Python:提取日期从一根绳子放进一个链子里

2024-10-02 18:15:05 发布

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我有这样一根线:

   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

我试着这样做:

  #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()

这个过程工作得很好,但我只想在一个操作中完成,而不是在两个步骤中完成(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.

因此,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']

Tags: thetxtsendlogstringresponserequestmatch
2条回答

使用^{},您可以实现如下所示:

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

如果每行中只需要第二个日期,请尝试:

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

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']

首先尝试按\n拆分,然后可以按行迭代并使用反向子字符串获取日期,然后使用.append()函数获取所需列表

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