Regex for extracting all complex dates formats from a string in python

Question

I have following string:

 dateEntries = "04-20-2009; 04/20/09; 4/20/09; 4/3/09; Mar 20, 2009; March 20, 2009; Mar. 20, 2009; Mar 20 2009; 20 Mar 2009; 20 March 2009; 2 Mar. 2009; 20 March, 2009; Mar 20th, 2009; Mar 21st, 2009; Mar 22nd, 2009; Feb 2009; Sep 2009; Oct 2010; 6/2008; 12/2009; 2009; 2010"

Here I want to extract all mentioned dates using regex. As an attempt I have written following regex:

import re

regEx = r'(?:\d{1,2}[-/th|st|nd|rd\s]*)?(?:Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec)[a-z\s,.]*(?:\d{1,2}[-/th|st|nd|rd)\s,]*)?(?:\d{2,4})'

re.findall(regEx, dateEntries)

I was expecting this to work but it only return subset of dates.

A = ['Mar 20, 2009',
 'March 20, 2009',
 'Mar. 20, 2009',
 'Mar 20 2009',
 '20 Mar 2009',
 '20 March 2009',
 '2 Mar. 2009',
 '20 March, 2009',
 'Mar 20th, 2009',
 'Mar 21st, 2009',
 'Mar 22nd, 2009',
 'Feb 2009',
 'Sep 2009',
 'Oct 2010']

I'm not getting why its not returning the dates:

B=[04-20-2009; 04/20/09; 4/20/09; 4/3/09; 6/2008; 12/2009; 2009; 2010"]

I created the regEx by extending the r'(?:\d{1,2}[-\s\/])?(?:\d{1,2}[-\/\s])?(?:\d{2,4})' which works good for set B. But regEx is not able to produce A+B

Can anyone help in making a regex for extracting all dates mentioned in my dateEntries ?

NOTE: I want to solve this using regex only.

Why do you want to use a regex? For your example you could just use dateEntries.split(";"). — Nils Schlüter
– Nils Schlüter, Commented Jul 1, 2018 at 10:49
Because my real data has text file in which set A categories dates are possible, and text file has other data apart from dates. — Amit Sharma
– Amit Sharma, Commented Jul 1, 2018 at 10:52
FYI [] matches single characters and character ranges, not strings like th or st. You should replace with () — DisappointedByUnaccountableMod
– DisappointedByUnaccountableMod, Commented Jul 1, 2018 at 11:01
Try (?:[\s]?\d{1,2}[-/th|st|nd|rd\s]*)?(?:Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec)?[a-z\s,./]*(?:\d{1,2}[-/th|st|nd|rd)\s,]*)?(?:\d{2,4}) here — Paolo
– Paolo, Commented Jul 1, 2018 at 11:10

Nils Schlüter · Accepted Answer · 2018-07-01 11:12:35Z

6

You are just missing a single ? after the (?:Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec) group to mark it as not necessary. Additionally I added a + behind the last two groups to make sure the regex doesn't split dates like "20 March 2009" into two different dates.

The full code:

import re

regEx = r'(?:\d{1,2}[-/th|st|nd|rd\s]*)?(?:Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec)?[a-z\s,.]*(?:\d{1,2}[-/th|st|nd|rd)\s,]*)+(?:\d{2,4})+'

dateEntries = "04-20-2009; 04/20/09; 4/20/09; 4/3/09; Mar 20, 2009; March 20, 2009; Mar. 20, 2009; Mar 20 2009; 20 Mar 2009; 20 March 2009; 2 Mar. 2009; 20 March, 2009; Mar 20th, 2009; Mar 21st, 2009; Mar 22nd, 2009; Feb 2009; Sep 2009; Oct 2010; 6/2008; 12/2009; 2009; 2010"
result = re.findall(regEx, dateEntries)
print(result)

If your date has leading whitespaces, the result will also have leading whitespaces. If you continue using the date string you could remove them for example with the .strip() method

edited Jul 1, 2018 at 11:12

answered Jul 1, 2018 at 10:43

Nils Schlüter

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3 Comments

Amit Sharma Over a year ago

Thanks @Nils. I'm also able to extract using this regex

r'(?:\d{1,2}[-/th|st|nd|rd\s.])?(?:(?:Jan|January|Feb|February|Mar|March|Apr|April|May|Jun|June|Jul|July|August|Sep|September|Oct|October|Nov|November|Dec|December)[\s,.]*)?(?:(?:\d{1,2})[-/th|st|nd|rd\s,.]*)?(?:\d{2,4})'

Matt Mateo Over a year ago

How to capture this format? 2020 January 1 2020 January 01 ; 2020 Jan. 1 ; 2020 Jan. 01 ; 2020 JAN. 1 ; 2020 JAN. 01

Barmar Dec 19, 2024 at 21:15

[-/th|st|nd|rd)\s] doesn't do what you think it does. I don't see how this can possibly work.

Sunitha · Accepted Answer · 2020-08-08 10:25:35Z

3

Your regex pattern is totally unreadable.. Please build your regex pattern with simple building blocks. That would make the code a lot more readable

import re
import calendar

full_months = [month for month in calendar.month_name if month]
short_months = [d[:3] for d in full_months]
months = '|'.join(short_months + full_months)

sep = r'[.,]?\s+'               # seperator
day = r'\d+'
year = r'\d+'
day_or_year = r'\d+(?:\w+)?'

r = re.compile(rf'(?:{day}{sep})?(?:{months}){sep}{day_or_year}(?:{sep}{year})?')
r.findall(dateEntries)
# ['Mar 20, 2009', 'March 20, 2009', 'Mar. 20, 2009', 'Mar 20 2009', '20 Mar 2009', '20 March 2009', '2 Mar. 2009', '20 March, 2009', 'Mar 20th, 2009', 'Mar 21st, 2009', 'Mar 22nd, 2009', 'Feb 2009', 'Sep 2009', 'Oct 2010']

edited Aug 8, 2020 at 10:25

answered Jul 1, 2018 at 12:09

Sunitha

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Comments

Matt.G · Accepted Answer · 2018-07-01 11:22:08Z

0

Try Regex:

^(?:\d{1,2}(?:(?:-|/)|(?:th|st|nd|rd)?\s))?(?:(?:(?:Jan(?:uary)?|Feb(?:ruary)?|Mar(?:ch)?|Apr(?:il)?|May|Jun(?:e)?|Jul(?:y)?|Aug(?:ust)?|Sep(?:tember)?|Oct(?:ober)?|Nov(?:ember)?|Dec(?:ember)?)(?:(?:-|/)|(?:,|\.)?\s)?)?(?:\d{1,2}(?:(?:-|/)|(?:th|st|nd|rd)?\s))?)(?:\d{2,4})$

Demo

answered Jul 1, 2018 at 11:22

Matt.G

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Comments

shuberman · Accepted Answer · 2020-07-08 08:35:07Z

-1

You can try the following regex

(?:\d{1,2}[-/th|st|nd|rd\s]*)?(?:Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec)?[a-z\s,.]*(?:\d{1,2}[-/th|st|nd|rd)\s,]*)+(?:\d{2,4})+

edited Jul 8, 2020 at 8:35

shuberman

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answered Jun 28, 2020 at 10:58

Raj

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