+\.([0-9]+)", re.VERBOSE)
number = "+33.300000000"
p.sub("0\1", number)

answered Oct 11, 2009 at 23:59

Natim

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