How to construct regex for this text [closed]

Question

Here's the input:

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.

And here's expected output:

[('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')]

I've tried this:

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

But of course it's not right solution - regex have to match (\d+) 1. but not PRub. 1 1..
What should I do to make it work?

Answer 1

How is this:

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']

For you exact output I'd do something like:

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')]

Might be a better way to do this but my python foo isn't as good as my regexp foo.

Regexplanation:

(?<=\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