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,
out=re.split('^(2[0-3]|[01]?[0-9]):([0-5]?[0-9]):([0-5]?[0-9])$',data)
what I get
["2018-03-14 06:08:18, he went on 2018-03-15 06:08:18, lets play"]
What I want:
["2018-03-14 06:08:18, he went on","2018-03-15 06:08:18, lets play"]
You want to split with at least 1 whitespace followed with a date like pattern, thus, you may use
re.split(r'\s+(?=\d{2}(?:\d{2})?-\d{1,2}-\d{1,2}\b)', s)
See the regex demo
Details
\s+ - 1+ whitespace chars(?=\d{2}(?:\d{2})?-\d{1,2}-\d{1,2}\b) - a positive lookahead that makes sure, that immediately to the left of the current location, there are
\d{2}(?:\d{2})? - 2 or 4 digits- - a hyphen\d{1,2} - 1 or 2 digits-\d{1,2} - again a hyphen and 1 or 2 digits\b - a word boundary (if not necessary, remove it, or replace with (?!\d) in case you may have dates glued to letters or other text)import re
rex = r"\s+(?=\d{2}(?:\d{2})?-\d{1,2}-\d{1,2}\b)"
s = "2018-03-14 06:08:18, he went on 2018-03-15 06:08:18, lets play"
print(re.split(rex, s))
# => ['2018-03-14 06:08:18, he went on', '2018-03-15 06:08:18, lets play']
NOTE If there can be no whitespace before the date, in Python 3.7 and newer you may use r"\s*(?=\d{2}(?:\d{2})?-\d{1,2}-\d{1,2}\b)" (note the * quantifier with \s* that will allow zero-length matches). For older versions, you will need to use a solution as @blhsing suggests or install PyPi regex module and use r"(?V1)\s*(?=\d{2}(?:\d{2})?-\d{1,2}-\d{1,2}\b)" with regex.split.
re.split is meant for cases where you have a certain delimiter pattern. Use re.findall with a lookahead pattern instead:
import re
data="2018-03-14 06:08:18, he went on \n2018-03-15 06:08:18, lets play"
d = r'\d{4}-\d?\d-\d?\d (?:2[0-3]|[01]?[0-9]):[0-5]?[0-9]:[0-5]?[0-9]'
print(re.findall(r'{0}.*?(?=\s*{0}|$)'.format(d), data, re.DOTALL))
This outputs:
['2018-03-14 06:08:18, he went on', '2018-03-15 06:08:18, lets play']
re.split(r'\s+(?=\d{2}(?:\d{2})?-\d{1,2}-\d{1,2}\b)', s) might be a better choice.
'\n' in the string you just need to add an re.DOTALL flag to findall. I've updated my answer accordingly then.'\n' before each date/time, you might as well use `str.split('\n')`` to get what you want.
r'\s+(?=(?:(?:20)?[01]?[0-9])-(?:1[0-2]|0?[0-9])-(?:[0-2]?[0-9]|3[01]))'with split.