# How to calculate moment Features for character recognition?

조회 수: 5(최근 30일)
mona 2017년 3월 3일
I implement this code to calculate 7 invarient moments but it always display I_1 with value and I_2:I_7 with zeros this is correct or not, please help me? 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]; end

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