Wav比较,同一个文件

Question

我现在很难过.我一直在环顾四周,试验音频比较.我已经找到了相当多的材料,并且大量引用了不同的库和方法.

到目前为止,我已经使用Audacity并导出一个名为"long.wav"的3分钟wav文件,然后将其前30秒分成一个名为"short.wav"的文件.我想在线上的某处我可以通过java在每个视觉上记录(log.txt)数据,并且应该能够在值中看到至少一些视觉上的相似性....这里是一些代码

主要方法:

        int totalFramesRead = 0;
        File fileIn = new File(filePath);
        BufferedWriter writer = new BufferedWriter(new FileWriter(outPath));
        writer.flush();
        writer.write("");
        try {
            AudioInputStream audioInputStream = 
                    AudioSystem.getAudioInputStream(fileIn);
            int bytesPerFrame = 
                    audioInputStream.getFormat().getFrameSize();
            if (bytesPerFrame == AudioSystem.NOT_SPECIFIED) {
                // some audio formats may have unspecified frame size
                // in that case we may read any amount of bytes
                bytesPerFrame = 1;
            } 
            // Set an arbitrary buffer size of 1024 frames.
            int numBytes = 1024 * bytesPerFrame; 
            byte[] audioBytes = new byte[numBytes];
            try {
                int numBytesRead = 0;
                int numFramesRead = 0;
                // Try to read numBytes bytes from the file.
                while ((numBytesRead = 
                        audioInputStream.read(audioBytes)) != -1) {
                    // Calculate the number of frames actually read.
                    numFramesRead = numBytesRead / bytesPerFrame;
                    totalFramesRead += numFramesRead;
                    // Here, do something useful with the audio data that's 
                    // now in the audioBytes array...

                    if(totalFramesRead <= 4096 * 100)
                    {                           

                    Complex[][] results = PerformFFT(audioBytes);
                    int[][] lines = GetKeyPoints(results);
                    DumpToFile(lines, writer);      

                    }   
                }
            } catch (Exception ex) { 
                // Handle the error...
            }
            audioInputStream.close();
        } catch (Exception e) {
            // Handle the error...
        }
        writer.close();

然后PerformFFT:

public static Complex[][] PerformFFT(byte[] data) throws IOException
    {
        final int totalSize = data.length;

        int amountPossible = totalSize/Harvester.CHUNK_SIZE;

        //When turning into frequency domain we'll need complex numbers:
        Complex[][] results = new Complex[amountPossible][];

        //For all the chunks:
        for(int times = 0;times < amountPossible; times++) {
            Complex[] complex = new Complex[Harvester.CHUNK_SIZE];
            for(int i = 0;i < Harvester.CHUNK_SIZE;i++) {
                //Put the time domain data into a complex number with imaginary part as 0:
                complex[i] = new Complex(data[(times*Harvester.CHUNK_SIZE)+i], 0);
            }
            //Perform FFT analysis on the chunk:
            results[times] = FFT.fft(complex);
        }
            return results;
}

此时我尝试了无处不在的日志:变换前的audioBytes,复杂值和FFT结果.

问题:无论我记录什么值,每个wav文件的log.txt都完全不同.我不明白.鉴于我从large.wav获取了small.wav(并且它们具有所有相同的属性),原始wav byte [] data ...或Complex [] [] fft数据之间应该存在非常大的相似性. ..或者到目前为止的东西......

如果数据在这些计算的任何一点都不接近相似,我怎么可能尝试比较这些文件.

我知道我在音频分析方面缺少相当多的知识,这就是我来到董事会寻求帮助的原因!感谢您提供的任何信息,帮助或修复!

Answer 1

对于 16 位音频，3e-05 与 0 并没有太大区别。因此，零文件与零文件几乎相同（可能由于一些微小的舍入误差而缺少相等性。）

添加：为了进行比较，使用一些 Java 绘图库读入并绘制两个波形中的每一个波形在超过大部分（接近）零的部分时的部分。