Question

我有yuv420数据的字节数组。

byte[] yuv420;//yuv data

如何将其转换为Image<Bgr, byte>？

我找到了一个转换为RGB然后转换为Image<Bgr, byte>的数学公式，但它非常慢。有没有办法更快地转换它？

Emgu中有一个用于转换的课程

COLOR_CONVERSION(enum CV_YUV2RGB    Convert YUV color to RGB)

但是我无法理解如何使用这个类。有人可以帮忙吗？

static Bitmap ConvertYUV2RGB(byte[] yuvFrame, byte[] rgbFrame, int width, int height)
{
    int uIndex = width * height;
    int vIndex = uIndex + ((width * height) >> 2);
    int gIndex = width * height;
    int bIndex = gIndex * 2;

    int temp = 0;


    //图片为pic1,RGB颜色的二进制数据转换得的int r,g,b;
    Bitmap bm = new Bitmap(width, height);

    int r = 0;
    int g = 0;
    int b = 0;


    for (int y = 0; y < height; y++)
    {
        for (int x = 0; x < width; x++)
        {
            // R分量
            temp = (int)(yuvFrame[y * width + x] + (yuvFrame[vIndex + (y / 2) * (width / 2) + x / 2] - 128) * YUV2RGB_CONVERT_MATRIX[0, 2]);
            rgbFrame[y * width + x] = (byte)(temp < 0 ? 0 : (temp > 255 ? 255 : temp));
            // G分量
            temp = (int)(yuvFrame[y * width + x] + (yuvFrame[uIndex + (y / 2) * (width / 2) + x / 2] - 128) * YUV2RGB_CONVERT_MATRIX[1, 1] + (yuvFrame[vIndex + (y / 2) * (width / 2) + x / 2] - 128) * YUV2RGB_CONVERT_MATRIX[1, 2]);
            rgbFrame[gIndex + y * width + x] = (byte)(temp < 0 ? 0 : (temp > 255 ? 255 : temp));
            // B分量
            temp = (int)(yuvFrame[y * width + x] + (yuvFrame[uIndex + (y / 2) * (width / 2) + x / 2] - 128) * YUV2RGB_CONVERT_MATRIX[2, 1]);
            rgbFrame[bIndex + y * width + x] = (byte)(temp < 0 ? 0 : (temp > 255 ? 255 : temp));
            Color c = Color.FromArgb(rgbFrame[y * width + x], rgbFrame[gIndex + y * width + x], rgbFrame[bIndex + y * width + x]);
            bm.SetPixel(x, y, c);
        }
    }
    return bm;

}

static double[,] YUV2RGB_CONVERT_MATRIX = new double[3, 3] { { 1, 0, 1.4022 }, { 1, -0.3456, -0.7145 }, { 1, 1.771, 0 } };
static byte clamp(float input)
{
    if (input < 0) input = 0;
    if (input > 255) input = 255;
    return (byte)Math.Abs(input);
}

Answer 1

你很幸运，因为我之前完全解决了这个问题。代码中有一些链接可以获得更多信息。

通常在执行图像处理时总是尝试使用指针，并避免在嵌套循环中调用函数。在我的代码中，大小比较是迄今为止最慢的部分，但不幸的是需要它（尝试使用预处理器开关将其关闭）。

我不得不说尽管最后我从未使用过这个函数，因为它太慢了，我选择用c ++实现它并使用p调用从c＃中调用它。

private static unsafe void YUV2RGBManaged(byte[] YUVData, byte[] RGBData, int width, int height)
    {

        //returned pixel format is 2yuv - i.e. luminance, y, is represented for every pixel and the u and v are alternated
        //like this (where Cb = u , Cr = y)
        //Y0 Cb Y1 Cr Y2 Cb Y3 

        /*http://msdn.microsoft.com/en-us/library/ms893078.aspx
         * 
         * C = Y - 16
         D = U - 128
         E = V - 128
         R = clip(( 298 * C           + 409 * E + 128) >> 8)
         G = clip(( 298 * C - 100 * D - 208 * E + 128) >> 8)
         B = clip(( 298 * C + 516 * D           + 128) >> 8)

         * here are a whole bunch more formats for doing this...
         * http://stackoverflow.com/questions/3943779/converting-to-yuv-ycbcr-colour-space-many-versions
         */


        fixed(byte* pRGBs = RGBData, pYUVs = YUVData)
        {
            for (int r = 0; r < height; r++)
            {
                byte* pRGB = pRGBs + r * width * 3;
                byte* pYUV = pYUVs + r * width * 2;

