如何按升序对文件名进行排序？

Question

我在一个文件夹中有一组文件，所有文件都以相似的名称开头，除了一个。这是一个例子：

 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，但该列表不是按光谱编号的升序排列。示例：我在执行程序时得到以下结果：

 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

但是这个顺序是不正确的。 Spectrum_999.txt 之后应该有Spectrum_1000.txt 文件。谁能帮忙？这是代码：

 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());
                }

            }
        }
    }

原文由 novicegeek 发布，翻译遵循 CC BY-SA 4.0 许可协议

Answer 1

您要的是数字排序。您需要实现一个 Comparator 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，然后比较这些数字。

你得到你现在得到的输出的原因是排序是按 字母数字顺序 进行的

这是一个非常基本的方法。此代码使用简单的 String 操作来提取数字。如果您知道文件名的格式，则此方法有效，在您的情况下 Spectrum_<number>.txt 。一种更好的提取方法是使用 正则表达式。

 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

原文由 A4L 发布，翻译遵循 CC BY-SA 3.0 许可协议

Answer 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，我在此处发布的博客，适用于任何文件名，将名称拆分为段并按数字（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;
    }
}

这也可以处理带有颠覆的版本，例如 version-1.2.3.txt

原文由 Lukas Eder 发布，翻译遵循 CC BY-SA 4.0 许可协议