for loop

Hi

I have matrix M(3,3)

I need to solve this equation

for i=1:3

    for j=1:3

        for k=1:3

            bi,j,k=(1/2)*(diff(M(k,j),theta1)+diff(M(k,i),theta1)-diff(M(i,j),theta1))
        end
    end
end

I need to have the answers in form

b111

b112

b121

....

b333

how i can do that thx

Answer 1

[Walter Roberson]

http://matlab.wikia.com/wiki/FAQ#How_can_I_create_variables_A1.2C_A2.2C....2CA10_in_a_loop.3F

Answer 2

[Andrei Bobrov]

eg

syms x
M = [x^4,2,x^(4/5)-4*x;x,2*x^3,log(x);exp(3*x),x,3*x];

solution

dM = diff(M,x);
idx = fliplr(fullfact(size(M,1)*ones(3,1)));
id = cellfun(@(x)sub2ind(size(M),idx(:,x(1)),idx(:,x(2))),{[3 2],[3 1],[1 2]},'un',0);
b = 1/2*(dM(id{1}) + dM(id{2}) - dM(id{3}));

ADD on Rami comment

eg

syms x y z
M = [x^4*z-y,z*2,x*z^(4/5)-4*x*y;x,2*z*y*x^3,log(x)/y;z*exp(3*x)-y,z*x*y^2,z*(y-z^3)*3*x];
theta = [x y z];

solution

dM = arrayfun(@(ii)diff(M,theta(ii)),1:numel(theta),'un',0);
dM = cat(3,dM{:});
[k j1 i1] = ndgrid(1:numel(theta));
id = {i1(:) j1(:) k(:)};
s = [3 2 1;3 1 2;1 2 3];
idx = cell2mat(arrayfun(@(x)sub2ind(size(dM),id{s(x,:)}),1:3,'un',0));
b = dM(idx)*[1;1;-1]/2

OR variant with loop for..end

for i1=1:3 
    for j1=1:3 
        for k=1:3 
            b1(i1,j1,k) = (1/2)*(diff(M(k,j1),theta(i1))...
                +diff(M(k,i1),theta(j1))-diff(M(i1,j1),theta(k))); 
        end 
    end 
end
b_ijk = reshape(permute(b1,[3 2 1]),[],1)

[rami]

Thx for your answer:

In the solution I can not know

b111

b112

b121

b122

...etc

and in my example I want to make some modify that there are 3 variables theta1, theta2, theta3

for i=1:3

for j=1:3

for k=1:3

bijk=(1/2)*(diff(M(k,j),theta(i))+diff(M(k,i),theta(j))-diff(M(i,j),theta(k)))

end

end

end

So i want to find the values

b111

b112

........

Answer 3

[Image Analyst]

Change "bi,j,k" to "b(i,j,k)"

[rami]

Thx

I try but wasn't useful

[Image Analyst]

Well like Andrei and I both suggested, you should use a 3D matrix and not individually named variables. In fact the answer you accepted, Walter's, also recommends using an indexed matrix and recommends against using what you say you wanted. So we're left confused. But whatever.....as long as you got something you like, even though it's not recommended.

Answer 4

[rami]

thx The link was so useful and the answer will be for one variable

for i=1:3 for j=1:3 for k=1:3

         eval(sprintf('b%d%d%d= (1/2)*(diff(M(k,j),theta1)+diff(M(k,i),theta1)-diff(M(i,j),theta1))', i,j,k));

        end
    end
end