Implementing a Harris corner detector

问题

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. 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 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

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

回答1:

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.

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

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).

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

[Ix, Iy] = intensityGradients(img);

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

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)

回答2:

The solution that I implemented with python, it works for me I hope you find what you are looking for

import numpy as np
import matplotlib.pyplot as plt
from PIL.Image import *
from scipy import ndimage

def imap1(im):
    print('testing the picture . . .')
    a = Image.getpixel(im, (0, 0))
    if type(a) == int:
        return im
    else:
        c, l = im.size
        imarr = np.asarray(im)
        neim = np.zeros((l, c))
        for i in range(l):
            for j in range(c):
                t = imarr[i, j]
                ts = sum(t)/len(t)
                neim[i, j] = ts
        return neim

def Harris(im):
    neim = imap1(im)
    imarr = np.asarray(neim, dtype=np.float64)
    ix = ndimage.sobel(imarr, 0)
    iy = ndimage.sobel(imarr, 1)
    ix2 = ix * ix
    iy2 = iy * iy
    ixy = ix * iy
    ix2 = ndimage.gaussian_filter(ix2, sigma=2)
    iy2 = ndimage.gaussian_filter(iy2, sigma=2)
    ixy = ndimage.gaussian_filter(ixy, sigma=2)
    c, l = imarr.shape
    result = np.zeros((c, l))
    r = np.zeros((c, l))
    rmax = 0
    for i in range(c):
        print('loking for corner . . .')
        for j in range(l):
            print('test ',j)
            m = np.array([[ix2[i, j], ixy[i, j]], [ixy[i, j], iy2[i, j]]], dtype=np.float64)
            r[i, j] = np.linalg.det(m) - 0.04 * (np.power(np.trace(m), 2))
            if r[i, j] > rmax:
                rmax = r[i, j]
    for i in range(c - 1):
        print(". .")
        for j in range(l - 1):
            print('loking')
            if r[i, j] > 0.01 * rmax and r[i, j] > r[i-1, j-1] and r[i, j] > r[i-1, j+1]\
                                     and r[i, j] > r[i+1, j-1] and r[i, j] > r[i+1, j+1]:
                result[i, j] = 1

    pc, pr = np.where(result == 1)
    plt.plot(pr, pc, 'r+')
    plt.savefig('harris_test.png')
    plt.imshow(im, 'gray')
    plt.show()
    # plt.imsave('harris_test.png', im, 'gray')

im = open('chess.png')
Harris(im)

回答3:

Proposed implementation is terribly inefficient. Lets' start after calculating gradients (which can be optimized too):

A = Ix.^2;
B = Iy.^2;
C = (Ix.*Iy).^4;
lambda = 0.04;

H = (A.*B - C) - lambda*(A+B).^2;

% if you really need max:
max(H(:))

No loops required, because Matlab hates loops.

回答4:

Basically, Harris corner detection will have 5 steps:

1. Gradient computation
2. Gaussian smoothing
3. Harris measure computation
4. Non-maximum suppression
5. Thresholding

If you are implementing in MATLAB, it will be easy to understand the algorithm and get the results.

The following code of MATLAB may help you to solve your doubts:

% Step 1: Compute derivatives of image
Ix = conv2(im, dx, 'same');
Iy = conv2(im, dy, 'same');

% Step 2: Smooth space image derivatives (gaussian filtering)
Ix2 = conv2(Ix .^ 2, g, 'same');
Iy2 = conv2(Iy .^ 2, g, 'same');
Ixy = conv2(Ix .* Iy, g, 'same');

% Step 3: Harris corner measure
harris = (Ix2 .* Iy2 - Ixy .^ 2) ./ (Ix2 + Iy2);

% Step 4: Find local maxima (non maximum suppression)
mx = ordfilt2(harris, size .^ 2, ones(size));

% Step 5: Thresholding
harris = (harris == mx) & (harris > threshold);

回答5:

There is a function for that in the Computer Vision System Toolbox called detectHarrisFeatures.

来源：https://stackoverflow.com/questions/3862225/implementing-a-harris-corner-detector