如何捕捉前瞻性的后瞻性正则表达式python

Question

Here is a string:

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"

I'd like to capture from "ADDTIONAL" to "Languages" so I wrote this regex:

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

However it only catches everything in between ([\\s\\S]+) . It does NOT catch ADDTIONAL & Languages . What am I missing here?

Answer 1

Your regex is

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

and your string is

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

So [\\s\\S]+? will contain the contents in between these two positions excluding ADDITIONAL and LANGUAGES .

You just have to find the starting position of ADDITIONAL and ending position of LANGUAGES . This can be done using the following regex

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

Further, if you want [\\s\\S]+? only to capture all contents, then you can use non capturing groups for Additional and Languages

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

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

Python Code

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))

Ideone Demo

Answer 2

It is being captured but it's not part of capture group 0 because group 0
contains only the consumed match, ie the match that moved the current
position.

Assertions don't move the position, so if you capture inside an assertion
it does not become part of the match.

However if the assertion were followed by some sub-expression that consumed
the ones referenced in the assertion, it would become part of the overall match.

Your current regex will not match your string. To match the string you have
to remove the newlines \\n references.

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

Answer 3

If you want to include them in the match, don't put them in lookarounds, since the purpose of those is to test for surrounding text without including it in the match result. Use ordinary non-capturing groups if you just need alternation.

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

BTW, your regexp requires newlines around ADDITIONAL and Languages , but there aren't any in your string.

Answer 4

Try this

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

Regex demo

Explanation:
(?<=…) : Positive lookbehind sample
\\s : "whitespace character": space, tab, newline, carriage return, vertical tab sample
. : Any character except line break sample
* : Zero or more times sample
? : Once or none sample
(?=…) : Positive lookahead 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 5

If you need to just capture content inclusive of ADDITIONAL and LANGUAGES , use simple regex like this.

\b(ADDITIONAL .* Languages)\b

Make sure you are including re.IGNORECASE flag when using in solution.

See demo at REGEX101

Answer 6

I guess you're complicating something easy, ie:

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

---

Regex explanation:

(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