[FIXED] Python regex for date and numbers, find the date format

Issue

How to extract dates alone from text file using regex in Python 3?

Below is my current code:

import datetime
from datetime import date
import re
s = "birthday on 20/12/2018 and wedding aniversry on 04/01/1997 and dob is on 
09/07/1897"
match = re.search(r'\d{2}/\d{2}/\d{4}', s)
date = datetime.datetime.strptime(match.group(), '%Y-%m-%d').date()
print (date)

Expected Output is

20/12/2018
04/01/1997
09/07/1897

Solution

You have an invalid date format near '%Y-%m-%d' since it should have been '%d/%m/%Y' looking at your provided date: birthday on 20/12/2018 (dd/mm/yyyy)

Change this:

date = datetime.datetime.strptime(match.group(), '%Y-%m-%d').date()

With this:

date = datetime.datetime.strptime(match.group(), '%d/%m/%Y').date()

Your Fix:

import datetime
from datetime import date
import re
s = "birthday on 20/12/2018"
match = re.search(r'\d{2}/\d{2}/\d{4}', s)
date = datetime.datetime.strptime(match.group(), '%d/%m/%Y').date()
print (date)

But:

Why get into all the trouble? When they're easier and elegant ways out there.

Using dparser:

import dateutil.parser as dparser
dt_1 = "birthday on 20/12/2018"
print("Date: {}".format(dparser.parse(dt_1,fuzzy=True).date()))

OUTPUT:

Date: 2018-12-20

EDIT:

With your edited question which now has multiple dates, you could extract them using regex:

import re
s = "birthday on 20/12/2018 and wedding aniversry on 04/01/1997 and dob is on 09/07/1897"
pattern = r'\d{2}/\d{2}/\d{4}'
print("\n".join(re.findall(pattern,s)))

OUTPUT:

20/12/2018
04/01/1997
09/07/1897

OR

Using dateutil:

from dateutil.parser import parse

for s in s.split():
    try:
        print(parse(s))
    except ValueError:
        pass

OUTPUT:

2018-12-20 00:00:00
1997-04-01 00:00:00
1897-09-07 00:00:00

Answered By - DirtyBit