Question

我有一项任务，要求我在txt文件中使用下面的数据。没有指定的分隔符，它可以使我更容易排序到数组列表中。我可以使用Scanner类来读取文本文件并将其排序为如下数组：

for (int rows; rows < array.length; rows++){
array[rows][0] = fileIn.next();
array[rows][1] = fileIn.next();

依此类推......然而，这些名字有点难，因为它们中有不同数量的空格，并且可能有不同数量的名称。我希望将“Allison，Hudson J C夫人（Bessie Waldo Daniels）”这个全名作为自己的元素。我不确定从哪里开始，但我认为一个解决方案是让程序检查是否存在“male”||“female”以便我们可以启动一个新元素。任何帮助将不胜感激。

1   1   Allen, Miss. Elisabeth Walton   female  29  211.3375
1   1   Allison, Master. Hudson Trevor  male    0.9167  151.5500
1   0   Allison, Miss. Helen Loraine    female  2   151.5500
1   0   Allison, Mr. Hudson Joshua Creighton    male    30  151.5500
1   0   Allison, Mrs. Hudson J C (Bessie Waldo Daniels) female  25  151.5500
1   1   Anderson, Mr. Harry male    48  26.5500
1   1   Andrews, Miss. Kornelia Theodosia   female  63  77.9583
1   0   Andrews, Mr. Thomas Jr  male    39  0.0000
1   1   Appleton, Mrs. Edward Dale (Charlotte Lamson)   female  53  51.4792
1   0   Artagaveytia, Mr. Ramon male    71  49.5042
1   0   Astor, Col. John Jacob  male    47  227.5250
1   1   Astor, Mrs. John Jacob (Madeleine Talmadge Force)   female  18  227.5250
1   1   Aubart, Mme. Leontine Pauline   female  24  69.3000

Answer 1

这非常适合正则表达式 - 请参阅here以获取数据示例。

([\d]) +([\d]) +(.+\S) +(female|male) +([\d.]+)  +([\d.]+)

Here Java中repl.it的完整示例

import java.util.regex.Matcher;
import java.util.regex.Pattern;

class Main {
    public static void main( String args[] ){
        String text = 
            "1   1   Allen, Miss. Elisabeth Walton   female  29  211.3375\n"+
            "1   1   Allison, Master. Hudson Trevor  male    0.9167  151.5500\n"+
            "1   0   Allison, Miss. Helen Loraine    female  2   151.5500\n"+
            "1   0   Allison, Mr. Hudson Joshua Creighton    male    30  151.5500\n"+
            "1   0   Allison, Mrs. Hudson J C (Bessie Waldo Daniels) female  25  151.5500\n"+
            "1   1   Anderson, Mr. Harry male    48  26.5500\n"+
            "1   1   Andrews, Miss. Kornelia Theodosia   female  63  77.9583\n"+
            "1   0   Andrews, Mr. Thomas Jr  male    39  0.0000\n"+
            "1   1   Appleton, Mrs. Edward Dale (Charlotte Lamson)   female  53  51.4792\n"+
            "1   0   Artagaveytia, Mr. Ramon male    71  49.5042\n"+
            "1   0   Astor, Col. John Jacob  male    47  227.5250\n"+
            "1   1   Astor, Mrs. John Jacob (Madeleine Talmadge Force)   female  18  227.5250\n"+
            "1   1   Aubart, Mme. Leontine Pauline   female  24  69.3000\n";

        String lines[] = text.split("\\r?\\n");

        String pattern = "([\\d]) +([\\d]) +(.+\\S) +(female|male) +([\\d.]+)  +([\\d.]+)";
        Pattern r = Pattern.compile(pattern);

        for (String l : lines) {
            Matcher m = r.matcher(l);
            if (m.find( )) {
                System.out.println(" ------------------- New Text Line -------------------");
                System.out.println("Group 1: " + m.group(1) );
                System.out.println("Group 2: " + m.group(2) );
                System.out.println("Group 3: " + m.group(3) );
                System.out.println("Group 4: " + m.group(4) );
                System.out.println("Group 5: " + m.group(5) );
                System.out.println("Group 6: " + m.group(6) );
            } else {
                System.out.println("Line did not match");
            }   
        }
    }
}

会产生类似的输出

 ------------------- New Text Line -------------------
Group 1: 1
Group 2: 1
Group 3: Allen, Miss. Elisabeth Walton
Group 4: female
Group 5: 29
Group 6: 211.3375
 ------------------- New Text Line -------------------
Group 1: 1
Group 2: 1
Group 3: Allison, Master. Hudson Trevor
Group 4: male
Group 5: 0.9167
Group 6: 151.5500
 ------------------- New Text Line -------------------
Group 1: 1
Group 2: 0
Group 3: Allison, Miss. Helen Loraine
Group 4: female
Group 5: 2
Group 6: 151.5500

Answer 2

我同意你自己的建议。您可以使用正则表达式来帮助解析最初的两个数字和＆＃34;男性|女性＆＃34;之间的所有内容。

您的代码可能类似于：

import java.util.regex.Pattern;
import java.util.regex.Matcher;

public class test {

    private String[] parseLine(String line) {
        String[] output = new String[6];

        Pattern nonWhitespace = Pattern.compile("\\S+");
        Pattern sex = Pattern.compile( "\\s*(male|female)" );

        Matcher m = sex.matcher( line );

        if ( ! m.find() ) {
            // Handle errors. Couldn't find "male" or "female"
        }

        String firstHalf = line.substring(0, m.start());
        String lastHalf = line.substring(m.start(), INPUT.length());

        Matcher firstHalfTokenizer = nonWhitespace.matcher(firstHalf);

        if ( ! firstHalfTokenizer.find() ) {
             // Handle errors. Couldn't find any non-whitespace characters
        }

         output[0] = firstHalf.substring(firstHalfTokenizer.start(), firstHalfTokenizer.end()).trim();

         if ( ! firstHalfTokenizer.find() ) {
             // Handle errors. Couldn't find a second non-whitespace token
         }
         output[1] = firstHalf.substring(firstHalfTokenizer.start(), firstHalfTokenizer.end()).trim();

         output[2] = firstHalf.substring(firstHalfTokenizer.end(), firstHalf.length()).trim();

         Matcher lastHalfTokenizer = nonWhitespace.matcher(lastHalf);

         int index = 3;

         // Need to catch index-out-of-bounds errors if file has too many columns
         while( lastHalfTokenizer.find() ) {
             output[ index ] = lastHalf.substring(lastHalfTokenizer.start(), lastHalfTokenizer.end()).trim();
             index++;
         }

         return output;
      }

      public static void main(String[] args) {
          List<String[]> array = new ArrayList<String[]>();

          for ( String line in file ) { //XXX: Replace this with actual code to loop through the file
              array.add( parseLine(line) );
          }

          // Do whatever you want with it
      }
   }

使用未指定的分隔符扫描数据并将数据添加到数组

2 个答案: