如何按升序对文件名进行排序？

file sorting java arraylist

2022-09-02 05:22:54

我在一个文件夹中有一组文件，除了一个之外，所有文件都以相似的名称开头。下面是一个示例：

Coordinate.txt
Spectrum_1.txt
Spectrum_2.txt
Spectrum_3.txt
.
.
.
Spectrum_11235

我能够列出指定文件夹中的所有文件，但列表不是按频谱号的升序排列的。示例：执行程序时，我得到以下结果：

Spectrum_999.txt
Spectrum_9990.txt
Spectrum_9991.txt
Spectrum_9992.txt
Spectrum_9993.txt
Spectrum_9994.txt
Spectrum_9995.txt
Spectrum_9996.txt
Spectrum_9997.txt
Spectrum_9998.txt
Spectrum_9999.txt

但这个顺序是不正确的。Spectrum_999.txt后应该有Spectrum_1000.txt文件。任何人都可以帮忙吗？代码如下：

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

    public class FileInput {

        public void userInput()
        {
            Scanner scanner = new Scanner( System.in );
            System.out.println("Enter the file path: ");
            String dirPath = scanner.nextLine(); // Takes the directory path as the user input

            File folder = new File(dirPath);
            if(folder.isDirectory())
            {
                File[] fileList = folder.listFiles();

                Arrays.sort(fileList);

                System.out.println("\nTotal number of items present in the directory: " + fileList.length );


                // Lists only files since we have applied file filter
                for(File file:fileList)
                {
                    System.out.println(file.getName());
                }

                // Creating a filter to return only files.
                FileFilter fileFilter = new FileFilter()
                {
                    @Override
                    public boolean accept(File file) {
                        return !file.isDirectory();
                    }
                };

                fileList = folder.listFiles(fileFilter);

                // Sort files by name
                Arrays.sort(fileList, new Comparator()
                {
                    @Override
                    public int compare(Object f1, Object f2) {
                        return ((File) f1).getName().compareTo(((File) f2).getName());
                    }
                });

                //Prints the files in file name ascending order
                for(File file:fileList)
                {
                    System.out.println(file.getName());
                }

            }   
        }
    }

答案 1

您所要求的是数字排序。您需要实现一个比较器并将其传递给 Arrays#sort 方法。在比较方法中，您需要从每个文件名中提取数字，然后比较数字。

您获得现在获得的输出的原因是排序以字母数字方式发生

这是一种非常基本的方法。此代码使用简单的 -操作来提取数字。如果您知道文件名的格式，这有效，在您的情况下。执行提取的更好方法是使用正则表达式。StringSpectrum_<number>.txt

public class FileNameNumericSort {

    private final static File[] files = {
        new File("Spectrum_1.txt"),
        new File("Spectrum_14.txt"),
        new File("Spectrum_2.txt"),
        new File("Spectrum_7.txt"),     
        new File("Spectrum_1000.txt"), 
        new File("Spectrum_999.txt"), 
        new File("Spectrum_9990.txt"), 
        new File("Spectrum_9991.txt"), 
    };

    @Test
    public void sortByNumber() {
        Arrays.sort(files, new Comparator<File>() {
            @Override
            public int compare(File o1, File o2) {
                int n1 = extractNumber(o1.getName());
                int n2 = extractNumber(o2.getName());
                return n1 - n2;
            }

            private int extractNumber(String name) {
                int i = 0;
                try {
                    int s = name.indexOf('_')+1;
                    int e = name.lastIndexOf('.');
                    String number = name.substring(s, e);
                    i = Integer.parseInt(number);
                } catch(Exception e) {
                    i = 0; // if filename does not match the format
                           // then default to 0
                }
                return i;
            }
        });

        for(File f : files) {
            System.out.println(f.getName());
        }
    }
}

输出

Spectrum_1.txt
Spectrum_2.txt
Spectrum_7.txt
Spectrum_14.txt
Spectrum_999.txt
Spectrum_1000.txt
Spectrum_9990.txt
Spectrum_9991.txt

答案 2

当前接受的答案仅对始终称为相同名称的文件的数字后缀执行此操作（即忽略前缀）。

一个更通用的解决方案，我在这里写博客，适用于任何文件名，将名称拆分为段，并以数字（如果两个段都是数字）或字典顺序对段进行排序，否则。灵感来自这个答案的想法：

public final class FilenameComparator implements Comparator<String> {
    private static final Pattern NUMBERS = 
        Pattern.compile("(?<=\\D)(?=\\d)|(?<=\\d)(?=\\D)");
    @Override public final int compare(String o1, String o2) {
        // Optional "NULLS LAST" semantics:
        if (o1 == null || o2 == null)
            return o1 == null ? o2 == null ? 0 : -1 : 1;

        // Splitting both input strings by the above patterns
        String[] split1 = NUMBERS.split(o1);
        String[] split2 = NUMBERS.split(o2);
        for (int i = 0; i < Math.min(split1.length, split2.length); i++) {
            char c1 = split1[i].charAt(0);
            char c2 = split2[i].charAt(0);
            int cmp = 0;

            // If both segments start with a digit, sort them numerically using 
            // BigInteger to stay safe
            if (c1 >= '0' && c1 <= '9' && c2 >= '0' && c2 <= '9')
                cmp = new BigInteger(split1[i]).compareTo(new BigInteger(split2[i]));

            // If we haven't sorted numerically before, or if numeric sorting yielded 
            // equality (e.g 007 and 7) then sort lexicographically
            if (cmp == 0)
                cmp = split1[i].compareTo(split2[i]);

            // Abort once some prefix has unequal ordering
            if (cmp != 0)
                return cmp;
        }

        // If we reach this, then both strings have equally ordered prefixes, but 
        // maybe one string is longer than the other (i.e. has more segments)
        return split1.length - split2.length;
    }
}

这也可以处理带有子版本的版本，例如version-1.2.3.txt