Convert international datestring to ISO-format

Question

The function below takes a datestring in the form d-mmm-yyyy and converts it to ISO date format (yyyy-mm-dd). Delimiters may be hyphen, space or /, and Dutch or English month abbreviations may be used.

Now, I know there is dateutil, but it returns unknown string format if you try to parse something with a non-English month in it. I haven't digested all its documentation, but I think dateutil is mainly intended for date calculation. I'm not doing that, I'm just cleaning up user input.

So I wrote my own.

import re
.
.

def ISOdate(date): 
    '''
        converts the following date string format to ISO (yyyy-mm-dd):  
        28-okt-1924 (dutch month abbreviations)
        28 oct 1924 (english..) 
         9/nov/2012 (single digit)
    '''

    shortmonths = [
        'jan', 'feb', 'mrt', 'apr', 'mei', 'jun', 
        'jul', 'aug', 'sep', 'okt', 'nov', 'dec', 
        'jan', 'feb', 'mar', 'apr', 'may', 'jun', 
        'jul', 'aug', 'sep', 'oct', 'nov', 'dec'
        ] 

    # Month abbrevs are only different march, may and october.             

    pat = r'(\d{1,2})\s?[-\/]?\s?(\w{3})\s?[-\/]?\s?(\d{4})'

    q = re.match(pat, date)
    if q: 
        year = q.group(3)
        day = int(q.group(1)) 
        month = shortmonths.index(q.group(2).lower()) % 12 + 1
        return u'{}-{:02d}-{:02d}'.format(year, month, day)
    else:
        # just return input, date fields may be empty
        return date

The regex match parses date month and year may not be pretty, but it works and it's easy to expand for other patterns. Likewise for the month number lookup with index, which is more concise than a chain of if, elif to match month strings to numbers.

Instead of the code between if q: and else:, I also had this, which uses datetime:

year = int(q.group(3))
day = int(q.group(1))
month = shortmonths.index(q.group(2).lower()) % 12 + 1
d = datetime.datetime(year, month, day)
return u'{:%YY-%m-%d}'.format(d)

This works too, but I think the first version is more to-the-point.

I'm always striving for optimal Pythonicness, but I'm by no means a seasoned coder.

Please comment.

Answer 1

Neat stuff.

Suggestions:

Changing shortmonths to a dictionary. This will allow for a pair between numerical months and alphabetical months. No need to repeat 'jan' for example, as you have it now.

Pythonic: unpack month, year, day in a one liner.

Use datetime's strftime to format dates...makes life easier in case you want to change the format down the road.

import re
import datetime

def ISOdate(date):

    month_d = {'01': 'jan',
               '02': 'feb',
               '03': ['mar', 'mrt'],
               '04': 'apr',
               '05': ['may', 'mei'],
               '06': 'jun',
               '07': 'jul',
               '08': 'aug',
               '09': 'sep',
               '10': ['oct', 'okt'],
               '11': 'nov',
               '12': 'dec'
               }

     pat = r'(\d{1,2})\s?[-\/]?\s?(\w{3})\s?[-\/]?\s?(\d{4})'
     q = re.match(pat, date)

     if q:
         day, month, year = [q.group(idx+1) for idx in range(3)]
         if month.isalpha(): # change from letters to numbers
             month = [k for k, v in month_d.items() if month in v][0]
         out_date = datetime.date(int(year), int(month), int(day))
         return datetime.datetime.strftime(out_date, '%Y-%m-%d')  

     else:
         return date    

Answer 2

A few comments/suggestions:

• Use longer variable names to improve readability. For example: pat -> pattern, q -> match, idx -> index.
• Try to follow pep257 for docstrings
• Why return the same input if it wasn't possible to parse a date? What about returning None or maybe raising a ValueError exception?

• Regular expressions will always be hard to read, but group names can be used to avoid hardcoding group numbers later:

  pattern = r'(?P<day>\d{1,2})\s?[-\/]?\s?(?P<month>\w{3})\s?[-\/]?\s?(?P<year>\d{4})'
  
  match = re.match(pat, date)
  if match:
      year = int(match.group('year'))
      month = match.group('month')
      day = int(match.group('day'))
  

Finally, if dateutil.parser.parse is already doing a good job for the strings you want to parse except for the month strings, what about replacing the dutch abbreviations with english ones an pass the string to dateutil.parser.parse?

import dateutil.parser

def ISO_date(date):
    """Convert date string format to ISO (yyyy-mm-dd).

    Examples:
        28-okt-1924 (dutch month abbreviations)
        28 oct 1924 (english..)
         9/nov/2012 (single digit)
    """
    months = {
        'mrt': 'mar',
        'mei': 'may',
        'okt': 'oct',
    }

    for month_dutch, month_english in months.items():
        date = date.replace(month_dutch, month_english)

    return dateutil.parser.parse(date)

print ISO_date('28-okt-1924')