在OpenCV中从RGB转换为YUYV

Pav*_*sky 5 c++ opencv colors video-streaming libyuv

有没有办法从RGB转换为YUYV(YUY 4:2:2)格式?我注意到OpenCV有反向操作,但由于某种原因没有RGB到YUYV.也许有人可以指向那样做的代码(即使在OpenCV库之外)?

UPDATE

我找到了libyuv库,它可以通过BGR到ARGB转换然后ARGB到YUY2格式来工作(希望这与YUYV 4:2:2相同).但它似乎没有用.你碰巧知道yuyv缓冲区尺寸/类型应该是什么样的吗?它的步伐是什么?

澄清YUYV和YUY2是相同的格式,如果它有帮助.

更新2 这是我使用libyuv库的代码:

  Mat frame;
  // Convert original image im from BGR to BGRA for further use in libyuv
  cvtColor(im, frame, CVX_BGR2BGRA);
  // Actually libyuv requires ARGB (i.e. reverse of BGRA), so I swap channels here
  int from_to[] = { 0,3, 1,2, 2,1, 3,0 };
  mixChannels(&frame, 1, &frame, 1, from_to, 4);
  // This is the most confusing part. Not sure what argb_stride suppose to be - length of a row in bytes or size of single value in the array?
  const uint8_t* argb_data = frame.data;
  int argb_stride = 8;

  // Also it is not clear what size of yuyv frame should be since we duplicate one Y 
  Mat yuyv(frame.rows, frame.cols, CVX_8UC2);
  uint8_t* yuyv_data = yuyv.data;
  int yuyv_stride = 16;
  // Do actual conversion
  libyuv::ARGBToYUY2(argb_data, argb_stride, yuyv_data, yuyv_stride, 
  frame.cols, frame.rows);
  // Then I feed yuyv_data to video stream buffer and see green or purple image instead of video stream.
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更新3

  Mat frame;
  cvtColor(im, frame, CVX_BGR2BGRA);

  // ARGB
  int from_to[] = { 0,3, 1,2, 2,1, 3,0 };
  Mat rgba(frame.size(), frame.type());
  mixChannels(&frame, 1, &rgba, 1, from_to, 4);
  const uint8_t* argb_data = rgba.data;
  int argb_stride = rgba.cols*4;

  Mat yuyv(rgba.rows, rgba.cols, CVX_8UC2);
  uint8_t* yuyv_data = yuyv.data;
  int yuyv_stride = width * 2;
  int res = libyuv::ARGBToYUY2(argb_data, argb_stride, yuyv_data, yuyv_stride, rgba.cols, rgba.rows);
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在此输入图像描述

Pav*_*sky 1

看起来虽然方法被称为 ARGBToYUY2,但它需要 BGRA 通道顺序(而不是反向)。