正则表达式在双引号之间提取不超过十个单词

Question

Could anyone please guide me to write a regex to find maximum of ten words in a quoted string?

string = "\"Michael Jackson is a great singer\". There were many rumours about his relationship with his girlfriend.  \"He won many national awards and one of the most famous pop singer in the late 80s and 90s\""
re.findall(r'"(.*)"', string)

The above regex extracts both the quoted string but I want to extract only the quoted string which has less than 10 words

Answer 1

try the following regex:

\"(\b\w+\b\s?){,10}\"

demo regex 101

explanation:

• \\" matches "

• \\"(\\b\\w+\\b\\s?) matches a word followed by space with space being optional

• {,10} quantifier specifies less than or equal to 10 words
• \\" matches the last "

if your sentences contain punctuation marks at the end, you can use to to match [\\.\\?\\!] and make it optional

\"(\b\w+\b\s?){,10}[\.\?\!]?\"

Answer 2

re.findall(r'"[^\s"]+(?:\s+[^\s"]+){,9}"', string)

Explanation:

You want to find up to 10 space separated words between double quotes. The first and the last " limit this expression to quoted phrases only.

(Not really, as it suggests using ".+" would work. But then you get the entire string from the first quote up to the last one, because GREP is Greedy. You can use ".+?" to find the shortest matches only, but then you cannot 'count' the words inside.)

After the first quote, you want to match the first 'entire word', which will necessarily consist of a sequence of non-space characters: \\S+ . However, that might eat up the closing double quote if you only have a single word and continue after that, so it is necessary to exclude that as well:

[^\s"]+

-- a sequence of one or more not (space character or double quote). This will match the first word. Then, zero or up to 9 sequences of "space -- word-like sequence" may follow:

\s+[^\s"]+

matches a single occurrence of these, and

(\s+[^\s"]+){,9}

matches 0 up to 9 occurrences.

You may not have noticed it but your own attempt discarded the double quotes at the start and end. That is because you used parentheses in your regex, and findall returns this as a group . To prevent this, I used ?: at the start of the group. (And without this, you will get just singer , the contents of the last group that matched!)

If you don't want the quotes, strip them off later or add a new explicit group around the entire regex:

>>> re.findall(r'"([^\s"]+(?:\s+[^\s"]*){,9})"', string)
['Michael Jackson is a great singer']

Answer 3

By default, regular expressions are greedy, which means that they will try to match as much as possible. What you need to do is then say that you want the non-greedy matcher by using .*? . But this will match the whole string.

So what you need to create is a regular expression that matches a word, but not spaces, and then at most 9 others (starting with spaces).

All the information required to build this is in the documentation ( https://docs.python.org/2/library/re.html ).

Answer 4

Your code can be written as follow:

string = "Michael Jackson is a great singer". There were many rumours about his relationship with his girlfriend.  "He won many national awards and one of the most famous pop singer in the late 80s and 90s"
re.findall(r'"(\w* ){0,9}\w*"', string)

"(\\w* ){0,9} --> to match 0 to 9 word(s) after a opened quote(")

\\w*" --> to match the last word comming before an ended quote(")