以下是使用MATLAB编写的代码,用于计算Zernike矩和边缘提取:

% 计算Zernike矩
function [Z00, Z11, Z20, Z31, Z40] = calculateZernikeMoments(image)
    [rows, cols] = size(image);
    [x, y] = meshgrid(1:cols, 1:rows);
    r = sqrt((2*x-cols-1).^2 + (2*y-rows-1).^2) / rows;
    theta = atan2((rows-1-2*y+2), (2*x-cols+1-2));
    
    Z00 = sum(sum(image));
    Z11 = sqrt(2) * sum(sum(image .* r .* exp(1i*theta)));
    Z20 = sqrt(2) * sum(sum(image .* ((2*r.^2) - 1)));
    Z31 = sqrt(2) * sum(sum(image .* r .* exp(1i*theta) .* ((3*r.^2) - 2)));
    Z40 = sqrt(2) * sum(sum(image .* ((6*r.^4) - (6*r.^2) + 1)));
end

% 边缘提取
function [edgeMap] = extractEdges(image, kt, lt)
    [Z00, Z11, Z20, Z31, Z40] = calculateZernikeMoments(image);
    
    [rows, cols] = size(image);
    [x, y] = meshgrid(1:cols, 1:rows);
    r = sqrt((2*x-cols-1).^2 + (2*y-rows-1).^2) / rows;
    theta = atan2((rows-1-2*y+2), (2*x-cols+1-2));
    
    Z11_prime = Z11 * sqrt(2) * r .* exp(1i*theta);
    Z20_prime = Z20 * sqrt(2) * ((2*r.^2) - 1);
    Z31_prime = Z31 * sqrt(2) * r .* exp(1i*theta) .* ((3*r.^2) - 2);
    Z40_prime = Z40 * sqrt(2) * ((6*r.^4) - (6*r.^2) + 1);
    
    phi = atan2(imag(Z31_prime), real(Z31_prime));
    l1 = sqrt((5*real(Z40_prime) + 3*real(Z20_prime)) ./ (8*real(Z20_prime)));
    l1_prime = -sqrt((5*real(Z40_prime) + 3*real(Z20_prime)) ./ (8*real(Z20_prime)));
    l2 = sqrt((5*real(Z31_prime) + real(Z11)) ./ (6*real(Z11)));
    l2_prime = -sqrt((5*real(Z31_prime) + real(Z11)) ./ (6*real(Z11)));
    l = (l1 + l2) / 2;
    
    k = (3*real(Z11)) ./ (2*(1 - l2.^2).^(3/2));
    h = (Z00 - (k*pi/2) + k.*asin(l2) + k.*l2.*sqrt(1-l2.^2)) / pi;
    
    edgeMap = zeros(rows, cols);
    edgeMap(k >= kt & abs(l2 - l1) <= lt) = 1;
end

% 示例用法
image = imread('image.jpg'); % 替换为实际图像的路径
image = rgb2gray(image); % 如果图像是彩色的,将其转换为灰度图像
edgeMap = extractEdges(image, 0.5, 0.1); % 替换kt和lt的值
imshow(edgeMap);

请注意,上述代码仅提供了一个基本的实现框架,具体的参数和阈值需要根据实际情况进行调整和优化。此外,代码中使用的Zernike矩计算方法仅适用于正方形图像,如果输入图像不是正方形,可能需要进行适当的调整。

MATLAB 亚像素边缘提取:基于Zernike矩的边缘检测算法

原文地址: http://www.cveoy.top/t/topic/fB7l 著作权归作者所有。请勿转载和采集!

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