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2Â¥2012-04-24 19:22:29
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3Â¥2012-04-25 21:58:11
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Сľ³æ: ½ð±Ò+0.5, ¸ø¸öºì°ü£¬Ð»Ð»»ØÌû
Сľ³æ: ½ð±Ò+0.5, ¸ø¸öºì°ü£¬Ð»Ð»»ØÌû
4Â¥2012-04-25 22:03:35
5Â¥2012-04-26 03:11:17
6Â¥2012-04-26 09:40:02
7Â¥2012-04-26 10:33:24
ÎÒ°ÑÓÃmatlab¼ÆËãÐý¶ÈµÄ´úÂëÌù³öÀ´£¬´ó¼Ò°ïæ¿´¿´£¬¶ÔÕâ¸öÖ÷ÌâÓÐûÓаïÖú![]() I = imread('a.jpg'); [rows cols dim]=size(I); %sobel operator used to calculate gradient image Grd=[ -1 -2 -1; 0 0 0; 1 2 1]; Emean=zeros(rows,cols); for i=1:dim Ev(:,:,i)=conv2(x(:,:,i),Grd,'same'); Eh(:,:,i)=conv2(x(:,:,i),Grd.','same'); E(:,:,i)=abs(Eh(:,:,i))+abs(Ev(:,:,i)); % E(:,:,i)=conv2(X(:,:,i),Grd1,'same'); end Emean=1/dim*sum(E,3); %finds average gradient image %count the "curl" Ecurl = zeros(rows, cols);%creat a 0 rows * cols matrix for i = 1:dim Ecurl(:,:,i) = conv2(Iy(:,:,i), Grd.', 'same') - conv2(Ix(:,:,i), Grd, 'same'); end theCurl = 1/dim*sum(Ecurl,3); |
8Â¥2012-04-26 10:49:19









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