                //process two pixels at a time
                for (int c = 0; c < width; c += 2)
                {
                    int C1 = pYUV[1] - 16;
                    int C2 = pYUV[3] - 16;
                    int D = pYUV[2] - 128;
                    int E = pYUV[0] - 128;

                    int R1 = (298 * C1 + 409 * E + 128) >> 8;
                    int G1 = (298 * C1 - 100 * D - 208 * E + 128) >> 8;
                    int B1 = (298 * C1 + 516 * D + 128) >> 8;

                    int R2 = (298 * C2 + 409 * E + 128) >> 8;
                    int G2 = (298 * C2 - 100 * D - 208 * E + 128) >> 8;
                    int B2 = (298 * C2 + 516 * D + 128) >> 8;
#if true
                    //check for overflow
                    //unsurprisingly this takes the bulk of the time.
                    pRGB[0] = (byte)(R1 < 0 ? 0 : R1 > 255 ? 255 : R1);
                    pRGB[1] = (byte)(G1 < 0 ? 0 : G1 > 255 ? 255 : G1);
                    pRGB[2] = (byte)(B1 < 0 ? 0 : B1 > 255 ? 255 : B1);

                    pRGB[3] = (byte)(R2 < 0 ? 0 : R2 > 255 ? 255 : R2);
                    pRGB[4] = (byte)(G2 < 0 ? 0 : G2 > 255 ? 255 : G2);
                    pRGB[5] = (byte)(B2 < 0 ? 0 : B2 > 255 ? 255 : B2);
#else
                    pRGB[0] = (byte)(R1);
                    pRGB[1] = (byte)(G1);
                    pRGB[2] = (byte)(B1);

                    pRGB[3] = (byte)(R2);
                    pRGB[4] = (byte)(G2);
                    pRGB[5] = (byte)(B2);
#endif

                    pRGB += 6;
                    pYUV += 4;
                }
            }
        }
    }

并且你决定在c ++中实现它

void YUV2RGB(void *yuvDataIn,void *rgbDataOut, int w, int h, int outNCh)
{

    const int ch2 = 2 * outNCh;

    unsigned char* pRGBs = (unsigned char*)rgbDataOut;
    unsigned char* pYUVs = (unsigned char*)yuvDataIn;

    for (int r = 0; r < h; r++)
    {
        unsigned char* pRGB = pRGBs + r * w * outNCh;
        unsigned char* pYUV = pYUVs + r * w * 2;

        //process two pixels at a time
        for (int c = 0; c < w; c += 2)
        {
            int C1 = pYUV[1] - 16;
            int C2 = pYUV[3] - 16;
            int D = pYUV[2] - 128;
            int E = pYUV[0] - 128;

            int R1 = (298 * C1 + 409 * E + 128) >> 8;
            int G1 = (298 * C1 - 100 * D - 208 * E + 128) >> 8;
            int B1 = (298 * C1 + 516 * D + 128) >> 8;

            int R2 = (298 * C2 + 409 * E + 128) >> 8;
            int G2 = (298 * C2 - 100 * D - 208 * E + 128) >> 8;
            int B2 = (298 * C2 + 516 * D + 128) >> 8;

            //unsurprisingly this takes the bulk of the time.
            pRGB[0] = (unsigned char)(R1 < 0 ? 0 : R1 > 255 ? 255 : R1);
            pRGB[1] = (unsigned char)(G1 < 0 ? 0 : G1 > 255 ? 255 : G1);
            pRGB[2] = (unsigned char)(B1 < 0 ? 0 : B1 > 255 ? 255 : B1);

            pRGB[3] = (unsigned char)(R2 < 0 ? 0 : R2 > 255 ? 255 : R2);
            pRGB[4] = (unsigned char)(G2 < 0 ? 0 : G2 > 255 ? 255 : G2);
            pRGB[5] = (unsigned char)(B2 < 0 ? 0 : B2 > 255 ? 255 : B2);

            pRGB += ch2;
            pYUV += 4;
        }
    }
}

Answer 2

该代码中最大的罪犯是使用Bitmap.SetPixel;在每个内循环迭代中执行此操作非常慢。相反，使用字节数组存储RGB值，填充后，将其复制到位图as a single step。

