Java比较两个音频文件与指纹

Question

我想知道，如果两个音频文件是相同的，或者一个包含另一个。

这里我使用Fingerprint of musicg

byte[] firstAudio = readAudioFileData("first.mp3");
byte[] secondAudio = readAudioFileData("second.mp3");
FingerprintSimilarityComputer fingerprint = 
            new FingerprintSimilarityComputer(firstAudio, secondAudio);
FingerprintSimilarity fingerprintSimilarity = fingerprint.getFingerprintsSimilarity();
System.out.println("clip is found at " + fingerprintSimilarity.getScore());

将音频转换为字节数组我使用声音API

public static byte[] readAudioFileData(final String filePath) {
    byte[] data = null;
    try {
        final ByteArrayOutputStream baout = new ByteArrayOutputStream();
        final File file = new File(filePath);
        final AudioInputStream audioInputStream = AudioSystem.getAudioInputStream(file);
        byte[] buffer = new byte[4096];
        int c;
        while ((c = audioInputStream.read(buffer, 0, buffer.length)) != -1) {
            baout.write(buffer, 0, c);
        }
        audioInputStream.close();
        baout.close();
        data = baout.toByteArray();
    } catch (Exception e) {
        e.printStackTrace();
    }
    return data;
}

但是当我执行它时，我变成了fingerprint.getFingerprintsSimilarity()和Exception。

Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 15999
at com.musicg.fingerprint.PairManager.getPairPositionList(PairManager.java:133)
at com.musicg.fingerprint.PairManager.getPair_PositionList_Table(PairManager.java:80)
at com.musicg.fingerprint.FingerprintSimilarityComputer.getFingerprintsSimilarity(FingerprintSimilarityComputer.java:71)
at Main.main(Main.java:42)

如何在Java中比较2个mp3文件与指纹?

Answer 1

我以前从未在Java中做过任何音频内容，但我简要地研究了一下您的代码。我认为musicg只适用于WAV文件，不适用于MP3。因此，您需要首先转换文件。一个网络搜索显示，你可以使用JLayer来实现这个目的。对应的代码如下所示:

package de.scrum_master.so;
import com.musicg.fingerprint.FingerprintManager;
import com.musicg.fingerprint.FingerprintSimilarity;
import com.musicg.fingerprint.FingerprintSimilarityComputer;
import com.musicg.wave.Wave;
import javazoom.jl.converter.Converter;
import javazoom.jl.decoder.JavaLayerException;
public class Application {
  public static void main(String[] args) throws JavaLayerException {
    // MP3 to WAV
    new Converter().convert("White Wedding.mp3", "White Wedding.wav");
    new Converter().convert("Poison.mp3", "Poison.wav");
    // Fingerprint from WAV
    byte[] firstFingerPrint = new FingerprintManager().extractFingerprint(new Wave("White Wedding.wav"));
    byte[] secondFingerPrint = new FingerprintManager().extractFingerprint(new Wave("Poison.wav"));
    // Compare fingerprints
    FingerprintSimilarity fingerprintSimilarity = new FingerprintSimilarityComputer(firstFingerPrint, secondFingerPrint).getFingerprintsSimilarity();
    System.out.println("Similarity score = " + fingerprintSimilarity.getScore());
  }
}

当然，你应该确保你不转换每个文件再次每当程序启动，也就是说，你应该检查WAV文件是否已经存在。我跳过了这一步，并将示例代码简化为最小的工作版本。

对于FingerprintSimilarityComputer(inputt1, input2)，它应该接收加载音频数据的指纹，而不是加载音频数据本身。

在你的例子中，它应该是:

// Convert your audio to wav using FFMpeg
Wave w1 = new Wave("first.wav");
Wave w2 = new Wave("second.wav");
FingerprintSimilarityComputer fingerprint = 
        new FingerprintSimilarityComputer(w1.getFingerprint(), w2.getFingerprint());
// print fingerprint.getFingerprintSimilarity()

也许我遗漏了一点，但如果我理解对了，应该这样做:

byte[] firstAudio = readAudioFileData("first.mp3");
byte[] secondAudio = readAudioFileData("second.mp3");
byte[] smaller = firstAudio.length <= secondAudio.length ? firstAudio : secondAudio;
byte[] bigger = firstAudio.length > secondAudio.length ? firstAudio : secondAudio;
int ixS = 0;
int ixB = 0;
boolean contains = false;
for (; ixB<bigger.length; ixB++) {
    
    if (smaller[ixS] == bigger[ixB]) {
        ixS++;
        if (ixS == smaller.length) {
            contains = true;
            break;
        }
    }
    else {
        ixS = 0;
    }
}
if (contains) {
    if (smaller.length == bigger.length) {
        System.out.println("Both tracks are equal");
    }
    else {
        System.out.println("The bigger track, fully contains the smaller track starting at byte: "+(ixB-smaller.lenght));
    }
}
else {
    System.out.println("No track completely contains the other track");
}