[英]Behaviour of Python non-greedy regular expression
I'm using python version 3.4.1 and I don't understand the result of the following regular expression: 我正在使用python版本3.4.1,但无法理解以下正则表达式的结果:
import re
print(re.match("\[{E=(.*?),Q=(.*?)}\]","[{E=KT,Q=P1.p01},{E=KT2,Q=P2.p02}]").groups())
('KT', 'P1.p01},{E=KT2,Q=P2.p02')
I would expect the result to be 我希望结果是
('KT', 'P1.p01')
but apparently the second .*? 但显然是第二个*。 'eats' all characters until '}]' at the end. “吃掉”所有字符,直到末尾的“}]”为止。 I would expect to stop at the first '}" character. 我希望在第一个'}“字符处停止。
If I leave out the '[' and ']' characters the behavior is as I expect: 如果我省略'['和']'字符,则行为符合我的预期:
print(re.match("{E=(.*?),Q=(.*?)}","{E=KT,Q=P1.p01},{E=KT2,Q=P2.p02}").groups())
('KT', 'P1.p01')
The \\]
forces a square bracket to be present in the match - and there only is one at the end of the string. \\]
强制在比赛中出现一个方括号-字符串的末尾只有一个。 The regex engine has to other option to match. 正则表达式引擎必须与其他选项匹配。 If you remove it or make it optional ( \\]?
), it stops at the closest }
. 如果将其删除或使其成为可选项( \\]?
则它停在最接近的}
。
What you seem to want is everything between '{E='
and the next comma ','
, then everything between 'Q='
and the next closing brace '}'
. 您似乎想要的是'{E='
和下一个逗号','
之间'Q='
所有内容,然后是'Q='
和下一个右括号'}'
。 One expression to do this would be: 一种表达方式是:
{E=([^,]*),Q=([^}]*)}
Here eg [^,]*
means "as many non-comma characters as possible" . 在这里,例如[^,]*
表示“尽可能多的非逗号字符” 。
Example usage: 用法示例:
>>> import re
>>> re.findall("{E=([^,]*),Q=([^}]*)}",
"{E=KT,Q=P1.p01},{E=KT2,Q=P2.p02}")
[('KT', 'P1.p01'), ('KT2', 'P2.p02')]
You can see the full explanation in this regex101 demo . 您可以在此regex101演示中看到完整的说明。
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