regex pattern to match datetime in python


I have a string contains datetimes, I am trying to split the string based on the datetime occurances,

data="2018-03-14 06:08:18, he went on \n2018-03-15 06:08:18, lets play"

what I am doing,


what I get

["2018-03-14 06:08:18, he went on 2018-03-15 06:08:18, lets play"]

What I want:

["2018-03-14 06:08:18, he went on","2018-03-15 06:08:18, lets play"]

You want to split with at least 1 whitespace followed with a date like pattern, thus, you may use

re.split(r'\s+(?=\d{2}(?:\d{2})?-\d{1,2}-\d{1,2}\b)', s)

See the regex demo


<ul><li>\s+ - 1+ whitespace chars</li> <li>(?=\d{2}(?:\d{2})?-\d{1,2}-\d{1,2}\b) - a <em>positive lookahead</em> that makes sure, that immediately to the left of the current location, there are <ul><li>\d{2}(?:\d{2})? - 2 or 4 digits</li> <li>- - a hyphen</li> <li>\d{1,2} - 1 or 2 digits</li> <li>-\d{1,2} - again a hyphen and 1 or 2 digits</li> <li>\b - a word boundary (if not necessary, remove it, or replace with (?!\d) in case you may have dates glued to letters or other text)</li> </ul></li> </ul>

Python demo:

import re rex = r"\s+(?=\d{2}(?:\d{2})?-\d{1,2}-\d{1,2}\b)" s = "2018-03-14 06:08:18, he went on 2018-03-15 06:08:18, lets play" print(re.split(rex, s)) # => ['2018-03-14 06:08:18, he went on', '2018-03-15 06:08:18, lets play']

NOTE If there can be no whitespace before the date, in Python 3.7 and newer you may use r"\s*(?=\d{2}(?:\d{2})?-\d{1,2}-\d{1,2}\b)" (note the * quantifier with \s* that will allow zero-length matches). For older versions, you will need to use a solution as @blhsing suggests or install PyPi regex module and use r"(?V1)\s*(?=\d{2}(?:\d{2})?-\d{1,2}-\d{1,2}\b)" with regex.split.


re.split is meant for cases where you have a certain delimiter pattern. Use re.findall with a lookahead pattern instead:

import re data="2018-03-14 06:08:18, he went on \n2018-03-15 06:08:18, lets play" d = r'\d{4}-\d?\d-\d?\d (?:2[0-3]|[01]?[0-9]):[0-5]?[0-9]:[0-5]?[0-9]' print(re.findall(r'{0}.*?(?=\s*{0}|$)'.format(d), data, re.DOTALL))

This outputs:

['2018-03-14 06:08:18, he went on', '2018-03-15 06:08:18, lets play']

An similar, but alternative solution using a group instead:

import re data="2018-03-14 06:08:18, he went on 2018-03-15 06:08:18, lets play" print(re.findall(r'(.*?\D{2,})', data))

Which gives:

['2018-03-14 06:08:18, he went on ', '2018-03-15 06:08:18, lets play']



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