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GUI3

于 2011-12-18 发布 文件大小:2KB
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  Its about pavement crack identification

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  • li9_23
    clear all I=imread( lena.bmp ) figure imshow(I) I2=imrotate(I,-4, bilinear ) 逆时针旋转4度 figure imshow(I2) I3=fliplr(I) 垂直镜像 figure imshow(I3) I4=imresize(I,0.5, bilinear ) 缩小为原图的1/2 figure imshow(I4) A=double(I) 转换为double类型 计算7个不变矩 [nc,nr]=size(A) [x,y]=meshgrid(1:nr,1:nc) 得到网格 x=x(:) y=y(:) A=A(:) m.m00=sum(A) if m.m00==0 m.m00=eps end m.m10=sum(x.*A) m.m01=sum(y.*A) 计算均值 xmean=m.m10/m.m00 ymean=m.m01/m.m00 计算中心矩(li9_23.rar cm.cm00=m.m00 cm.cm02=(sum((y-ymean).^2.*A))/(m.m00^2) cm.cm03=(sum((y-ymean).^3.*A))/(m.m00^2.5) cm.cm11=(sum((x-xmean).*(y-ymean).*A))/(m.m00^2) cm.cm12=(sum((x-xmean).*(y-ymean).^2.*A))/(m.m00^2.5) cm.cm20=(sum((x-xmean).^2.*A))/(m.m00^2) cm.cm21=(sum((x-xmean).^2.*(y-ymean).*A))/(m.m00^2.5) cm.cm30=(sum((x-xmean).^3.*A))/(m.m00^2.5) im(1)=cm.cm20+cm.cm02 im(2)=(cm.cm20-cm.cm02)^2+4*cm.cm11^2 im(3)=(cm.cm30-3*cm.cm12)^2+(3*cm.cm21-cm.cm03)^2 im(4)=(cm.cm30+cm.cm12)^2+(cm.cm21+cm.cm03)^2 im(5)=(cm.cm30-3*cm.cm12)*(cm.cm30+cm.cm12)... *((cm.cm30+cm.cm12)^2-3*(cm.cm21+cm.cm03)^2)... +(3*cm.cm21-cm.cm03)*(cm.cm21+cm.cm03)... +(3*(cm.cm30+cm.cm12)^2-(cm.cm21+cm.cm03)^2) im(6)=(cm.cm20-cm.cm02)*((cm.cm30+cm.cm12)^2-(cm.cm21+cm.cm03)^2)... +4*cm.cm11*(cm.cm30+cm.cm12)*(cm.cm21+cm.cm03) im(7)=(3*cm.cm21-cm.cm03)*(cm.cm30+cm.cm12)... *((cm.cm30+cm.cm12)^2-3*(cm.cm21+cm.cm03)^2)... +(3*cm.cm12-cm.cm30)*(cm.cm21+cm.cm03)... *()
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