如何为这个tex构造regex

Question

以下是输入：

7. Data 1 1. STR1 STR2 3. 12345 4. 0876 9. NO 2 1. STR 2. STRT STR 3. 9909090 5. YES 6. NO 7. YES 8. NO 9. YES 10. 5000 XX 11. 1000 ZŁ 12. NO PRub. 1 1. 1000 XX 2. NO 3 1. STRT 2. STRT 3. 63110300291 5. YES 6. NO 7. NO 8. NO 9. YES 10. 5000 XX 11. 1000 ZŁ 12. NO PRub. 1 1. 1000 XX 2. NO 4 1. QWERET 2. IOSTR9 3. 76012509879 5. YES 6. NO 7. NO 8. NO 9. YES 10. 5000 XX 11. 1000 XX 12. NO PRub. 1 1. 1000 XX 2. NO 0 1.

下面是预期的结果：

[('1', '1. STR1 STR2 3. 12345 4. 0876 9. NO'),
('2', '1. STR 2. STRT STR 3. 9909090 5. YES 6. NO 7. YES 8. NO 9. YES 10. 5000 XX 11. 1000 ZŁ 12. NO PRub. 1 1. 1000 XX 2. NO'),
('3', '1. STRT 2. STRT 3. 63110300291 5. YES 6. NO 7. NO 8. NO 9. YES 10. 5000 XX 11. 1000 ZŁ 12. NO PRub. 1 1. 1000 XX 2. NO'),
('4', '1. QWERET 2. IOSTR9 3. 76012509879 5. YES 6. NO 7. NO 8. NO 9. YES 10. 5000 XX 11. 1000 XX 12. NO PRub. 1 1. 1000 XX 2. NO')]

我试过这个：

re.findall(r'(?=\s(\d+)\s(1\..*?)\s\d+\s1\.)', txt, re.DOTALL)

当然，这不是正确的解决方案-正则表达式必须匹配(\d+) 1.，而不是PRub. 1 1.。
我该怎么做才能让它工作？你知道吗

Answer 1

这是怎么回事：

In [1]: s='7. Data 1 1. STR1 STR2 3. 12345 4. 0876 9. NO 2 1. STR 2. STRT STR 3. 9909090 5. YES 6. NO 7. YES 8. NO 9. YES 10. 5000 XX 11. 1000 ZŁ 12. NO PRub. 1 1. 1000 XX 2. NO 3 1. STRT 2. STRT 3. 63110300291 5. YES 6. NO 7. NO 8. NO 9. YES 10. 5000 XX 11. 1000 ZŁ 12. NO PRub. 1 1. 1000 XX 2. NO 4 1. QWERET 2. IOSTR9 3. 76012509879 5. YES 6. NO 7. NO 8. NO 9. YES 10. 5000 XX 11. 1000 XX 12. NO PRub. 1 1. 1000 XX 2. NO 0 1.'

In [2]: import re

In [3]: re.findall('(?<=\s)\d.*?(?=\s\d\s\d[.](?=$|\s[A-Z]))',s)
Out[3]: 
['1 1. STR1 STR2 3. 12345 4. 0876 9. NO',
 '2 1. STR 2. STRT STR 3. 9909090 5. YES 6. NO 7. YES 8. NO 9. YES 10. 5000 XX 11. 1000 Z\xc5\x81 12. NO PRub. 1 1. 1000 XX 2. NO',
 '3 1. STRT 2. STRT 3. 63110300291 5. YES 6. NO 7. NO 8. NO 9. YES 10. 5000 XX 11. 1000 Z\xc5\x81 12. NO PRub. 1 1. 1000 XX 2. NO',
 '4 1. QWERET 2. IOSTR9 3. 76012509879 5. YES 6. NO 7. NO 8. NO 9. YES 10. 5000 XX 11. 1000 XX 12. NO PRub. 1 1. 1000 XX 2. NO']

对于您的确切输出，我会做如下操作：

In [4]: ns = re.findall('(?<=\s)\d.*?(?=\s\d\s\d[.](?=$|\s[A-Z]))',s)

In [5]: [tuple(f.split(' ',1)) for f in ns]
Out[5]: 
[('1', '1. STR1 STR2 3. 12345 4. 0876 9. NO'),
 ('2', '1. STR 2. STRT STR 3. 9909090 5. YES 6. NO 7. YES 8. NO 9. YES 10. 5000 XX 11. 1000 Z\xc5\x81 12. NO PRub. 1 1. 1000 XX 2. NO'),
 ('3', '1. STRT 2. STRT 3. 63110300291 5. YES 6. NO 7. NO 8. NO 9. YES 10. 5000 XX 11. 1000 Z\xc5\x81 12. NO PRub. 1 1. 1000 XX 2. NO'),
 ('4', '1. QWERET 2. IOSTR9 3. 76012509879 5. YES 6. NO 7. NO 8. NO 9. YES 10. 5000 XX 11. 1000 XX 12. NO PRub. 1 1. 1000 XX 2. NO')]

可能是更好的方法，但是我的python foo不如regexp foo好。你知道吗

规则解释：

(?<=\s) # Use positive look-behind to match a leading space but don't include it
\d      # match digit    
.*?     # Match everything up till the next record (lazy)
        # The following positive look-behinds is the key. It matches the start of
        # each new record i.e
        # 2 1. S
        # 3 1. S
        # 4 1. Q
        # 0 1.$ 
        # look-arounds match but don't seek past.  
(?=\s\d\s\d[.](?=$|\s[A-Z]))
(?=     # positive look-ahead 1
\s      # space
\d      # digit
\s      # space
\d      # digit
[.]     # period
(?=     # postive look-ahead 2 
$       # end of string
|       # OR
\s[A-Z] # space followed by uppercase letter
)       # close look-ahead 1
)       # close look-ahead 2