正则表达式模式以匹配python中的日期时间

Question

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"]

Answer 1

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)

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 .

Answer 2

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']

Answer 3

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']