## how to write matlab code for moments?

### Sachin (view profile)

on 11 Apr 2013
Latest activity Answered by Sparsh Garg

on 15 Jun 2018

### Iman Ansari (view profile)

i have this code for calculating moments function [M]= moments(I) [r c]=size(I); m=zeros(r,c); % geometric moments for i=0:1 for j=0:1 for x=1:r for y=1:c m(i+1,j+1)=m(i+1,j+1)+(x^i*y^j*I(x,y)); end end end end
xb=m(2,1)/m(1,1); yb=m(1,2)/m(1,1);
% central moments u=[ 0 0 0 0;0 0 0 0;0 0 0 0;0 0 0 0]; for i=0:3 for j=0:3 for x=1:r for y=1:c u(i+1,j+1)=u(i+1,j+1)+(x-xb)^i*(y-yb)^j*I(x,y); end end end end
% scale invariant moments n=[ 0 0 0 0;0 0 0 0;0 0 0 0;0 0 0 0]; for i=0:3 for j=0:3 n(i+1,j+1)=u(i+1,j+1)/(u(1,1)^(1+(i+j)/2)); end end
%rotation invariant moments I_1= n(3,1)+ n(1,3); I_2=(n(3,1)- n(1,3) )^2+ (2*n(2,2))^2; I_3=(n(4,1)-3*n(2,3))^2+ (3*n(3,2)-n(1,4))^2; I_4=(n(4,1)+n(2,3))^2+ (n(3,2)+n(1,4))^2; I_5=(n(4,1)-3*n(2,3))*(n(4,1)+n(2,3))*((n(4,1)+n(2,3))^2-3*(n(3,2)+n(1,4))^2)+(3*n(3,2)-n(1,4))*(n(3,2)+n(1,4))*(3*(n(4,1)+n(2,3))^2-(n(3,2)+n(1,4))^2); I_6=(n(3,1)-n(1,3))*((n(4,1)+n(2,3))^2-(n(3,2)+n(1,4))^2)+ 4*n(2,2)*(n(4,1)+n(2,3))*(n(3,2)+n(1,4)); I_7=(3*n(3,2)-n(1,4))*(n(4,1)+n(2,3))*((n(4,1)+n(2,3))^2- 3*(n(3,2)+n(1,4))^2 )- (n(1,4)-3*n(2,3))*(n(3,2)+n(1,4))*(3*(n(4,1)+n(2,3))^2-(n(3,2)+n(1,4))^2);
M= [I_1 I_2 I_3 I_4 I_5 I_6 I_7];
but it gives same values for all images is this correct way to calculate moments?

### Iman Ansari (view profile)

on 12 Apr 2013
Edited by Iman Ansari

### Iman Ansari (view profile)

on 12 Apr 2013

You may need to convert your input image to double. In uint8 numbers greater than 255 became 255.
function [M]= moments(I)
I=double(I);
[r c]=size(I);
m=zeros(r,c);
% geometric moments
for i=0:1
for j=0:1
for x=1:r
for y=1:c
m(i+1,j+1)=m(i+1,j+1)+(x^i*y^j*I(x,y));
end
end
end
end
xb=m(2,1)/m(1,1);
yb=m(1,2)/m(1,1);
% central moments
u=[ 0 0 0 0;0 0 0 0;0 0 0 0;0 0 0 0];
for i=0:3
for j=0:3
for x=1:r
for y=1:c
u(i+1,j+1)=u(i+1,j+1)+(x-xb)^i*(y-yb)^j*I(x,y);
end
end
end
end
% scale invariant moments
n=[ 0 0 0 0;0 0 0 0;0 0 0 0;0 0 0 0];
for i=0:3
for j=0:3
n(i+1,j+1)=u(i+1,j+1)/(u(1,1)^(1+(i+j)/2));
end
end
%rotation invariant moments
I_1= n(3,1)+ n(1,3);
I_2=(n(3,1)- n(1,3) )^2+ (2*n(2,2))^2;
I_3=(n(4,1)-3*n(2,3))^2+ (3*n(3,2)-n(1,4))^2;
I_4=(n(4,1)+n(2,3))^2+ (n(3,2)+n(1,4))^2;
I_5=(n(4,1)-3*n(2,3))*(n(4,1)+n(2,3))*((n(4,1)+n(2,3))^2-3*(n(3,2)+n(1,4))^2)...
+(3*n(3,2)-n(1,4))*(n(3,2)+n(1,4))*(3*(n(4,1)+n(2,3))^2-(n(3,2)+n(1,4))^2);
I_6=(n(3,1)-n(1,3))*((n(4,1)+n(2,3))^2-(n(3,2)+n(1,4))^2)+ 4*n(2,2)*(n(4,1)...
+n(2,3))*(n(3,2)+n(1,4));
I_7=(3*n(3,2)-n(1,4))*(n(4,1)+n(2,3))*((n(4,1)+n(2,3))^2- 3*(n(3,2)+n(1,4))^2 )...
- (n(1,4)-3*n(2,3))*(n(3,2)+n(1,4))*(3*(n(4,1)+n(2,3))^2-(n(3,2)+n(1,4))^2);
M= [I_1 I_2 I_3 I_4 I_5 I_6 I_7];

Sachin

on 12 Apr 2013
thnx a lot iman
fateme

on 28 May 2014
thanks
Ricki Chindra

### Ricki Chindra (view profile)

on 9 Nov 2014
Iman how to call the function? I'm trying it, no error but when I run it, There is no image in the figure

### Akshit Vekariya (view profile)

on 6 May 2015
Edited by Akshit Vekariya

### Akshit Vekariya (view profile)

on 6 May 2015

will this code work on moment invariant...?