GREP line counting

Well, if it's like that you don't need a script at all. This GREP query breaks a text after every fourth dot:

Find what: (.+?\.){4}\K\x20
Change to: \r

(\x20 denotes the space character.)

The number 4 determines the number of dots after which to place the paragraph break, so if you want more or fewer paragraph breaks, change that number.

This one breaks after dots only. To break after exclamation points and question marks as well, replace \. with [.?!].

P.

Hi @_doctor7, I totally agree with the sentiment expressed by others here about the wisdom of using an automated approach to solve your problem—deciding on what consitutes a paragraph is a job for a writer or editor. Paragraphs have meaning, made of sentences conveying a common idea or theme.

But, for the sake of fun, I will assume that the above doesn't apply in your particular case, and here is a script that adds to Rob's idea but with a few extras. It is still very dumb, so be careful using it.

- Mark

/**
 * Splits paragraphs by trying to attempting to find
 * paragraphs of active document with many sentences.
 * Known Limitations:
 * 1. Are you sure this is right for your text? Deciding how many sentences are in a paragraph is the job of a writer or editor, not a script! :)
 * 2. Script tries to parse sentences, but is pretty dumb. You must check carefully. Although it leaves numbers alone.
 * 3. Script makes no attempt to evaluate the *length* of sentences, just the *number* of sentences.
 * by @m1b
 * @discussion https://community.adobe.com/t5/indesign-discussions/grep-line-counting/m-p/13850733
 */
var settings = {
    targetNumberOfSentencesInParagraph: 3,
}

function main() {

    if (app.documents.length == 0)
        return;

    // A capital letter followed by anything except a period, exclamation or question mark and ending with a word character and then a period, exclamation or question mark.
    var findWhat = '((([A-Z][^\\.!?]*\\w[\\.!?]) ){' + settings.targetNumberOfSentencesInParagraph + '})';
    app.findGrepPreferences = NothingEnum.NOTHING;
    app.findGrepPreferences.findWhat = findWhat;

    // find in every paragraph
    // (this breaks up the results by paragraph)
    var found = app.activeDocument.stories.everyItem().paragraphs.everyItem().findGrep();
    var breaks = [];

    paragraphLoop:
    for (var i = found.length - 1; i >= 0; i--) {

        if (
            found[i] == undefined
            || found[i].length < 2
        )
            // ignore short paragraphs
            continue paragraphLoop;

        foundLoop:
        for (var j = found[i].length - 1; j >= 0; j--)
            // collect these insertion points
            breaks.push(found[i][j].insertionPoints.lastItem());

    }

    var counter = 0;
    for (var i = 0; i < breaks.length; i++) {
        // add the carriage return
        breaks[i].contents = '\r';
        counter++;
    }

    alert(counter + ' carriage returns added.');

};

app.doScript(main, ScriptLanguage.JAVASCRIPT, undefined, UndoModes.ENTIRE_SCRIPT, 'Split Paragraphs');

Apart from the problems noted by Peter S., dots are used also in abbreviations, initials, ellipses, etc. etc.  What you're after looks like a manual job to me. 

OK, first off, ciounting full stops does not equal counting lines -- lines wrap according to the column width and if you cahnge that you change the line count.

That said, GREP can count sentences. Expression that seems to work in quick testing is

Find ^((.+?\.){n}) where n= the number of sentences you want to find.

Change to $0\r

This is flawed, though, in that it finds not only the first n sentences of a long paragraph, but also paragraphs of only n sentence length and adds an unwanted extra break at the ends of those. This can be remedied by adding a negative lookahead for a paragrph break:

Find ^((.+?\.){n})(?!\r)

There are still some issues, though. First it fails to delete empty space at what will become the start of the next paragraph.It also doesn't find sentences that end with a question mark or exclamation point, nor does it pick up and keep any quotation marks, footonote references or other things like that which might follow the end-of-sentence punctuation mark. 

The first spaces can be eliminated , I think by adding \s* to the xpression:

Find ^((.+?\.){n})(?!\r)\s*

but that sill leaves the problems with sentences that don't end with a full stop. I believe those could be solved using classes, but I haven't tested.

Finally, this expression doesn't care if you break leaving only a single sentence by itself in the next paragraph.

Scripting can probably give you a more nuanced count and vary how many lines to include in each broken paragraph depending on how many it starts with, but that might not be an issue for you.

Script would be probably easiest solution. 

I'm not a GREP expert - but you can search for exact number of occurrences of the specified character /phrase - so I think it would be possible?