20

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,但该列表不是按频谱编号的升序排列的。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 后面应该有 Spectrum_1000.txt 文件。任何人都可以帮忙吗?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());
                }

            }   
        }
    }
4

11 回答 11

35

你要求的是数字排序。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,然后比较数字。

你得到你现在得到的输出的原因是排序发生在字母数字上

这是一种非常基本的方法。此代码使用简单String的 - 操作来提取数字。如果您知道文件名的格式(在您的情况下),则此方法有效Spectrum_<number>.txt。进行提取的更好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
于 2013-06-03T13:58:38.947 回答
6

Commons IO 库中可用的NameFileComparator类,具有按名称、上次修改日期、大小等对文件数组进行排序的功能。文件可以按升序和降序排序,区分大小写或不区分大小写。

进口 :

org.apache.commons.io.comparator.NameFileComparator

代码 :

File directory = new File(".");
File[] files = directory.listFiles();
Arrays.sort(files, NameFileComparator.NAME_COMPARATOR)
于 2016-03-03T04:41:49.353 回答
6

当前接受的答案仅适用于始终称为相同名称的文件的数字后缀(即忽略前缀)。

一个更通用的解决方案,我在博客上写过,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)或按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

于 2018-02-23T11:23:39.353 回答
3

您可以在上面的评论中找到您的问题的解决方案,但考虑到仅发布了链接,我将提供该站点的代码。工作得很好。

  1. 您需要创建自己的 AlphanumericalComparator。

     import java.io.File;
     import java.util.Comparator;
    
    public class AlphanumFileComparator implements Comparator
    {
    
       private final boolean isDigit(char ch)
       {
        return ch >= 48 && ch <= 57;
       }
    
    
    private final String getChunk(String s, int slength, int marker)
    {
        StringBuilder chunk = new StringBuilder();
        char c = s.charAt(marker);
        chunk.append(c);
        marker++;
        if (isDigit(c))
        {
            while (marker < slength)
            {
                c = s.charAt(marker);
                if (!isDigit(c))
                    break;
                chunk.append(c);
                marker++;
            }
        } else
        {
            while (marker < slength)
            {
                c = s.charAt(marker);
                if (isDigit(c))
                    break;
                chunk.append(c);
                marker++;
            }
        }
        return chunk.toString();
    }
    
    public int compare(Object o1, Object o2)
    {
        if (!(o1 instanceof File) || !(o2 instanceof File))
        {
            return 0;
        }
        File f1 = (File)o1;
        File f2 = (File)o2;
        String s1 = f1.getName();
        String s2 = f2.getName();
    
        int thisMarker = 0;
        int thatMarker = 0;
        int s1Length = s1.length();
        int s2Length = s2.length();
    
        while (thisMarker < s1Length && thatMarker < s2Length)
        {
            String thisChunk = getChunk(s1, s1Length, thisMarker);
            thisMarker += thisChunk.length();
    
            String thatChunk = getChunk(s2, s2Length, thatMarker);
            thatMarker += thatChunk.length();
    
            /** If both chunks contain numeric characters, sort them numerically **/
    
            int result = 0;
            if (isDigit(thisChunk.charAt(0)) && isDigit(thatChunk.charAt(0)))
            {
                // Simple chunk comparison by length.
                int thisChunkLength = thisChunk.length();
                result = thisChunkLength - thatChunk.length();
                // If equal, the first different number counts
                if (result == 0)
                {
                    for (int i = 0; i < thisChunkLength; i++)
                    {
                        result = thisChunk.charAt(i) - thatChunk.charAt(i);
                        if (result != 0)
                        {
                            return result;
                        }
                    }
                }
            } else
            {
                result = thisChunk.compareTo(thatChunk);
            }
    
            if (result != 0)
                return result;
        }
    
        return s1Length - s2Length;
    }
    }
    

2. 根据这个类对文件进行排序。

     File[] listOfFiles = rootFolder.listFiles();
     Arrays.sort(listOfFiles, new AlphanumFileComparator() );
     ...to sth with your files.

希望能帮助到你。它对我有用,就像一个魅力。

解决方案来自:http ://www.davekoelle.com/files/AlphanumComparator.java 这里

于 2016-05-29T12:34:18.687 回答
2
Arrays.sort(fileList, new Comparator()
{
    @Override
    public int compare(Object f1, Object f2) {
        String fileName1 = ((File) f1).getName();
        String fileName2 = ((File) f1).getName();

        int fileId1 = Integer.parseInt(fileName1.split("_")[1]);
        int fileId2 = Integer.parseInt(fileName2.split("_")[1]);

        return fileId1 - fileId2;
    }
});

确保处理名称中没有 _ 的文件

于 2013-06-03T14:15:30.620 回答
2

只需使用:

  1. 对于升序: Collections.sort(List)

  2. 对于降序: Collections.sort(List,Collections.reverseOrder())

