How to retrieve month with optional separator?

Question

I tried to write python script to retrieve year and month from string.

The requirements are,

year is fixed at 4 characters
month is allowed to be two consecutive characters, or one or two characters when followed by a non-numeric value

"""
This is the "search year and month" module.

>>> search_year_month('202301')
True
>>> search_year_month('2023-1')
True
>>> search_year_month('2023-01')
True
>>> search_year_month('20231')
False
"""

import re

_re = re.compile(
    r"(?P<year>d{4})"
    r"(?P<month>d{2}|(?<=[^d])d{1,2})"
)

def search_year_month(v):
    match = _re.search(v)
    return match is not None

if __name__ == "__main__":
    import doctest
    doctest.testmod()

But, 2023-1 and 2023-01 are failed…
Is there a better way to build regular expressions?

I tried only month part, I got expected result.

"""
This is the "single lookbehind sample" module.

>>> search_lookbehind('01')
True
>>> search_lookbehind('-1')
True
>>> search_lookbehind('-01')
True
>>> search_lookbehind('1')
False
"""

import re

_re = re.compile(
    r"(?P<month>d{2}|(?<=[^d])d{1,2})"
)

def search_lookbehind(v):
    match = _re.search(v)
    return match is not None

if __name__ == "__main__":
    import doctest
    doctest.testmod()

Asked By: Kohei Sugimura

||

Source

Answer 1

option 1: search and replace strategy

import re
test_string=['202301','2023-1','2023-01','20231']
year=r"^d{4}"
sep=r"W+"
for e in test_string:
    year=re.findall(r"^d{4}",'202301')[0]
    e=re.sub(year,"",e)
    e=re.sub(sep,"",e)
    print(f"year:{year},month:{e}")

option 2: regex (all in one)

re_all=r"(^d{4})(?:W+){0,1}(d+)"
for e in test_string:
    search_results=re.search(re_all,e)
    print(f"year:{search_results.group(1)},month:{search_results.group(2)}")

same outcome:

year:2023,month:01
year:2023,month:1
year:2023,month:01
year:2023,month:1

Answered By: simpleApp

Answer 2

You could use a conditional checking for group 2 that holds an optional - char.

If there is a value for group 2, then match 1 or 2 digits for the month, else match 2 digits for the month.

b(?P<year>d{4})(-)?(?P<month>(?(2)dd?|dd))b

Explanation

b A word boundary to prevent a partial word match
(?P<year>d{4}) Group year matching 4 digits
(-)? Optional capture group 2 matching a -
(?P<month> Group month
- (? Start a conditional
  - (2)dd? If we have group 2, match 1 or 2 digits
  - | Or
  - dd Match 2 digits
- ) Close the conditional
) Close group month
b A word boundary

See a regex101 demo and a Python demo.

For example

import re

pattern = r"b(?P<year>d{4})(-)?(?P<month>(?(2)dd?|dd))b"
s = ("202301n"
            "2023-1n"
            "2023-01n"
            "20231")

for m in re.finditer(pattern, s):
    print(m.groupdict())

Output

{'year': '2023', 'month': '01'}
{'year': '2023', 'month': '1'}
{'year': '2023', 'month': '01'}

Answered By: The fourth bird

How to retrieve month with optional separator?

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