Separating a paragraph by sentences

Question

I need to break a paragraph into a set of sentences.

For example:

var paragrafo = "Sou Dr. José. Meu passatempo é assistir séries. Adoro animais!! E você?";
var frases = paragrafo.split('.')

But he ends up breaking the word Dr. also:

var array = [
      "Sou Dr."
      " José."
      "Meu passatempo é assistir séries."
      "Adoro animais!! E você?"
    ];

What I hope will return is:

var array = [
  "Sou Dr. José."
  "Meu passatempo é assistir séries."
  "Adoro animais!!"
  "E você?"
];

Answer 1

Solution (Ecmascript 2018 / ES9):

.*?[.!?](?![.!?])(?<!\b\w\w.)

Demonstration:

var paragrafo = "Sou Dr. José. Meu passatempo é assistir séries. Adoro animais!! E você?";
var frases = paragrafo.match(/.*?[.!?](?![.!?])(?<!\b\w\w\.)/g);
console.log(frases);

Explanation:

First of all, I warn you that this regex has the limitation of letting go abbreviations of only two letters (e.g. "Dr.", "Mr.", "Fr.", "Mr." etc).

• .*?[.!?] - Here we are capturing any text piece finished with endpoint, exclamation or interrogation. I use a Lazy quantifier to capture each part separately.
• (?![.!?]) - This is a Negative Lookahead. Here we are saying not to accept a match if in front of him there is one of these scores (I used to capture also the repeated scores, as in the excerpt Adoro animais!!).
• (?<!\b\w\w.) - This is a Negative lookbehind. Here we are saying not to capture when our match ends with a \b (represents a word separator) and two more characters of the type \w (which means the same as [a-zA-Z0-9_]). This will make texts like Dr. José are still considered within the same sentence, but will continue to separate if something like Dra. Maria.

That is the idea of this regular expression. However, if we want to improve, such as removing the remaining spaces at the beginning of the separations, we can add another Negative Lookahead to ignore spaces:

(?! ).*?[.!?](?![.!?])(?<!\b\w\w.)

And instead of trying to generalize all cases of abbreviations, you might want to insert each specific case into that Negative lookbehind from before:

(?! ).*?[.!?](?![.!?])(?<!\bDr\.|Dra\.|Srs\.|Sras\.)

Final result:

var paragrafo = "Sras. e Srs., eu sou Dr. José. Minha esposa é a Dra. Maria. Meu passatempo é assistir séries. Adoro animais!! E vocês?";
var frases = paragrafo.match(/(?! ).*?[.!?](?![.!?])(?<!\bDr\.|Dra\.|Srs\.|Sras\.)/g);
console.log(frases);

I hope I’ve helped.

Updating:

The solution presented above uses the new lookbehind implemented in ES9. As the OP, in comment below, said that it is using a browser that does not yet support this implementation, also I present a solution that does not use lookbehind:

(?! )(.*?(\b\w\w\.))*.*?[.?!](?![.?!])

Explanation:

• (?! ) - Is a Lookahead that I used to not capture the spaces left behind the sentences.
• (.*?(\b\w\w\.))* - Here I capture any character until I reach the exceptions. I defined as a general rule the same pattern explained before (\b\w\w\.), but you also have the option to add the exceptions separately as in the example with the lookbehind. This pattern is placed in a capture group, and I place a quantifier * after it, to say that it can repeat itself zero or more times.
• .*?[.?!] - Here all characters are captured using a Lazy quantifier, to the nearest endpoint, exclamation or interrogation.
• (?![.?!]) - This is a Negative Lookahead. I mean I don’t want no match followed by a score. I use to capture phrases like Adoro animais!!.

Demonstration:

var paragrafo = "Sou Dr. José. Meu passatempo é assistir séries. Adoro animais!! E você?";
var frases = paragrafo.match(/(?! )(.*?(\b\w\w\.))*.*?[.?!](?![.?!])/g);
console.log(frases);