Question

这是一个字符串：

str = "Academy \nADDITIONAL\nAwards and Recognition: Greek Man of the Year 2011 Stanford PanHellenic Community, American Delegate 2010 Global\nEngagement Summit, Honorary Speaker 2010 SELA Convention, Semi-Finalist 2010 Strauss Foundation Scholarship Program\nComputer Skills: Competency: MATLAB, MySQL/PHP, JavaScript, Objective-C, Git Proficiency: Adobe Creative Suite, Excel\n(highly advanced), PowerPoint, HTML5/CSS3\nLanguages: Fluent English, Advanced Spanish\n\x0c"

我想从“ADDTIONAL”捕获到“Languages”所以我写了这个正则表达式：

regex = r'(?<=\n(ADDITIONAL|Additional)\n)[\s\S]+?(?=\n(Languages|LANGUAGES)\n*)'

然而，它只捕获([\s\S]+)之间的所有内容。它没有抓住ADDTIONAL＆amp; Languages。我在这里缺少什么？

Answer 1

你的正则表达式是

regex = r'(?<=\n(ADDITIONAL|Additional)\n)[\s\S]+?(?=\n(Languages|LANGUAGES)\n*)'

，你的字符串是

Academy \nADDITIONAL\nAwards and Recognition: ... \nLanguages:
                     ^^                          ^^
                     ||                          ||
Match Position:-(?<=\n(ADDITIONAL|Additional)\n)(?=\n(Languages|LANGUAGES)\n*)

所以[\s\S]+?将包含这两个排名之间的内容，不包括ADDITIONAL和LANGUAGES。

您只需要找到ADDITIONAL的起始位置和LANGUAGES的结束位置。这可以使用以下正则表达式

完成

(?=\n(ADDITIONAL|Additional)\n)([\s\S]+?)(?<=\n(Languages|LANGUAGES)\b)

此外，如果您希望[\s\S]+?仅捕获所有内容，则可以对Additional和Languages使用非捕获组

(?=\n(?:ADDITIONAL|Additional)\n)[\s\S]+?(?<=\n(?:Languages|LANGUAGES)\b)

Academy \nADDITIONAL\nAwards and Recognition: ... \nLanguages:
        ^^                                                  ^^
        ||                                                  ||
(?=\n(ADDITIONAL|Additional)\n)             (?<=\n(Languages|LANGUAGES))

Python代码

p = re.compile(r'(?=\n(?:ADDITIONAL|Additional)\n)[\s\S]+?(?<=\n(?:Languages|LANGUAGES)\b)', re.MULTILINE)
test_str = "Academy \nADDITIONAL\nAwards and Recognition: Greek Man of the Year 2011 Stanford PanHellenic Community, American Delegate 2010 Global\nEngagement Summit, Honorary Speaker 2010 SELA Convention, Semi-Finalist 2010 Strauss Foundation Scholarship Program\nComputer Skills: Competency: MATLAB, MySQL/PHP, JavaScript, Objective-C, Git Proficiency: Adobe Creative Suite, Excel\n(highly advanced), PowerPoint, HTML5/CSS3\nLanguages: Fluent English, Advanced Spanish\n\x0c"
print(re.findall(p, test_str))

<强> Ideone Demo

Answer 2

它被捕获但它不是捕获组0的一部分，因为组0为仅包含消耗的匹配，即移动当前
的匹配位置。

断言不会移动位置，所以如果你在一个断言内捕获它不会成为比赛的一部分。

然而，如果断言之后是一些消耗了的子表达式在断言中引用的那些，它将成为整体匹配的一部分。

您当前的正则表达式与您的字符串不匹配。要匹配你的字符串删除换行符\n。

 (?<=
      ( ADDITIONAL | Additional )   # (1)
 )
 [\s\S]+? 
 (?=
      ( Languages | LANGUAGES )     # (2)
 )

Answer 3

如果要将它们包含在匹配中，请不要将它们放在外观中，因为它们的目的是测试周围的文本而不将其包含在匹配结果中。如果您只是需要更换，请使用普通的非捕获组。

regex = r'\n(?:ADDITIONAL|Additional)\n[\s\S]+?\n(?:Languages|LANGUAGES)\n*'

顺便说一句，你的正则表达式需要ADDITIONAL和Languages周围的换行符，但你的字符串中没有任何换行符。

Answer 4

试试这个

(?<=ADDITIONAL\s).*?(?=\sLanguages)

Regex demo

<强>解释
(?<=…)：正面观察sample
\s：“空格字符”：空格，制表符，换行符，回车符，垂直制表符sample
.：除了换行符sample之外的任何字符 *：零次或多次sample
?：一次或无sample
(?=…)：积极前瞻sample

的Python：

import re
p = re.compile(ur'(?<=ADDITIONAL\s).*?(?=\sLanguages)', re.IGNORECASE)
test_str = u"the companys direction ADDITIONAL Awards: 2010 Global Engagement Summit, Languages: Fluent Japanese"

g = re.findall(p, test_str)
print g //[u'Awards: 2010 Global Engagement Summit,']

Answer 5

如果您需要捕获包含ADDITIONAL和LANGUAGES的内容，请使用这样的简单正则表达式。

\b(ADDITIONAL .* Languages)\b

确保在解决方案中使用时包含re.IGNORECASE标志。

请参阅REGEX101

上的演示

Answer 6

我猜你会让事情变得复杂，即：

match = re.search("(ADDITIONAL.*?Languages)", subject, re.MULTILINE)

正则表达式解释：

(ADDITIONAL.*?Languages)


Match the regex below and capture its match into backreference number 1 «(ADDITIONAL.*?Languages)»
   Match the character string “ADDITIONAL” literally (case sensitive) «ADDITIONAL»
   Match any single character that is NOT a line break character (line feed) «.*?»
      Between zero and unlimited times, as few times as possible, expanding as needed (lazy) «*?»
   Match the character string “Languages” literally (case sensitive) «Languages»

Regex101 Demo

如何捕捉前瞻性的后瞻性正则表达式python

6 个答案: