Python regEx for datetime mining

Question

I'm trying to mine a text into a list using re.

Here is what I've written:

dateStr = "20 Mar 2009; 20 March 2009; 20 Mar. 2009; 20 March, 2009" regex = r'(?:\d{1,2}[/-]*)?(?:Mar)?[a-z\s,.]*(?:\d{1,2}[/-]*)+(?:\d{2,4})+' result = re.findall(regex, dateStr)

Even if I stated (?:\d{1,2}[/-]*) at the beginning of the expression, I'm missing the days digits. Here is what I get :

['Mar 2009', 'March 2009', 'Mar. 2009', 'March, 2009']

Could you help? Thanks

Edit:
This question was solved through the comments.

Original assignment string: dateStr = "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"

You missed the whitespace, r'(?:\d{1,2}[\s/-]*)?(?:Mar)?[a-z\s,.]*(?:\d{1,2}[\s/-]*)+(?:\d{2,4})+', see demo. From the looks of it, you want to match many more date formats, what are your pattern requirements? — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Oct 1, 2021 at 11:15
That's it! @WiktorStribiżew . I'm trying to match the pattern below for a Coursera Assignment and your answer greatly solves my problem. Thanks! — Girolamo
– Girolamo, Commented Oct 1, 2021 at 12:12

Tim Biegeleisen · Accepted Answer · 2021-10-01 10:53:15Z

I would use:

dateStr = "20 Mar 2009; 20 March 2009; 20 Mar. 2009; 20 March, 2009" dt = re.findall(r'\d{1,2} \w+[,.]? \d{4}', dateStr) print(dt) # ['20 Mar 2009', '20 March 2009', '20 Mar. 2009', '20 March, 2009']

The one size fits all regex pattern used above says to match:

\d{1,2} a one or two digit day [ ] space \w+ month name or abbreviation [,.]? possibly followed by comma or period [ ] space \d{4} four digit year

Bhagyesh Dudhediya · Accepted Answer · 2021-10-01 10:59:46Z

One of the many approaches:

import re dateStr = "20 Mar 2009; 20 March 2009; 20 Mar. 2009; 20 March, 2009" regex = r'[0-9]{1,2}\s[a-zA-Z]+[.,]*\s[0-9]{4}' result = re.findall(regex, dateStr) print (result)

Output:

['20 Mar 2009', '20 March 2009', '20 Mar. 2009', '20 March, 2009']

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