Python 3 replace part of string with data from dict - regex

Question

I need to turn phonenumbers into international format. I have a list of phone number examples
rows = [
(datetime.time(20, 35, 30), '0707262078',),
(datetime.time(20, 38, 18), '[phone]',),
(datetime.time(20, 56, 35), '065017063'),
(datetime.time(21, 45, 1), '[phone]',),
(datetime.time(22, 13, 47), '0046733165812')
]
I need to replace all numbers starting with ex 07 with +467, all 06 with +466 and 00 with +. For above example I need the number to turn out 0707262078 to [phone], 065017063 to [phone] and 0046733165812 to [phone].
Dont know if its possible to do this in regex only or if I need to do it with other code.
Been trying with re.sub combined with lamda, my thought is to make a dictionary with the matching replaces like this:
repl_dict = {
'01': '+461',
'02': '+462',
'03': '+463',
'04': '+464',
'05': '+465',
'06': '+466',
'07': '+467',
'08': '+468',
'09': '+469',
'00': '+'
}
My try so far:
import re
rows = [
(datetime.time(20, 35, 30), '0707262078',),
(datetime.time(20, 38, 18), '[phone]Ring via Mitel ',),
(datetime.time(20, 56, 35), '065017063'),
(datetime.time(21, 45, 1), '[phone]Ring via Mitel ',),
(datetime.time(22, 13, 47), '0046733165812')
]
repl_dict = {
'01': '+461',
'02': '+462',
'03': '+463',
'04': '+464',
'05': '+465',
'06': '+466',
'07': '+467',
'08': '+468',
'09': '+469',
'00': '+'
}
for row in rows:
regex = re.compile(r'^\d{1}[0-9](\d*)'), re.S
DialedNumber = regex.sub(lambda match: repl_dict.get(match.group(0), row[1]), row[1], row[1])

Answer 1

Your regex, ending in \d*, will match the entire number, and hence no entry is found in the dict. Also, there seems to be an unmatched parens and one too many row[1] in the call to sub.
You can simplify your regex to ^00? and your replacements dict to {'00': '+', '0': '+46'}. This will check whether the number starts with either one or two 0, making the replacement dict much simpler and less repetetive.
rows = [(datetime.time(20, 35, 30), '0707262078',), (datetime.time(20, 38, 18), '[phone]Ring via Mitel ',), (datetime.time(20, 56, 35), '065017063'), (datetime.time(21, 45, 1), '[phone]Ring via Mitel ',), (datetime.time(22, 13, 47), '0046733165812')]
repl_dict = {'00': '+', '0': '+46'}
regex = re.compile(r'^00?')
for date, number in rows:
print(regex.sub(lambda match: repl_dict.get(match.group(0)), number))
Output:
[phone]
[phone]Ring via Mitel
[phone]
[phone]Ring via Mitel
[phone]
If you only want the numeric part, you can pre- or postprocess the numbers with a second regex like [0-9+]*.

Answer 2

This is the naive approach based on repl_dict as given in your question.
def repl(match):
return repl_dict[match.group(0)]
pat = '^(' + '|'.join(repl_dict) + ')'
new_rows = [(tm, re.sub(pat, repl, ph)) for (tm, ph) in rows]
tobias_k's answer gives a better approach by improving your repl_dict and pattern.

Answer 3

Regex: ^[0-9]{2}
Details:
^ Asserts position at start of a line
[] Match a single character present in the list
{n} Matches exactly n times
Python code:
By #tobias_k you can use repl_dict.get(m.group(), m.group()) instead of repl_dict.get(m.group()) or m.group().
regex = re.compile(r'^[0-9]{2}')
for i in range(len(rows)):
rows[i] = (rows[i][0], regex.sub(lambda m: repl_dict.get(m.group()) or m.group(), rows[i][1]))
Output:
[(datetime.time(20, 35, 30), '[phone]'), (datetime.time(20, 38, 18), '[phone]Ring via Mitel '), (datetime.time(20, 56, 35), '[phone]'), (datetime.time(21, 45, 1), '[phone]Ring via Mitel '), (datetime.time(22, 13, 47), '[phone]')]
Code demo

Answer 4

You can do it without a regex:
for row in rows:
for repl in repl_dict:
if row[1].startswith(repl):
print repl_dict[repl]+row[1][len(repl):]
break