Regular Expression: is it possible to get numbers in optional parts by one regex - regex

Question

I have one string, it will be like: 1A2B3C or 2B3C or 1A2B or 1A3C.
The string is comprised by serval optional parts of number + [A|B|C].
It is possible to get the numbers before every character with one regex?
For example:
1A2B3C => (1, 2, 3)
1A3C => (1, 0, 3) There is no 'B', so gives 0 instead.
=> Or just (1, 3) but should show that the 3 is in front of 'C'.

Answer 1

Assuming Python because of your tuple notation, and because that's what I feel like using.
If the only allowed letters are A, B and C, you can do it with an extra processing step:
pattern = re.compile(r'(?:(\d+)A)(?:(\d+)B)?(?:(\d+)C)?')
match = pattern.fullmatch(some_string)
if match:
result = tuple(int(g) for g in match.groups('0'))
else:
raise ValueError('Bad input string')
Each option is surrounded by a non-capturing group (?:...) so the whole thing gets treated as a unit. Inside the unit, there is a capturing group (\d+) to capture the number, and an uncaptured character.
The method Matcher.groups returns a tuple of all the groups in the regex, with unmatched ones set to '0'. The generator then converts to an int for you. You could use tuple(map(int, match.groups('0'))).
You can also use a dictionary to hold the numbers, keyed by character:
pattern = re.compile(r'(?:(?P<A>\d+)A)(?:(?P<B>\d+)B)?(?:(?P<C>\d+)C)?')
match = pattern.fullmatch(some_string)
if match:
result = {k: int(v) for k, v in match.groupdict('0').items()}
else:
raise ValueError('Bad input string')
Matcher.groupdict is just like groups except that it returns a dictionary of the named groups: capture groups marked (?P<NAME>...).
Finally, if you don't mind having the dictionary, you can adapt this approach to parse any number of groups with arbitrary characters:
pattern = re.compile(r'(\d+)([A-Z])')
result = {}
while some_string:
match = pattern.match(some_string)
if not match:
raise ValueError('Bad input string')
result[match.group(2)] = int(match.group(1))
some_string = some_string[match.end():]