Question

我正在尝试编写一个层次结构块，它将元素放入一个ArrayList中，而不是更大的ArrayList。该块的作用是获取现有的文本输入，并将每行文本添加到ArrayList的元素中。然后将每一行创建为字符串的ArrayList，每个String都是该行上的一个单词（我在空格（“”）处使用字符串拆分来执行此操作）。

我的问题是在尝试创建时我需要使用Arrays.asList（因为String Split返回一个List）

Action Syllables = new AbstractAction("Syllables") {
    public void actionPerformed(ActionEvent e) {
        ArrayList<String> text = LinetoList(area);
        //"text" is a String ArrayList of every "line" in a piece of text
        //LinetoList is a method that returns an ArrayList based on each new line
        ArrayList<ArrayList> words = new ArrayList();
        for (int i = 0;  i < text.size(); i++) {
            ArrayList sentence =  (ArrayList) Arrays.asList(text.get(i).split(" "));
            /*Sentence is currently a list, however, changing the type to an Array 
             * or Arraylist Changes nothing */
            words.add(sentence);
            }
        for (int k = 0; k < words.size(); k++) {
            for (int i= 0; i < words.get(k).size(); i ++) {
                System.out.println(words.get(k).get(i));
            }
        }
    }
};

这是我原来的返回错误的方法。我已经稍微调整过，不再返回错误但是没有返回任何内容。

Action Syllables = new AbstractAction("Syllables") {
    public void actionPerformed(ActionEvent e) {
        ArrayList<String> text = LinetoList(area);
        //"text" is a String ArrayList of every "line" in a piece of text
        //LinetoList is a method that returns an ArrayList based on each new "line"
        ArrayList<ArrayList> words = new ArrayList();
        for (int i = 0;  i < text.size(); i++) {
            ArrayList <String> sentences = new ArrayList();
            String sentence[] =  text.get(i).split(" ");
            sentence = sentences.toArray(sentence);
            /*Sentence is currently a list, however, changing the type to an Array 
             * or Arraylist Changes nothing */
            words.add(sentences);
            }
        if (words.size() ==  0)
        {System.out.println("Theres nothing here"); }
        else {
        for (int k = 0; k < words.size(); k++) {
            for (int i= 0; i < words.get(k).size(); i ++) {

                System.out.println(words.get(k).get(i));}

            }
        }
    }
};

非常感谢有关如何处理解决方案的任何反馈或想法。

编辑：有些人要求使用LinetoList功能。该程序的大部分使用字符串的ArrayLists，这就是为什么它在这里如此大量使用。

private static ArrayList<String> LinetoList(JTextArea textArea) {

  String s[] = textArea.getText().split("\\r?\\n");
    ArrayList<String>arrList = new ArrayList<>(Arrays.asList(s)) ;
    return arrList;
}

Answer 1

我实际上并没遵循：

   String sentence[] =  text.get(i).split(" "); // here you have array
    sentence = sentences.toArray(sentence); // WTF?

现在关于WTF标记 - 您正在填充内容为sencence的{{1}}数组。什么是更好，我没有看到sentences的声明，但我想它只是空的对吗？这就是你为什么一直看空的原因。至于我，很多奇怪的代码和无用的评论。

Answer 2

1）Arrays.asList不返回java.util.ArrayList，如果你想让它成为ArrayList，那么你需要做ArrayList sentence = new ArrayList( Arrays.asList(text.get(i).split(" ")));

2）toArray方法将集合中的所有元素放入数组中。在你的情况下，在你调用方法

之前几乎没有创建集合

Answer 3

因此，在看了关于toArray和asList如何工作的评论之后，我尝试了一种更基于我的linetolist函数的不同方法

结果是

Action Syllables = new AbstractAction("Syllables") {
    public void actionPerformed(ActionEvent e) {
        ArrayList<ArrayList> words = new ArrayList();
        ArrayList<ArrayList> splitLines = new ArrayList();
        ArrayList<String> characters = new ArrayList();
        //^^goes to (As hierarchey )
        ArrayList<String> text = LinetoList(area);
        //text is a String ArrayList of every line in a text
        //LinetoList is a method that returns an ArrayList based on each new line
        for (int i = 0;  i < text.size(); i++) {
            //for each line we have
            String sentence[] =  text.get(i).split(" ");
            ArrayList<String> splitText = new ArrayList<>(Arrays.asList(sentence));
            words.add(splitText);
            }
        for (int j = 0; j < words.size(); j++) {
            String sentence [] = text.get(j).split(" ");
            ArrayList<String> splitText = new ArrayList<>(Arrays.asList(sentence));
            ArrayList<ArrayList> SplitWords = new ArrayList();
            for (int i =0; i < sentence.length; i++) {
                ArrayList<Character> SplitCharacters = new ArrayList<Character>();
                for (int k = 0; k < splitText.get(i).length(); k ++) {

                SplitCharacters.add(splitText.get(i).charAt(k));

                }
                SplitWords.add(SplitCharacters);
            }
            splitLines.add(SplitWords);

        if (words.size() ==  0)
        {System.out.println("Theres nothing here"); }
        else {
        for (int k = 0; k < words.size(); k++) {
            System.out.println("\n");
            for (int i= 0; i < words.get(k).size(); i ++) {
                System.out.println(words.get(k).get(i));
                }

            }
        System.out.println("That was the original method \n");
        for (int k = 0; k < splitLines.size(); k++) {
            System.out.println(splitLines.get(k).toString() + "\n");

            }
        }
    }       
}
};

我在这个问题中加入了两种不同的方法，一种是将字数简化为单词的ArrayList，另一种是给出ArrayList的ArrayList（或者是一个包含字母的数组的单词列表）谢谢你的帮助。希望其他任何人都可以从我的错误中拿走一些东西。

ArrayList无法强制转换为java.util.ArrayList，toArray不返回任何内容

3 个答案: