10

我正在尝试从 TCP 套接字在 C# 中进行“流式传输”语音识别。我遇到的问题是 SpeechRecognitionEngine.SetInputToAudioStream() 似乎需要一个可以寻找的定义长度的流。现在，我能想到的唯一方法是随着更多输入的进入，在 MemoryStream 上重复运行识别器。

这里有一些代码来说明：

            SpeechRecognitionEngine appRecognizer = new SpeechRecognitionEngine();

            System.Speech.AudioFormat.SpeechAudioFormatInfo formatInfo = new System.Speech.AudioFormat.SpeechAudioFormatInfo(8000, System.Speech.AudioFormat.AudioBitsPerSample.Sixteen, System.Speech.AudioFormat.AudioChannel.Mono);

            NetworkStream stream = new NetworkStream(socket,true);
            appRecognizer.SetInputToAudioStream(stream, formatInfo);
            // At the line above a "NotSupportedException" complaining that "This stream does not support seek operations."

有谁知道如何解决这个问题？它必须支持某种类型的流输入，因为它可以与使用 SetInputToDefaultAudioDevice() 的麦克风一起正常工作。

谢谢，肖恩

4

5 回答 5

16

我通过覆盖流类来进行实时语音识别：

class SpeechStreamer : Stream
{
    private AutoResetEvent _writeEvent;
    private List<byte> _buffer;
    private int _buffersize;
    private int _readposition;
    private int _writeposition;
    private bool _reset;

    public SpeechStreamer(int bufferSize)
    {
        _writeEvent = new AutoResetEvent(false);
         _buffersize = bufferSize;
         _buffer = new List<byte>(_buffersize);
         for (int i = 0; i < _buffersize;i++ )
             _buffer.Add(new byte());
        _readposition = 0;
        _writeposition = 0;
    }

    public override bool CanRead
    {
        get { return true; }
    }

    public override bool CanSeek
    {
        get { return false; }
    }

    public override bool CanWrite
    {
        get { return true; }
    }

    public override long Length
    {
        get { return -1L; }
    }

    public override long Position
    {
        get { return 0L; }
        set {  }
    }

    public override long Seek(long offset, SeekOrigin origin)
    {
        return 0L;
    }

    public override void SetLength(long value)
    {

    }

    public override int Read(byte[] buffer, int offset, int count)
    {
        int i = 0;
        while (i<count && _writeEvent!=null)
        {
            if (!_reset && _readposition >= _writeposition)
            {
                _writeEvent.WaitOne(100, true);
                continue;
            }
            buffer[i] = _buffer[_readposition+offset];
            _readposition++;
            if (_readposition == _buffersize)
            {
                _readposition = 0;
                _reset = false;
            }
            i++;
        }

        return count;
    }

    public override void Write(byte[] buffer, int offset, int count)
    {
        for (int i = offset; i < offset+count; i++)
        {
            _buffer[_writeposition] = buffer[i];
            _writeposition++;
            if (_writeposition == _buffersize)
            {
                _writeposition = 0;
                _reset = true;
            }
        }
        _writeEvent.Set();

    }

    public override void Close()
    {
        _writeEvent.Close();
        _writeEvent = null;
        base.Close();
    }

    public override void Flush()
    {

    }
}

...并将其实例用作 SetInputToAudioStream 方法的流输入。一旦流返回一个长度或返回的计数小于请求的计数，识别引擎就认为输入已经完成。这设置了一个永远不会完成的循环缓冲区。

于 2012-08-05T01:29:14.680 回答
2

您是否尝试过将网络流包装在 System.IO.BufferedStream 中？

NetworkStream netStream = new NetworkStream(socket,true);
BufferedStream buffStream = new BufferedStream(netStream, 8000*16*1); // buffers 1 second worth of data
appRecognizer.SetInputToAudioStream(buffStream, formatInfo);
于 2009-11-05T19:24:42.647 回答
2

显然它不能完成（“按设计”！）。请参阅http://social.msdn.microsoft.com/Forums/en/netfxbcl/thread/fcf62d6d-19df-4ca9-9f1f-17724441f84e

于 2010-11-20T05:49:54.540 回答
2

这是我的解决方案。

class FakeStreamer : Stream
{
    public bool bExit = false;
    Stream stream;
    TcpClient client;
    public FakeStreamer(TcpClient client)
    {
        this.client = client;
        this.stream = client.GetStream();
        this.stream.ReadTimeout = 100; //100ms
    }
    public override bool CanRead
    {
        get { return stream.CanRead; }
    }

    public override bool CanSeek
    {
        get { return false; }
    }

    public override bool CanWrite
    {
        get { return stream.CanWrite; }
    }

    public override long Length
    {
        get { return -1L; }
    }

    public override long Position
    {
        get { return 0L; }
        set { }
    }
    public override long Seek(long offset, SeekOrigin origin)
    {
        return 0L;
    }

    public override void SetLength(long value)
    {
        stream.SetLength(value);
    }
    public override int Read(byte[] buffer, int offset, int count)
    {
        int len = 0, c = count;
        while (c > 0 && !bExit)
        {
            try
            {
                len = stream.Read(buffer, offset, c);
            }
            catch (Exception e)
            {
                if (e.HResult == -2146232800) // Timeout
                {
                    continue;
                }
                else
                {
                    //Exit read loop
                    break;
                }
            }
            if (!client.Connected || len == 0)
            {
                //Exit read loop
                return 0;
            }
            offset += len;
            c -= len;
        }
        return count;
    }

    public override void Write(byte[] buffer, int offset, int count)
    {
        stream.Write(buffer,offset,count);
    }

    public override void Close()
    {
        stream.Close();
        base.Close();
    }

    public override void Flush()
    {
        stream.Flush();
    }
}

如何使用：

//client connect in
TcpClient clientSocket = ServerSocket.AcceptTcpClient();
FakeStreamer buffStream = new FakeStreamer(clientSocket);
...
//recognizer init
m_recognizer.SetInputToAudioStream(buffStream , audioFormat);
...
//recognizer end
if (buffStream != null)
    buffStream.bExit = true;
于 2017-06-09T06:04:46.107 回答
1

我最终缓冲了输入，然后将其以连续更大的块发送到语音识别引擎。例如，我可能首先发送前 0.25 秒，然后是前 0.5 秒，然后是前 0.75 秒，以此类推，直到我得到结果。我不确定这是否是解决此问题的最有效方法，但它对我来说产生了令人满意的结果。

祝你好运，肖恩

于 2010-01-09T13:17:43.243 回答