匹配 3-D 数组中矩阵的值

Question

我正在尝试匹配图像中的 RGB 值。

         % R  G   B
RGBset = [ 3  9  12;
           4  8  13;
          11 13  13;
           8  3   2]

img(:,:,1) = [1   2   3
              6   5   4
              7   9   8
             10  11  12];

img(:,:,2) = [3  4  8;
              6  7  8;
             11 10  9;
             12 13 14];

img(:,:,3)= [3  7  2;
             4  9 10;
             5 11 12;
             6 13 14]

在这张图片中，只有一个 RGB 值与 RGBset 匹配，即[11,13,13]，因此预期输出为：

[0  0  0;
 0  0  0;
 0  0  0;
 0  1  0]; % reshape(img(4,2,:),1,3) = [11, 13 13] is available in RGBset
           % no other RGB value is present in the image

我已经编写了这段代码，但是对于较大的图像它非常慢。

matched= zeros(img(:,:,1));
for r=1:size(img(:,:,1),1)
    for c=1:size(img(:,:,2),2)
     matched(r,c)=ismember(reshape(img(r,c,:),1,3),RGBset,'rows');
    end
end

更快的解决方案是什么？

Answer 1

我们可以将每个 RGB 三元组简化为一个标量，我们将为 RGBset 和 img 执行此操作。这会将它们分别减少到 2D 和 1D 矩阵。我们称之为降维。通过这些减少的数据，我们实现了内存效率，并有望提升性能。

因此，包含这些基础的解决方案看起来像这样 -

% Scaling array for dim reduction
s = [256^2, 256, 1].';

% Reduce dims for RGBset and img
RGBset1D = RGBset*s;
img1D = reshape(img,[],3)*s;

% Finally use find membership and reshape to 2D
out = reshape(ismember(img1D, RGBset1D), size(img,1), []);

矢量化解决方案的基准测试

基准代码 -

         % R  G   B
RGBset = [ 3  9  12;
           4  8  13;
          11 13  13;
           8  3   2]

% Setup inputs
img = randi(255, 2000, 2000, 3);
img(3,2,:) = RGBset(4,:);

% Luis's soln
disp('--------------------- Reshape + Permute ------------------')
tic
img2 = reshape(permute(img, [3 1 2]), 3, []).';
matched = ismember(img2, RGBset, 'rows');
matched = reshape(matched, size(img,1), []);
toc

% Proposed in this post
disp('--------------------- Dim reduction ------------------')
tic
s = [256^2, 256, 1].';
RGBset1D = RGBset*s;
img1D = reshape(img,[],3)*s;
out = reshape(ismember(img1D, RGBset1D), size(img,1), []);
toc

基准测试输出 -

--------------------- Reshape + Permute ------------------
Elapsed time is 3.101870 seconds.
--------------------- Dim reduction ------------------
Elapsed time is 0.031589 seconds.

Answer 2

您可以将循环替换为permute 和reshape：

img2 = reshape(permute(img, [3 1 2]), 3, []).';
matched = ismember(img2, RGBset, 'rows');
matched = reshape(matched, size(img,1), []);

这将创建一个 3 列矩阵 img2，其中每一行对应于来自 img 的一个像素。这样ismember(..., 'rows') 可以以矢量化方式应用。然后根据需要对获得的结果进行整形。

Answer 3

您可以只循环颜色，这比循环每个像素要快得多。

% Set up your colours into the 3rd dimension, so they match along the same axis
RGB3D = reshape(RGBset,[],1,3);
% Loop over them
for ii = 1:size(RGB3D, 1)
    % See if all 3 members of the colour match any pixel
    matched = all(ismember(img, RGB3D(ii,:,:)),3)
    if any(matched)
        disp(matched) 
        disp(['matched color: ' num2str(ii)]); 
        % do something else with the matched pixels
    end
end