实现一个Harris角检测器 [英] Implementing a Harris corner detector

问题描述

我采取以教育为目的Harris角器，但我被困在哈里斯响应的一部分。基本上，我在做什么，就是：

I am implementing a Harris corner detector for educational purposes but I'm stuck at the harris response part. Basically, what I am doing, is:

1. 在x方向和y方向计算图像强度梯度
2. （1）
模糊输出
3. 在计算哈里斯反应过度（2）
输出
4. 在燮preSS非极大值的输出（3）在一个3×3邻域和阈值输出

1. Compute image intensity gradients in x- and y-direction
2. Blur output of (1)
3. Compute Harris response over output of (2)
4. Suppress non-maximas in output of (3) in a 3x3-neighborhood and threshold output

1和2似乎做工精细;不过，我得到非常小的值作为哈里斯的反应，并没有一点确实达到阈值。输入是一个标准的户外摄影。

1 and 2 seem to work fine; however, I get very small values as the Harris response, and no point does reach the threshold. Input is a standard outdoor photography.

[...]
[Ix, Iy] = intensityGradients(img);
g = fspecial('gaussian');
Ix = imfilter(Ix, g);
Iy = imfilter(Iy, g);
H = harrisResponse(Ix, Iy);
[...]

function K = harrisResponse(Ix, Iy)
    max = 0;
    [sy, sx] = size(Ix);
    K = zeros(sy, sx);
    for i = 1:sx,
        for j = 1:sy,
            H = [Ix(j,i) * Ix(j,i), Ix(j,i) * Iy(j,i)
                Ix(j,i) * Iy(j,i), Iy(j,i) * Iy(j,i)];
            K(j,i) = det(H) / trace(H);
            if K(j,i) > max,
                max = K(j,i);
            end
        end
    end
    max
end

有关的样品图片，最大最终被6.4163e-018似乎太低了。

For the sample picture, max ends up being 6.4163e-018 which seems far too low.

推荐答案

在Harris角点检测一个角落被定义为在一个地区的最高值像素（通常 3X3 或 5×5 ），所以你对没有点达到门槛的评论似乎很奇怪我。只要收集有更高的价值比 5x5的身边附近其他所有像素的所有像素。

A corner in Harris corner detection is defined as "the highest value pixel in a region" (usually 3X3 or 5x5) so your comment about no point reaching a "threshold" seems strange to me. Just collect all pixels that have a higher value than all other pixels in the 5x5 neighborhood around them.

除此之外： 我不是100％肯定，但我认为你应该有：

Apart from that: I'm not 100% sure, but I think you should have:

K（J，I）= DET（H） - 拉姆达*（跟踪（H）^ 2） 其中的λ是一个正的常数，在你的情况下工作（和哈里斯建议值是0.04）。

K(j,i) = det(H) - lambda*(trace(H)^2) Where lambda is a positive constant that works in your case (and Harris suggested value is 0.04).

在一般的唯一明智的时刻来过滤输入了这一点之前：

In general the only sensible moment to filter your input is before this point:

[IX，IY = intensityGradients（IMG）;

过滤 Ix2的， IY2 和积Ixy 没有按'牛逼多大意义了我。

Filtering Ix2, Iy2 and Ixy doesn't make much sense to me.

另外，我觉得你的样品code是错在这里（不函数 harrisResponse 有两个或三个输入变量）：

Further, I think your sample code is wrong here (does function harrisResponse have two or three input variables?):

H = harrisResponse(Ix2, Ixy, Iy2);
[...]

function K = harrisResponse(Ix, Iy)

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