Question

我编写了java代码，用于从具有200k记录的文件中删除初始字符，该文件正在删除初始字符，但是它逐行读取文件并删除字符。程序执行速度非常慢。可以对下面的代码进行任何调整以更快地执行它吗？

程序正在执行并将输出写入文件，但速度非常慢

import java.io.*;
import java.util.Scanner;

public class truncate {
    public static void main(String [] args) {

        // The name of the file to open.
        String inputfile = "C:\\Program Files\\eclipse\\twfiles.txt";
        String outputfile = "C:\\Program Files\\eclipse\\rename.txt";

        // This will reference one line at a time
        String line = "";
        int number_of_char_to_erased =19;

        try {
            // FileReader reads text files in the default encoding.
            FileReader fileReader = 
                new FileReader(inputfile);

            // Always wrap FileReader in BufferedReader.
            BufferedReader bufferedReader = 
                new BufferedReader(fileReader);

            while((line = bufferedReader.readLine()) != null) {
                System.out.println(line);

                File input = new File(inputfile);
                Scanner scan = new Scanner(input);
                File output = new File(outputfile);
                PrintStream print = new PrintStream(output);

                while (scan.hasNext()) {
                 line = scan.nextLine();
                 line = line.substring(number_of_char_to_erased);
                 print.println(line);
                }

                scan.close();
                print.close();
            }   

            // Always close files.
            bufferedReader.close();         
        }
        catch(FileNotFoundException ex) {
            System.out.println(
                "Unable to open file '" + 
                        inputfile + "'");               
        }
        catch(IOException ex) {
            System.out.println(
                "Error reading file '" 
                + inputfile + "'");                 
            // Or we could just do this: 
            // ex.printStackTrace();
        }
    }
}

Answer 1

这里的问题似乎是你刚刚创建了一个缓冲读卡器来读取文件。然后，它读取文件的第一行。然后，创建一个扫描程序以读取文件中的所有行，省略某些字符。然后你的BufferedReader读取文件中的下一行。这个过程重演。所以你要做的就是：

  File output = new File(outputfile);
  PrintStream print = new PrintStream(output);   

  while((line = bufferedReader.readLine()) != null) {

            print.println(line.substring(number_of_char_to_erased);

   }
   print.close();

这应该快得多。基本上，由于您已经从文件中为读取行分配了行，因此您只需将该行（减去字符数）打印到输出文件即可。使用扫描仪的整个for循环是完全没必要的，关闭和打开每行的打印流也是不必要的。

编辑：删除了println语句，因为它会减慢它的速度。

Answer 2

尝试此操作（删除扫描仪和Println，在循环外重构输出文件）：

import java.io.*;

public class truncate {
public static void main(String [] args) {

    // The name of the file to open.
    String inputfile = "C:\\Program Files\\eclipse\\twfiles.txt";
    String outputfile = "C:\\Program Files\\eclipse\\rename.txt";

    // This will reference one line at a time
    String line = "";
    int number_of_char_to_erased =19;

    try {
        // FileReader reads text files in the default encoding.
        FileReader fileReader = new FileReader(inputfile);

        // Always wrap FileReader in BufferedReader.
        BufferedReader bufferedReader = new BufferedReader(fileReader);

        File output = new File(outputfile);
        PrintStream print = new PrintStream(output);

        while((line = bufferedReader.readLine()) != null) {

             String trimmedLine = line.substring(number_of_char_to_erased);
             print.println(trimmedLine);


        }   

        // Always close files.
        bufferedReader.close();
        print.close();        
    }
    catch(FileNotFoundException ex) {
        System.out.println(
            "Unable to open file '" + 
                    inputfile + "'");               
    }
    catch(IOException ex) {
        System.out.println(
            "Error reading file '" 
            + inputfile + "'");                 
        // Or we could just do this: 
        // ex.printStackTrace();
    }
}
}

用于删除执行速度非常慢的文件的初始行的Java代码

2 个答案: