小波矩特征提取matlab代码

这是我上研究生时写的小波矩特征提取代码：



%新归一化方法小波矩特征提取----------------------------------------------------------

F=imread('a1.bmp');
F=im2bw(F);
F=imresize(F,[128 128]);
%求取最上点
for i=1:128
for j=1:128
if (F(i,j)==1)
ytop=i;
xtop=j;
break;
end
end
if(F(i,j)==1)
break;
end
end
%求取最下点
for i=1:128
for j=1:128
if (F(129-i,j)==1)
ybottom=129-i;
xbottom=j;
break;
end
end
if(F(129-i,j)==1)
break;
end
end
%求取最左点
for i=1:128
for j=1:128
if (F(j,i)==1)
yleft=j;
xleft=i;
break;
end
end
if(F(j,i)==1)
break;
end
end
%求取最右点
for i=1:128
for j=1:128
if (F(j,129-i)==1)
yright=j;
xright=129-i;
break;
end
end
if(F(j,129-i)==1)
break;
end
end
%求取中心点
x0=(xright-xleft)/2+xleft;
y0=(ybottom-ytop)/2+ytop;
x0=round(x0);
y0=round(y0);
%图像平移
F=double(F);
[M,N]=size(F);
F1=zeros(M,N);
M0=M/2;
N0=N/2;
for i=1:M
for j=1:N
if F(i,j)==1
F1(i+M0-y0,j+N0-x0)=1;
end
end
end
%figure,imshow(F1);
%图像缩放
max=0;
for i=1:128
for j=1:128
if(F(i,j)==1)
d=sqrt((i-y0)^2+(j-x0)^2);
if(max<d)
max=d;
end
end
end
end
%max=round(max);
a=200.0/(max*2);
F2=imresize(F1,a);
%figure,imshow(F2);
%将所有图像均复制到500*500的图像的大小
[M,N]=size(F2);
m0=M/2;
m0=round(m0);
n0=N/2;
n0=round(n0);
f3=zeros(500,500);
y1=round((500-M)/2);
x1=round((500-N)/2);
for i=1:M
for j=1:N
if(F2(i,j)==1)
f3(y1+i,x1+j)=1;
end
end
end
%figure,imshow(f3);
%图像从笛卡儿坐标转换为极坐标------------------------------------------------------------
%角度间隔为2*pi/(128*128),128个像素长设为图像的单位圆半径
%f1矩阵里面放着对应极半径和角度的值
f1=zeros(128,16384);
%直角坐标与极坐标建立起一一对应的关系
for i=1:128
for j=1:16384
a=j*2*pi/16384.0;
r=i;
y=round(r*sin(a));
x=round(r*cos(a));
if (f3(250+x,250+y)==1)
f1(i,j)=1;
end
end
end
F3=zeros(512,512);
%将极坐标转换后的图像显示出来
for i=1:128
for j=1:16384
if f1(i,j)==1
a=j*2*pi/16384.0;
x=round(i*cos(a));
y=round(i*sin(a));
F3(256+x,256+y)=1;
end
end
end
F3(256,256)=1;
%figure,imshow(F3);
%小波矩特征提取-------------------------------------------------------------------
%进行角度积分得到Sq(r)------------------
N=16384;
Sq=zeros(128,4);
for r=1:128
for q=1:4
for m=1:N
Sq(r,q)=(f1(r,m)*exp(-j*2.0*pi*(m-1)*(q-1)/N))+Sq(r,q);
end
Sq(r,q)=1.0/N*Sq(r,q);
end
end
%小波矩特征提取-------------------------
x=3;
a=0.697066;
f0=0.409177;
w=sqrt(0.561145);
F1=zeros(3,9,4);
tt=4*a^(x+1)/sqrt(2*pi*(x+1))*w;
for q=1:4
for m=1:3
rr=2^m+1;
for n=1:rr
for r=1:128
%pp=2*(2^(m-1)*(r-1)/128.0-(n-1))-1;
pp=2*(2^(m-1)*r/128.0-(n-1))-1;
cc=cos(2*pi*f0*pp)*exp(-1.0*pp^2/(2*(w^2)*(x+1)));
fan=2^((m-1)/2)*tt*cc;
%Fmnq(m,n,q)=abs(Sq(r,q)*fan*(r-1)/128)+Fmnq(m,n,q);
F1(m,n,q)=abs(Sq(r,q)*fan*r)+F1(m,n,q);
end
end
end
end