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。也许有人可以指出这样做的代码（甚至在 OpenCV 库之外）？
更新
我发现 libyuv 库可以通过将 BGR 转换为 ARGB，然后将 ARGB 转换为 YUY2 格式（希望这与 YUYV 4:2:2 相同）来实现此目的。但这似乎不起作用。你碰巧知道 yuyv 缓冲区的尺寸/类型应该是什么样子吗？它的步伐是多少？
如果有帮助，澄清 YUYV 和 YUY2 是相同的格式。
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.
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);
