Question

我想在一个文件中读取，并将每行的内容写入维度数组[三] [不确定，但很长]

到目前为止，我有以下代码，它能够使用模式匹配器来梳理我正在寻找的输入文件的组件，但是，这个问题卡在第一行输入上，只是添加了一遍又一遍，如何使输入文件进展并每次向数组写一个新行。

到目前为止，我的代码看起来像这样：

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

    BufferedReader br_0 = new BufferedReader(new FileReader("file.txt"));
    String line_0;

    //while the file is still reading
    while ((line_0 = br_0.readLine()) != null) 
    {           

        int i = 0;
        Pattern p = Pattern.compile("'(.*?)'(?![a-zA-Z])");
        //count from zero
        String[][] arr = new String[262978][3];

        for (int count = 0; count < 262978; count++) 
        {

            Matcher m = p.matcher(line_0);

            int j = 0;
            while (m.find()) 
            {
                arr[i][j++] = m.group(1);
            }
            i++;

        }
    }
    br_0.close();
}

输入文件如下所示：

'end with'('the playing of the british national anthem', 'hong kong').
'follow at'('the stroke of midnight', 'this').
'take part in'('the ceremony', 'both countries').
'start at about'('# pm', 'the ceremony').
'end about'('# am', 'the ceremony').
'lower'('the british hong kong flag', '# royal hong kong police officers').
'raise'('the sar flag', 'another #').
'leave for'('the royal yacht britannia', 'the #').
'hold by'('the chinese and british governments', 'the handover of hong kong').
'rise over'('this land', 'the regional flag of the hong kong special administrative region of the people \'s republic of china').
'cast eye on'('hong kong', 'the world').
'hold on'('schedule', 'the # governments').
'be festival for'('the chinese nation', 'this').
'go in'('the annals of history', 'july # , #').
...

理想情况下，数组索引如下所示：

[0] [0] end with

[0] [1] the playing of the british national anthem

[0] [2] hong kong

[1] [0] follow at

[1] [1] the stroke of midnight

[1] [2] this

[2] [0] take part in

[3] [1] the ceremony

[2] [2] both countries

重要的是，数组长度能够容纳很长的文件，但也可以容纳很短的文件。

此时的输出如下所示：

[45993][2] the president of the people \'s republic of china he mr jiang zemin
[45994][0] speak at
[45994][1] the ceremony
[45994][2] the president of the people \'s republic of china he mr jiang zemin
[45995][0] speak at
[45995][1] the ceremony
[45995][2] the president of the people \'s republic of china he mr jiang zemin
[45996][0] speak at
[45996][1] the ceremony
[45996][2] the president of the people \'s republic of china he mr jiang zemin
[45997][0] speak at
[45997][1] the ceremony
[45997][2] the president of the people \'s republic of china he mr jiang zemin
[45998][0] speak at
[45998][1] the ceremony
[45998][2] the president of the people \'s republic of china he mr jiang zemin
[45999][0] speak at

Answer 1

这会处理第一行262978次。

for (int count = 0; count < 262978; count++)

更好的是：

int count = 0;
String[][] arr = new String[262978][3];
Pattern p = Pattern.compile("'(.*?)'(?![a-zA-Z])"); 
//while the file is still reading
while ((line_0 = br_0.readLine()) != null) {           
     Matcher m = p.matcher(line_0);
     int j = 0;
     while (m.find()) {
         arr[count][j++] = m.group(1);
     }
     count++;
}

br_0.close();

但是，不应使用幻数262978，也不应使用数组。显然也是最大的假设。每行三个字符串不正确。

将其替换为

List<List<String>> arr = new ArrayList<>();
Pattern p = Pattern.compile("'(.*?)'(?![a-zA-Z])"); 
//while the file is still reading
while ((line_0 = br_0.readLine()) != null) {
     List<String> three = new ArrayList<>();         
     Matcher m = p.matcher(line_0);
     int j = 0;
     while (m.find()) {
         three.add( m.group(1) );
     }
     arr.add( three );
}

br_0.close();

要打印，

for( List<String> three: arr ){
    for( String s: three ){
        System.out.print( s  + " " );
    }
    System.out.println();
}

Answer 2

您正在此处阅读文件中的数据：while ((line_0 = br_0.readLine()) != null)，但是，您正在使用相同的行进行262978次迭代：for (int count = 0; count < 262978; count++)。

你可以做的，就是用这样的东西取代它：

int i = 0;
Pattern p = Pattern.compile("'(.*?)'(?![a-zA-Z])");
//count from zero
String[][] arr = new String[262978][3];
while (((line_0 = br_0.readLine()) != null) && (i < 262978))
{
    Matcher m = p.matcher(line_0);

    int j = 0;
    while (m.find()) 
    {
        arr[i][j++] = m.group(1);
    }
    i++;    
}

填充动态2D java数组

2 个答案: