5

入力は次のとおりです。

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.

そして、期待される出力は次のとおりです。

[('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')]

私はこれを試しました：

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

しかしもちろん、それは正しい解決策ではありません。正規表現は一致する必要がありますが、一致する必要は(\d+) 1.ありませんPRub. 1 1.。
機能させるにはどうすればよいですか？

4

1 に答える 1

4

これはどのように：

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']

正確な出力については、次のようにします。

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')]

これを行うにはより良い方法かもしれませんが、私の python foo は私の正規表現 foo ほど良くありません。

再説明:

(?<=\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
于 2013-04-18T08:21:23.190 に答える