其次，要了解y，u和v是字节，因此只能有256个可能的值。因此，为r，g和b构建查找表是完全可行的，因此您不必在内循环中执行任何计算。

最后，如果你真的想要性能，你必须使用指针算法在C ++中编写它，并在所有优化的基础上进行编译。由于每次迭代都对独立数据进行操作，因此该循环也是parallel for的非常好的候选者。也可以使用SSE intrinsics进一步优化，每个指令转换几个像素。

希望这可以让你开始。

Answer 3

我刚刚发现了一段可能对您有帮助的旧代码。使用OpenCVSharp进行YUV转换（免责声明：我删除了一些不必要的代码，但没有对此进行测试！）

IplImage yuvImage = new IplImage(w, h, BitDepth.U8, 3);
IplImage rgbImage = new IplImage(w, h, BitDepth.U8, 3);

Cv.CvtColor(yuvImage, rgbImage, ColorConversion.CrCbToBgr);

回答你的另一个问题 - 将byte []转换为Bitmap使用此

int w= 100;
int h = 200;
int ch = 3;

byte[] imageData    = new byte[w*h*ch]; //you image data here
Bitmap bitmap       = new Bitmap(w,h,PixelFormat.Format24bppRgb);
BitmapData bmData   = bitmap.LockBits(new System.Drawing.Rectangle(0, 0, bitmap.Width, bitmap.Height), ImageLockMode.ReadWrite, bitmap.PixelFormat);
IntPtr pNative      = bmData.Scan0;
Marshal.Copy(imageData,0,pNative,w*h*ch);
bitmap.UnlockBits(bmData);

Answer 4

一种更快的模式。两个mutiplication和两个每像素添加更少：

private static unsafe void YUV2RGBManaged(byte[] YUVData, byte[] RGBData, int width, int height)
{
    //returned pixel format is 2yuv - i.e. luminance, y, is represented for every pixel and the u and v are alternated
    //like this (where Cb = u , Cr = y)
    //Y0 Cb Y1 Cr Y2 Cb Y3 

    /*http://msdn.microsoft.com/en-us/library/ms893078.aspx
     * 
     C = 298 * (Y - 16) + 128
     D = U - 128
     E = V - 128
     R = clip(( C           + 409 * E) >> 8)
     G = clip(( C - 100 * D - 208 * E) >> 8)
     B = clip(( C + 516 * D          ) >> 8)

     * here are a whole bunch more formats for doing this...
     * http://stackoverflow.com/questions/3943779/converting-to-yuv-ycbcr-colour-space-many-versions
     */


    fixed(byte* pRGBs = RGBData, pYUVs = YUVData)
    {
        for (int r = 0; r < height; r++)
        {
            byte* pRGB = pRGBs + r * width * 3;
            byte* pYUV = pYUVs + r * width * 2;

            //process two pixels at a time
            for (int c = 0; c < width; c += 2)
            {
                int C1 = 298 * (pYUV[1] - 16) + 128;
                int C2 = 298 * (pYUV[3] - 16) + 128;
                int D = pYUV[2] - 128;
                int E = pYUV[0] - 128;

                int R1 = (C1 + 409 * E) >> 8;
                int G1 = (C1 - 100 * D - 208 * E) >> 8;
                int B1 = (C1 + 516 * D) >> 8;

                int R2 = (C2 + 409 * E) >> 8;
                int G2 = (C2 - 100 * D - 208 * E) >> 8;
                int B2 = (298 * C2 + 516 * D) >> 8;

                //check for overflow
                //unsurprisingly this takes the bulk of the time.
                pRGB[0] = (byte)(R1 < 0 ? 0 : R1 > 255 ? 255 : R1);
                pRGB[1] = (byte)(G1 < 0 ? 0 : G1 > 255 ? 255 : G1);
                pRGB[2] = (byte)(B1 < 0 ? 0 : B1 > 255 ? 255 : B1);

                pRGB[3] = (byte)(R2 < 0 ? 0 : R2 > 255 ? 255 : R2);
                pRGB[4] = (byte)(G2 < 0 ? 0 : G2 > 255 ? 255 : G2);
                pRGB[5] = (byte)(B2 < 0 ? 0 : B2 > 255 ? 255 : B2);

                pRGB += 6;
                pYUV += 4;
            }
        }
    }
}

从yuv 420转换为image <bgr，byte> </bgr，byte>

4 个答案: