Python regex to match dates

I built my solution on top of @aditya Prakash appraoch:

 print(re.search("^([1-9]|0[1-9]|1[0-9]|2[0-9]|3[0-1])(\.|-|/)([1-9]|0[1-9]|1[0-2])(\.|-|/)([0-9][0-9]|19[0-9][0-9]|20[0-9][0-9])$|^([0-9][0-9]|19[0-9][0-9]|20[0-9][0-9])(\.|-|/)([1-9]|0[1-9]|1[0-2])(\.|-|/)([1-9]|0[1-9]|1[0-9]|2[0-9]|3[0-1])$",'01/01/2018'))

The first part (^([1-9]|0[1-9]|1[0-9]|2[0-9]|3[0-1])(\.|-|/)([1-9]|0[1-9]|1[0-2])(\.|-|/)([0-9][0-9]|19[0-9][0-9]|20[0-9][0-9])$) can handle the following formats:

  • 01.10.2019
  • 1.1.2019
  • 1.1.19
  • 12/03/2020
  • 01.05.1950

The second part (^([0-9][0-9]|19[0-9][0-9]|20[0-9][0-9])(\.|-|/)([1-9]|0[1-9]|1[0-2])(\.|-|/)([1-9]|0[1-9]|1[0-9]|2[0-9]|3[0-1])$) can basically do the same, but in inverse order, where the year comes first, followed by month, and then day.

  • 2020/02/12

As delimiters it allows ., /, -. As years it allows everything from 1900-2099, also giving only two numbers is fine.

If you have suggestions for improvement please let me know in the comments, so I can update the answer.


I find the below RE working fine for Date in the following format;

  1. 14-11-2017
  2. 14.11.2017
  3. 14|11|2017

It can accept year from 2000-2099

Please do not forget to add $ at the end,if not it accept 14-11-201 or 20177

date="13-11-2017"

x=re.search("^([1-9] |1[0-9]| 2[0-9]|3[0-1])(.|-)([1-9] |1[0-2])(.|-|)20[0-9][0-9]$",date)

x.group()

output = '13-11-2017'


Instead of using regex, it is generally better to parse the string as a datetime.datetime object:

In [140]: datetime.datetime.strptime("11/12/98","%m/%d/%y")
Out[140]: datetime.datetime(1998, 11, 12, 0, 0)

In [141]: datetime.datetime.strptime("11/12/98","%d/%m/%y")
Out[141]: datetime.datetime(1998, 12, 11, 0, 0)

You could then access the day, month, and year (and hour, minutes, and seconds) as attributes of the datetime.datetime object:

In [143]: date.year
Out[143]: 1998

In [144]: date.month
Out[144]: 11

In [145]: date.day
Out[145]: 12

To test if a sequence of digits separated by forward-slashes represents a valid date, you could use a try..except block. Invalid dates will raise a ValueError:

In [159]: try:
   .....:     datetime.datetime.strptime("99/99/99","%m/%d/%y")
   .....: except ValueError as err:
   .....:     print(err)
   .....:     
   .....:     
time data '99/99/99' does not match format '%m/%d/%y'

If you need to search a longer string for a date, you could use regex to search for digits separated by forward-slashes:

In [146]: import re
In [152]: match = re.search(r'(\d+/\d+/\d+)','The date is 11/12/98')

In [153]: match.group(1)
Out[153]: '11/12/98'

Of course, invalid dates will also match:

In [154]: match = re.search(r'(\d+/\d+/\d+)','The date is 99/99/99')

In [155]: match.group(1)
Out[155]: '99/99/99'

To check that match.group(1) returns a valid date string, you could then parsing it using datetime.datetime.strptime as shown above.

Tags:

Python

Regex

Date

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