Python-单行与多行正则表达式

考虑到以下文本模式

#目标：流程报告时间戳，例如2011-09-21 15:45:00和succ中的前两个统计数据。统计行，例如：14381439

input_text = '''
# Process_Name     ( 23387) Report at 2011-09-21 15:45:00.001    Type:  Periodic    #\n
some line 1\n
some line 2\n
some other lines\n
succ. statistics |     1438     1439  99 |   3782245    3797376  99 |\n
some lines\n
Process_Name     ( 23387) Report at 2011-09-21 15:50:00.001    Type:  Periodic    #\n
some line 1\n
some line 2\n
some other lines\n
succ. statistics |     1436     1440  99 |   3782459    3797523  99 |\n
repeat the pattern several hundred times...
'''

我在一行一行迭代的时候让它工作了

def parse_file(file_handler, patterns):

    results = []
    for line in file_handler:
        for key in patterns.iterkeys():
            result = re.match(patterns[key], line)
            if result:
                results.append( result )

return results

patterns = {
    'report_date_time': re.compile('^# Process_Name\s*\(\s*\d+\) Report at (.*)\.[0-9]   {3}\s+Type:\s*Periodic\s*#\s*.*$'),
    'serv_term_stats': re.compile('^succ. statistics \|\s+(\d+)\s+   (\d+)+\s+\d+\s+\|\s+\d+\s+\d+\s+\d+\s+\|\s*$'),
    }
results = parse_file(fh, patterns)

返回

[('2011-09-21 15:40:00',),
('1425', '1428'),
('2011-09-21 15:45:00',),
('1438', '1439')]

但我的目标是输出一个元组列表

[('2011-09-21 15:40:00','1425', '1428'),
('2011-09-21 15:45:00', '1438', '1439')]

我尝试了几个带有初始模式和它们之间的惰性量词的组合，但不知道如何使用多行正则表达式捕获模式

# .+?   Lazy quantifier "match as few characters as possible (all characters allowed) until reaching the next expression"
pattern = '# Process_Name\s*\(\s*\d+\) Report at (.*)\.[0-9]{3}\s+Type:\s*Periodic.*?succ. statistics) \|\s+(\d+)\s+(\d+)+\s+\d+\s+\|\s+\d+\s+\d+\s+\d+\s+\|\s'
regex = re.compile(pattern, flags=re.MULTILINE)

data = file_handler.read()    
for match in regex.finditer(data):
    results = match.groups()

---

如何实现这一点？

使用

re.DOTALL

这样

将匹配任何字符，包括换行符：

import re

data = '''
# Process_Name     ( 23387) Report at 2011-09-21 15:45:00.001    Type:  Periodic    #\n
some line 1\n
some line 2\n
some other lines\n
succ. statistics |     1438     1439  99 |   3782245    3797376  99 |\n
some lines\n
repeat the pattern several hundred times...
'''

pattern = r'(\d{4}-\d{2}-\d{2} \d{2}:\d{2}:\d{2}).*?succ. statistics\s+\|\s+(\d+)\s+(\d+)'
regex = re.compile(pattern, flags=re.MULTILINE|re.DOTALL)

for match in regex.finditer(data):
    results = match.groups()
    print(results)

    # ('2011-09-21', '1438', '1439')

---

我没有答案，但为什么要将\n嵌入这样的多行字符串中？字符串中实际的换行符是换行符。对，Wooble，这是Linux中的换行符，所以只需添加它们来表示换行符（试图避免使用通常的\n或\r或\r\n？）哇。你跑得很快。谢谢你的回答和改进，谢谢你这样的大师！编辑：一个小碰撞，我确实需要保证一个非贪婪的量词，否则正则表达式将只捕获第一个时间戳，最后的统计数据，忽略中间的一千多行。因此，pattern=r'（\d{4}-\d{2}-\d{2}\d{2}:\d{2}:\d{2}）。*？成功。统计信息\s+\\124;\ s+（\d+）\s+（\d+）