正则表达式模式以匹配python中的日期时间 [英] regex pattern to match datetime in python

问题描述

我有一个包含日期时间的字符串，我正在尝试根据日期时间出现次数来拆分字符串，

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"

我在做什么

out=re.split('^(2[0-3]|[01]?[0-9]):([0-5]?[0-9]):([0-5]?[0-9])$',data)

我会得到什么

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

我想要什么:

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

推荐答案

您希望使用至少1个空格和后跟日期(如pattern)进行拆分，因此，您可以使用

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)

请参见 regex演示

详细信息

• \s+-1个以上空格字符
• (?=\d{2}(?:\d{2})?-\d{1,2}-\d{1,2}\b)-一个正向超前，可确保在当前位置的左侧紧随其后
  • \d{2}(?:\d{2})?-2位或4位数字
  • --连字符
  • \d{1,2}-1或2位数字
  • -\d{1,2}-再加上一个连字符和1或2位数字
  • \b-单词边界(如有必要，请将其删除，或将其替换为(?!\d)，以防您将日期粘贴到字母或其他文本上)
  • \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演示:

    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']
    

    注意如果日期前不能有空格，则在Python 3.7及更高版本中，您可以使用r"\s*(?=\d{2}(?:\d{2})?-\d{1,2}-\d{1,2}\b)"(请注意，将*量词与\s*配合使用，将允许零长度匹配).对于较旧的版本，您需要按照@blhsing的建议使用解决方案或安装与regex.split一起使用.

    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.

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