用 Python re 挖掘信息

Question

我想从日志文件中挖掘信息并编写以下脚本：

import re

file = '''Date,Time,Type,User,Message

Thu Jul 18, 2019 14:18:41.945,EFM,201202      ,Robot picked

Thu Jul 18, 2019 14:18:51.486,DS ,201202      ,Module 1

Thu Jul 18, 2019 14:19:07.747,DS ,201202      ,Door opened

Thu Jul 18, 2019 14:20:08.231,EFM,203204205206,Robot picked

Thu Jul 18, 2019 14:20:08.231,DS ,203204      ,Module 2

Thu Jul 18, 2019 14:20:10.282,DS ,203204      ,Door opened

...

'''

p1 = re.compile(r'\w{3} \w{3} \d\d, \d{4} (\d\d:\d\d:\d\d.\d{3}),EFM,(\d+?\s*?),Robot picked')

p2 = re.compile(r'\w{3} \w{3} \d\d, \d{4} (\d\d:\d\d:\d\d.\d{3}),DS ,(\d+?\s*?),Module 1')

p3 = re.compile(r'\w{3} \w{3} \d\d, \d{4} (\d\d:\d\d:\d\d.\d{3}),DS ,(\d+?\s*?),Door opened')

w_file = r'D:\sample.txt'

lines = file.readlines()

t_file =open(w_file,'w')

info = ['User','Time1','Time2','Time3' ]

t_file.write('{}\n'.format(','.join(item for item in info)))

for line in lines:

    p1_line = re.findall(p1, line.strip())

    p2_line = re.findall(p2, line.strip())

    p3_line = re.findall(p3, line.strip())

    if p1_line and p2_line and p3_line:

        if p1_line[0][1][:3] == p2_line[0][1][:3] and p1_line[0][1][:3] == p5_line[0][1][:3]:

            t_file.write('{},{},{},{}\n'.format(p1_line[0][1].strip(),p1_line[0][0],p2_line[0][0],p3_line[0][0])

t_file.close()

当我打开 sample.txt 文件时，只有 'User,Time1,Time2,Time3' 行。任何人都可以在我的脚本中找到问题吗？

我想要的如下：

User,Time1,Time2,Time3

201202,14:18:41.945,14:18:51.486,14:19:07.747

203204205206,14:20:08.231,14:20:08.231,14:20:10.282

Answer 1

您的脚本的问题是您试图将所有正则表达式匹配到同一行，然后执行一个和条件，这当然会失败。每个正则表达式都有效，但仅适用于特定行，因此 3 个中的 2 个将返回[]其计算结果为False.例如，给定： line = 'Thu Jul 18, 2019 14:18:41.945,EFM,201202      ,Robot picked'你将会有：p1_line = [('14:18:41.945', '201202      ')]  # matchp2_line = []                                  # no matchp3_line = []                                  # no match一旦你and有了这三个值，条件将评估为False，因此不会将任何内容写入文件：if p1_line and p2_line and p3_line:  # this evaluates to False因此，根据您要实现的确切逻辑，您可能必须存储和记住过去的匹配并以此为基础。