please check it ..this is another eqn<http://latex.codecogs.com/gif.latex?P1v(i,j)=\frac{{\sum_{u=1}^{^{S_{u}-1}}\sum_{v=1}^{S_{v}-1}\delta%20({F_{h}(u,v)=i,{F_{h}(u,v+1)=j}})}}{\sum_{u=1}^{{S_{u}-1}}\sum_{v=1}^{S_{v}-1}\delta%20(F_{h}(u,v)=i)}>..... i wrote code for this eqn is p1v=zeros(9); for ii=1:9 p1=Yh(1:end-1,:)==ii; for jj=1:9 p2=Yh(:,2:end-1)==jj; p1v(ii,jj) = nnz(p1 & p2)./nnz(p1); end end ...is it correct ?
this code is not working for transition probability matrix ....make it correct?
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my transition probability matrix equation is given under link. http://latex.codecogs.com/gif.latex?p1_{h}(i,j)=\frac{\sum_{u=1}^{S_{u}-2}\sum_{v=1}^{S_{v}}\delta%20(F_{h}(u,v)=i,{F_{h}}(u+1,v)=j)}{\sum_{u=1}^{^{S_{u}-2}}%20\sum_{v=1}^{^{S_{v}}}\delta%20({F_{h}(u,v)=i})}
for this eqn i wrote code given below
sv=128;
su=128;
for i=1:9
for j=1:9
a(i,j)=0;
for u=1:su-2
for v=1:sv
if (Yh(u,v)==i && Yh(u+1,v)==j)
dell=1;
else
dell=0;
end
a(i,j)=a(i,j)+dell;
b(i,j)=0;
if(Yh(u,v)==i)
dell1=1;
else
dell1=0;
end
b(i,j)=b(i,j)+dell1;
end
end
end
end
.....the final result want to know for above code is p=a./b.... and here the dell value as considered as dell(.)=1 if and only if its argumens are satisfied, otherwise dell(.)=0.
2 Commenti
Jan
il 17 Gen 2013
What does "is not working" mean? Do you get an error message or do the results differ from your expectations?
Risposte (2)
Walter Roberson
il 11 Gen 2013
Why are you setting b(i,j)=0 inside each "for v" iteration ?
0 Commenti
Andrei Bobrov
il 11 Gen 2013
Modificato: Andrei Bobrov
il 11 Gen 2013
p1h = zeros(9);
for ii = 1:9
p1 = Yh(1:end-2,:) == ii;
for jj = 1:9
p2 = Yh(2:end-1,:) == jj;
p1h(ii,jj) = nnz(p1 & p2)./nnz(p1);
end
end
2 Commenti
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