Question

我有一个这样的字符串：

   old_ActNacd_2016-12-21_07-09-08.txt:100:2016-12-21 07:08:20 - [HSM   ]Handle Identity Request. Send Identity Response. timeout: 1550s
   old_ActNacd_2016-12-21_08-21-04.txt:52:2016-12-21 07:21:42 - [HSM ]Handle Identity Request. Send Identity Response. timeout: 1550s
   old_ActNacd_2016-12-21_08-37-50.txt:49:2016-12-21 08:23:34 - [HSM ]Handle Identity Request. Send Identity Response. timeout: 1550s
   old_ActNacd_2016-12-21_15-00-47.txt:49:2016-12-21 08:39:16 - [HSM ]Handle Identity Request. Send Identity Response. timeout: 1550s

我试过这样做：

  #creating list after taking the string out
  log_list = ostring.split('Handle Identity Request. Send Identity Response. timeout: 1550s')
    for itr in log_list:
        #getting the dates from the log_list
        match = re.search(r'\d{4}-\d{2}-\d{2}', itr)
        if match:
            date = datetime.strptime(match.group(), '%Y-%m-%d').date()

此过程运行正常，但我只想在一个操作中执行，而不是分两步执行（拆分和匹配）

  Note:-I want to create a list of dates from the string present between ":" and "space" in the string. I don't want the date present with "_ActNacd_" string.

所以我将创建一个包含日期的列表：

['2016-12-21','2016-12-21', '2016-12-21', '2016-12-21']

Answer 1

使用re.findall()，您可以实现以下目标：

re.findall(r'(\d{4}\-\d{2}\-\d{2})', s)

如果您只需要每行中的第二个日期，请尝试：

re.findall(r':(\d{4}\-\d{2}\-\d{2})', s)

<强>输出：

>>> import re
>>> 
>>> s = '''old_ActNacd_2016-12-21_07-09-08.txt:100:2016-12-21 07:08:20 - [HSM   ]Handle Identity Request. Send Identity Response. timeout: 1550s
... old_ActNacd_2016-12-21_08-21-04.txt:52:2016-12-21 07:21:42 - [HSM ]Handle Identity Request. Send Identity Response. timeout: 1550s'''
>>>
>>> re.findall(r':(\d{4}\-\d{2}\-\d{2})', s)
['2016-12-21', '2016-12-21']

Answer 2

首先尝试按\ n分割，然后你可以逐行迭代并使用反向子字符串获取日期并使用.append（）函数获取所需的列表

Python：从字符串中提取日期并放入列表中

2 个答案: