Question

我正在尝试从intel realsense相机发送一个帧到matlab。 d通过imshow（...）或图像（...）显示图像没有成功，因为图像是rgba图像... 我已将图像作为对象从C＃发送：

matlab.Feval("getImage", 1, out result, bitmap_data);

是否有可以显示框架的功能？

Answer 1

你必须使用精确的Feval实现，但是如果你直接在matlab中强制它有两个选项

1.完全忽略alpha通道

%this says show all rows, all cols, for channels 1 through 3 or R,G,B
imshow(rgba_im(:,:,1:3));

2.使用alpha channel

%this says show all rows, all cols, for channels 1 through 3 or R,G,B
%also saves the handle to the newly drawn image
hndl = imshow(rgba_im(:,:,1:3));

%isolates alpha channel
alpha = rgba_im(:,:,4);

%displays alpha channel
set(hndl , 'AlphaData', alpha);

修改

现在我知道你的数据不是标准的rgba格式，这里是修复它的代码，评论应该告诉你所需要的一切

[num_rows,num_cols_x4]=size(rgba_matrix); %we have to transpose the array to get it to unfold properly, which is %first by columns, then rows at = rgba_matrix.'; %converts data from [r1 g1 b1 a1 r2 g2 b2 a2 r3 g3 b3 a3....] to % red_chan = [r1 r2 r3...] % green_chan = [g1 g2 g3...] % blue_chan = [b1 b2 b3...] % alpha_chan = [a1 a2 a3...] % it says start at some index and grab every 4th element till the end of the % matrix red_chan = at(1:4:end); grn_chan = at(2:4:end); blu_chan = at(3:4:end); alp_chan = at(4:4:end); % reshape each channel from one long vector into a num_rows x (num_cols_x4/4) red_chan = reshape(red_chan, num_cols_x4/4, num_rows).'; grn_chan = reshape(grn_chan, num_cols_x4/4, num_rows).'; blu_chan = reshape(blu_chan, num_cols_x4/4, num_rows).'; alp_chan = reshape(alp_chan, num_cols_x4/4, num_rows).'; % we concatenate the channels into a num_rows x (num_cols_x4/4) x 4 matrix standard_rgba = cat(3,red_chan,grn_chan,blu_chan,alp_chan);

从这一点开始，您可以使用standard_rgba数组进行处理。可能有一种更有效的方式来编写这段代码，但我希望尽可能清晰易懂，希望这有帮助

从intel realsense到matlab的RGBa图像

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