CODEWITHSUNDEEP
pythonregex
Kohei Sugimura
Kohei Sugimura

Reputation: 346

How to retrieve month with optional separator?

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

The requirements are,

"""
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()

Upvotes: 0

Views: 74

Answers (2)

The fourth bird
The fourth bird

Reputation: 163217

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)\d\d?|\d\d))\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)\d\d? If we have group 2, match 1 or 2 digits
      • | Or
      • \d\d 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)\d\d?|\d\d))\b"
s = ("202301\n"
            "2023-1\n"
            "2023-01\n"
            "20231")

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

Output

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

Upvotes: 1

simpleApp
simpleApp

Reputation: 3158

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

Upvotes: 1

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