将Java编程中的​​段落拆分为句子

问题描述

我正在执行一项需要将​​段落拆分为句子的任务。例如给定一个段落：

"This is a long string with some numbers 123.456,78 or 100.000 and e.g. some abbreviations in it, which shouldn't split the sentence. Sometimes there are problems, i.e. in this one. here and abbr at the end x.y.. cool."

我需要下面4个句子：

This is a long string with some numbers 123.456,78 or 100.000 and e.g. some abbreviations in it, which shouldn't split the sentence.

Sometimes there are problems, i.e. in this one.

here and abbr at the end x.y..

cool

现在它非常类似于用JavaScript实现的this task。

var re = /\b(\w\.\w\.)|([.?!])\s+(?=[A-Za-z])/g; 
var str = 'This is a long string with some numbers 123.456,78 or 100.000 and e.g. some abbreviations in it, which shouldn\'t split the sentence. Sometimes there are problems, i.e. in this one. here and abbr at the end x.y.. cool.';
var result = str.replace(re, function(m, g1, g2){
  return g1 ? g1 : g2+"\r";
});
var arr = result.split("\r");
document.body.innerHTML = "<pre>" + JSON.stringify(arr, 0, 4) + "</pre>";

[我试图在this link的帮助下用Java实现此功能，但在Java代码中卡住了如何从上方使用replace函数。

public static void main(String[] args) {
    String content = "This is a long string with some numbers 123.456,78 or 100.000 and e.g. some abbreviations in it, which shouldn't split the sentence. Sometimes there are problems, i.e. in this one. here and abbr at the end x.y.. cool.";
    Pattern p = Pattern.compile("/\\b(\\w\\.\\w\\.)|([.?!])\\s+(?=[A-Za-z])/g");
    Matcher m = p.matcher(content);
    List<String> tokens = new LinkedList<String>();
    while (m.find()) {
        String token = m.group(1); // group 0 is always the entire match
        tokens.add(token);
    }

    System.out.println(tokens);
}

如何在Java编程中做同样的事情？对于这种给定的示例文本，是否有比这更好的方法来将段落拆分为Java中的句子？

思路：

public static void main(String[] args) {

    String content = "This is a long string with some numbers 123.456,78 or 100.000 and e.g. some abbreviations in it, which shouldn't split the sentence. Sometimes there are problems, i.e. in this one. here and abbr at the end x.y.. cool.";
    BreakIterator bi = BreakIterator.getSentenceInstance();
    bi.setText(content);
    int index = 0;
    while (bi.next() != BreakIterator.DONE) {
        String sentence = content.substring(index, bi.current());
        System.out.println(sentence);
        index = bi.current();
    }
}

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