Select all lines of maximum 12 words that end that do not have any punctuation at the end

Question

I want to select all lines of maximum 12 words that end that do not have any punctuation at the end.

Example:

Love's Equal On Earth

In a moving "in" and "out of focus" image, what you see is what you want to believe about yourself.

I love the feeling.

I must confess every day for better feeling

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The Output must find:

Love's Equal On Earth
I must confess every day for better feeling

My regex is not too good:

^(.*?)(\w+\s?){1,12}(?!\S|$)

I want to avoid all the punctuation marks at the end of the lines, that is to find those lines of a maximum of 12 words that have nothing after the last word, no punctuation marks, nothing. My regex stops of the line, before the ' punctuation mark.

Answer 1

• Ctrl+H
• Find what: ^(?:(?:\w+\W){13}.+|.*[.!?]|)\R
• Replace with: LEAVE EMPTY
• TICK Wrap around
• SELECT Regular expression
• UNTICK . matches newline
• Replace all

Explanation:

^           # beginning of line
    (?:         # non capture group
        (?:         # non capture group
            \w+         # 1 or more word characters
            \W          # non word character
        ){13}       # end group, must appear 13 times
        .+          # 1 or more any character
      |           # OR
        .*          # 0 or more any character
        [.!?]       # a punctuation, you can add all punctuation you want
      |           # OR
                    # EMPTY
    )           # end group
    \R          # any kind of linebreak

Screenshot (before):

Screenshot (after):

Answer 2

^(\S+ ?){1,12}(?<!\.)$

Flags: gm

• ^: start of line (\S+ ?) => \S+ any character that isn't a whitespace, repeated 1 to n times
• followed by a optional whitespace ?
• All repeated 1 to 12 times {1,12}
• (?<!\.)$ : $ end of line not preceded by a . (feel free to put more like ?,!;)

Try it here.

What's interesting in this one is that you can have punctuation between your words, it will check only the last punctuation at line's end.

Answer 3

• Ctrl+H
• Find what: ^(.*?)(\w+\s){1,12}(\w+)\s*$
• Replace with: (Leave EMpty)
• CHECK Wrap around
• CHECK Regular expression
• UNCHECK . matches newline
• Replace all

Answer 4

Not sure if this is optimal, but try the following:

• Find what: ^(\w+[^\.,\:!\?]){1,12}$
• Search mode: Regular expression

Note: I have only considered a few punctuation characters (. , , , : , ! and ?).