Question

我有一个数据语料库，其中包含以下形式的实例：

'be in'('force', 'the closed area').
'advise'('coxswains', 'mr mak').
'be'('a good', 'restricted area').
'establish from'('person \'s id', 'the other').

我想从.txt文件中读取此数据，并仅使用单引号内的信息填充2D数组，即

be in          [0][0], force         [0][1], the closed area [0][2]
advise         [1][0], coxswains     [1][1], mr mak          [1][2]
be             [2][0], a good        [2][1], restricted area [2][2]
establish from [3][0], person \'s id [3][1], the other       [3][2]

^那些数组索引就像一个概念性参考一样，如上所述，只需要单引号中的信息，例如index [0] [0]为be in，index [3] [1]为person \'s id

但是与示例索引[3] [1]一样，我们可能有单引号，前面加一个反斜杠，不应该被解释为分隔符。

这就是我到目前为止：

BufferedReader br_0 = new BufferedReader(new FileReader("/home/matthias/Workbench/SUTD/2_January/Prolog/horn_data_test.pl"));
    String line_0;
    while ((line_0 = br_0.readLine()) != null) 
    {

        String[] items = line_0.split("'");
        String[][] dataArray = new String [3][262978];
        int i;
        for (String item : items) 
        {
            for (i = 0; i<items.length; i++)
            {
                if (i == 0) 
                {
                    System.out.println("first arg: " + items[i]);
                } 
                if (i == 1) 
                {
                    System.out.println("first arg: " + items[i]);
                }
                if (i == 2)
                {
                    System.out.println("second arg: " + items[i]);
                }
            }
        }           
    }
    br_0.close();

我知道我需要这样的东西：

if (the character under consideration == ' && the one before it is not \)
put it into first index, etc. etc.

但是如何让它在下一个分隔符之前停止？填充该数组的最佳方法是什么？输入文件非常大，所以我正在尝试优化效率。

Answer 1

您可以将{regex与Pattern和Matcher一起使用，如下所示：

public static void main(String[] args) throws IOException {

    String[] stringArr = { "'be in'('force', 'the closed area').",
            "'advise'('coxswains', 'mr mak').",
            "'be'('a good', 'restricted area').",
            "'establish from'('person \'s id', 'the other')." };
    int i = 0;
    Pattern p = Pattern.compile("'(.*?)'(?![a-zA-Z])");
    String[][] arr = new String[4][3];
    for (int count = 0; count < stringArr.length; count++) {
        Matcher m = p.matcher(stringArr[count]);
        int j = 0;
        while (m.find()) {

            arr[i][j++] = m.group(1);
        }
        i++;

    }

    for (int k = 0; k < arr.length; k++) {
        for (int j = 0; j < arr[k].length; j++) {
            System.out.println("arr[" + k + "][" + j + "] " + arr[k][j]);
        }
    }

}

O / P：

arr[0][0] be in
arr[0][1] force
arr[0][2] the closed area
arr[1][0] advise
arr[1][1] coxswains
arr[1][2] mr mak
arr[2][0] be
arr[2][1] a good
arr[2][2] restricted area
arr[3][0] establish from
arr[3][1] person 's id
arr[3][2] the other

Answer 2

您可以使用此正则表达式匹配单引号字符串并支持转义引号：

'(.*?)(?<!\\)'

使用matcher.group(1)作为引号内的字符串。

从输入文本文件填充2D数组

2 个答案:

RegEx Demo