于 2018-02-27T10:05:50.660 回答
1

在这里找到了 Ivan Gerasimiv 的智能 AlphaDecimal 比较器的最佳实现。它被提议作为标准 JDK 比较器的扩展,并在此处的线程中进行了讨论
不幸的是,这种变化仍然没有进入 JDK(据我所知)。但是您可以使用第一个链接中的代码作为解决方案。对我来说,它就像一种魅力。

于 2020-11-24T18:57:59.463 回答
0

我的 kotlin 版本算法。此算法用于按名称对文件和目录进行排序。

import kotlin.Comparator

const val DIFF_OF_CASES = 'a' - 'A'

/** File name comparator. */
class FileNameComparator(private val caseSensitive: Boolean = false) : Comparator<String> {

    override fun compare(left: String, right: String): Int {
        val csLeft = left.toCharArray()
        val csRight = right.toCharArray()
        var isNumberRegion = false
        var diff = 0; var i = 0; var j = 0
        val lenLeft = csLeft.size; val lenRight = csRight.size
        while (i < lenLeft && j < lenRight) {
            val cLeft = getCharByCaseSensitive(csLeft[i])
            val cRight = getCharByCaseSensitive(csRight[j])
            val isNumericLeft = cLeft in '0'..'9'
            val isNumericRight = cRight in '0'..'9'
            if (isNumericLeft && isNumericRight) {
                // Number start!
                if (!isNumberRegion) {
                    isNumberRegion = true
                    // Remove prefix '0'
                    while (i < lenLeft && cLeft == '0') i++
                    while (j < lenRight && cRight == '0') j++
                    if (i == lenLeft || j == lenRight) break
                }
                // Diff start: calculate the diff value.
                if (cLeft != cRight && diff == 0) diff = cLeft - cRight
            } else {
                if (isNumericLeft != isNumericRight) {
                    // One numeric and one char: the longer one is backwards.
                    return if (isNumberRegion) { if (isNumericLeft) 1 else -1 } else cLeft - cRight
                } else {
                    // Two chars: if (number) diff don't equal 0, return it (two number have same length).
                    if (diff != 0) return diff
                    // Calculate chars diff.
                    diff = cLeft - cRight
                    if (diff != 0) return diff
                    // Reset!
                    isNumberRegion = false
                    diff = 0
                }
            }
            i++; j++
        }
        if (i == lenLeft) i--
        if (j == lenRight) j--
        return csLeft[i] - csRight[j]
    }

    private fun getCharByCaseSensitive(c: Char): Char
            = if (caseSensitive) c else if (c in 'A'..'Z') (c + DIFF_OF_CASES) else c
}

例如,对于输入,

          "A__01__02",
            "A__2__02",
            "A__1__23",
            "A__11__23",
            "A__3++++",
            "B__1__02",
            "B__22_13",
            "1_22_2222",
            "12_222_222",
            "2222222222",
            "1.sadasdsadsa",
            "11.asdasdasdasdasd",
            "2.sadsadasdsad",
            "22.sadasdasdsadsa",
            "3.asdasdsadsadsa",
            "adsadsadsasd1",
            "adsadsadsasd10",
            "adsadsadsasd3",
            "adsadsadsasd02"

它将它们排序为,

1.sadasdsadsa 1_22_2222 2.sadsadasdsad 3.asdasdsadsadsa 11.asdasdasdasdasd 12_222_222 22.sadasdasdsadsa 2222222222 A__01__02 A__1__23 A__2__02 A__3++++ A__11__23 adsadsadsasd02 adsadsadsasd1 adsadsadsasd3 adsadsadsasd10 B__1__02 B__22_13 
于 2021-07-02T01:55:58.710 回答
0

实现这一点的最简单方法是使用NameFileComparator.NAME_COMPARATOR。以下是代码格式:

File[] fileNamesArr = filePath.listFiles();
if(fileNamesArr != null && fileNamesArr.length > 0) {
    Arrays.sort(fileNamesArr, NameFileComparator.NAME_COMPARATOR);
}

为了实现NameFileComparator您需要添加以下 maven 库:

实现'commons-io:commons-io:2.6'

还进口:

导入 org.apache.commons.io.comparator.NameFileComparator;

就这样...

于 2021-12-09T13:17:27.237 回答
-1

只是另一种方法,但使用 java8 的力量

List<Path> x = Files.list(Paths.get("C:\\myPath\\Tools"))
            .filter(p -> Files.exists(p))
            .map(s -> s.getFileName())
            .sorted()
            .collect(Collectors.toList());

x.forEach(System.out::println);
于 2017-07-04T08:29:28.547 回答
-2

您可以使用Collections.sort(fileList); 对arraylist 进行排序。

然后使用

 for(File file:fileList)                
         System.out.println(file.getName());

Collections.sort()

于 2013-06-03T13:35:19.150 回答