如何捕获先行后向正则表达式 python

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*)'

但是它只捕获 ([\s\S]+) 之间的所有内容。它不会捕获 ADDTIONAL & Languages。我在这里错过了什么？

Answer 1

如果您想将它们包含在匹配中，请不要将它们放在环视中，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。如果你只是需要交替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，您的正则表达式需要在 ADDITIONAL 和 Languages 周围换行，但您的字符串中没有换行符。

Answer 2

它正在被捕获，但它不是捕获组 0 的一部分，因为组 0
仅包含 consumed 匹配，即移动当前
的匹配位置。

断言不会移动位置，因此如果您在断言内部进行捕获
它不会成为比赛的一部分。

然而，如果断言后面跟着一些消耗的子表达式
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。匹配您拥有的字符串
删除换行符 \n 引用。

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

Answer 3

试试这个

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

Regex demo

解释：
(?<=…)：正面回顾 sample
\s: "whitespace character": space, tab, newline, carriage return, vertical tab sample
.：除换行符外的任何字符sample
*：零次或多次sample
?：一次或none sample
(?=…)：正面前瞻 sample

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 4

如果您只需要捕获包含 ADDITIONAL 和 LANGUAGES 的内容，请使用像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。

在 REGEX101

查看演示

Answer 5

我猜你把简单的事情复杂化了，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

Answer 6

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]+?将包含这两个位置之间的内容，不包括ADDITIONAL和LANGUAGES。

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

来完成

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

此外，如果您只想[\s\S]+?捕获所有内容，那么您可以对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))

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