图像之间的欧几里德距离 [英] Euclidean distance between images
问题描述
我有两张图片，比如 P 和 S ，大小8192×200，我想计算它们之间的自定义欧几里德距离。目前我使用以下步骤：
I have two images, say P and S, of size 8192×200, and I want to calculate a custom "Euclidean distance" between them. Currently I use the following steps:
-
将图像重塑为一对列和行向量：
Reshape the images into a pair of column and row vectors:
Ip = Ip(:).';
Is = Is(:);
计算指标矩阵， G ，其条目由公式给出
G(i,j) = 1/(2*pi*r*r) * exp((-d*d)/(2*r*r));
其中 r 是一个全局参数，不同从0到20，比如说， d 是像素 i 和像素 j之间的距离。例如，如果像素 i 是（k，l）并且像素 j 是（k1，l1），然后 d = sqrt（（k-k1）^ 2 +（l-l1）^ 2） ; 。像素1将是（1,1），像素2将是（1,2），依此类推。因此，矩阵 G 的大小将为 1638400×1638400 。
where r is a global parameter that varies from 0 to 20, say, and d is the distance between pixel i and pixel j. E.g., if pixel i is (k,l) and pixel j is (k1,l1), then d = sqrt((k-k1)^2 + (l-l1)^2);. Pixel 1 will be (1,1), Pixel 2 will be (1,2), and so on. Therefore, the size of matrix G will be 1638400×1638400.
计算两张图像之间的最终（标量）欧几里德距离，使用：
Compute the final (scalar) Euclidean distance between two images, using:
ImEuDist = sqrt( (Ip-Is) * G * (Ip-Is).' );
我已经使用mex编写了一些代码功能，但在给出结果（5-6小时）之前需要很长时间 - 参见这个问题以及对此的更多讨论。
I have already written some code using a mex function, but it is taking too long before giving the results (5-6 Hours) - see this SO question for code and more discussion on this.
请帮我优化一下;理想情况下，我希望它能在几秒钟内完成。请注意，我对涉及GPU的解决方案不感兴趣。
Please help me to optimize this; I would ideally like it to run in a matter of seconds. Note that I am not interested in solutions involving the GPU.
推荐答案
如果我理解正确，您应该能够执行以下操作，并让它在2s以下运行：
If I've understood correctly, you should be able to do the following, and have it run in under 2s:
样本数据：
s1 = 8192; s2 = 200;
img_a = rand(s1, s2);
img_b = rand(s1, s2);
r = 2;
和计算本身：
img_diff = img_a - img_b;
kernel = bsxfun(@plus, (-s1:s1).^2.', (-s2:s2).^2);
kernel = 1/(2/pi/r^2) * exp(-kernel/ (2*r*2));
g = conv2(img_diff, kernel, 'same');
res = g(:)' * img_diff(:);
res = sqrt(res);
以上约需25秒。要降低到2秒，您需要使用更快的基于fft的卷积替换标准的 conv2 。请参阅此和这个：
The above takes about 25s. To get down to 2s, you need to replace the standard conv2 with a faster, fft based convolution. See this and this:
function c = conv2fft(X, Y)
% ignoring small floating-point differences, this is equivalent
% to the inbuilt Matlab conv2(X, Y, 'same')
X1 = [X zeros(size(X,1), size(Y,2)-1);
zeros(size(Y,1)-1, size(X,2)+size(Y,2)-1)];
Y1 = zeros(size(X1));
Y1(1:size(Y,1), 1:size(Y,2)) = Y;
c = ifft2(fft2(X1).*fft2(Y1));
c = c(size(X,1)+1:size(X,1)+size(X,1), size(X,2)+1:size(X,2)+size(X,2));
end
顺便提一下，如果你仍然希望它更快，你可以利用事实上 exp（-d ^ 2 / r ^ 2）将非常接近变为零，因为相当小的d：所以你实际上可以裁剪你的内核只是一个小矩形，而不是上面提到的巨大的东西。较小的内核意味着 conv2fft （或特别是 conv2 ）将运行得更快。
Incidentally, if you still want it to go faster, you could make use of the fact that exp(-d^2/r^2) gets very close to zero for fairly small d: so you can actually crop your kernel to just a tiny rectangle, rather than the huge thing suggested above. A smaller kernel means conv2fft (or especially conv2) will run faster.
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