How do you store solutions to a loop in an array?

for r = 10:10:150
      for t =.1:.1:2
          for h = 45:.1:50
              v = pi*(r^2)*h
              hoop_max = (gamma_molasses*(h*12)*(r*12))/(t)
                  if v >= volume & hoop_max <= allow_stress;
                      x= [r,t,j]
                  else
                  end
          end
      end
  end

This is a basic question, but I can't figure out how to store the solutions to this in the array x, such that each solution takes up one row.

Answer 1

[Rik]

%replace
x=[r,t,j];
%with
k=k+1;x(k,:)=[r,t,j];

Set k to 0 before the loops and pre-allocate at least one row of x, but preferably as close to the number of solutions as you can.

A much better idea would be to use meshgrid to generate a 3D matrix for each loop value, which will enable you to calculate all of this at once and remove all loops. If you get findND, you can use the code below.

%just some random numbers:
gamma_molasses=1;volume=50000;allow_stress=10^5; 

[r,t,h]=meshgrid(10:10:150,.1:.1:2,45:.1:50);
v=pi*(r.^2).*h;
hoop_max = (gamma_molasses.*(h*12).*(r*12))./(t);
[r2,t2,h2]=findND(v >= volume & hoop_max <= allow_stress);
x=[r2,t2,h2];

[Ivan Arkhipov]

How do I add findND?

[Rik]

You can download it from the File Exchange with the link in my post ( here it is again). To use it, just make sure the m-file is in the current folder, or in another folder on the Matlab path.

Answer 2

[Star Strider]

In MATLAB, subscripts can only be integers greater than 0, or logical variables.

I do not follow what you are doing. This will at least provide the correct indices:

r = 10:10:150;
t =.1:.1:2;
h = 45:.1:50;
for ri = 1:length(r)
    for ti = 1:length(t)
        for hi = 1:length(h)
            v = pi*(r(ri)^2)*h(hi);
            hoop_max = (gamma_molasses*(h(hi)*12)*(r(ri)*12))/(t(ti));
            if v >= volume & hoop_max <= allow_stress;
                x(ri, ti, j) = [r(ri),t(ti),j]
            else
            end
        end
    end
end

NOTE — Because I do not know what you are calculating, this is UNTESTED CODE. The indexing should work.

Also, you did not tell us what ‘j’ is.

[Ivan Arkhipov]

j is meant to be h, I messed up the code. I am trying to find all of the possible solutions to the loop (where v>= volume and hoop max is <= allow_stress) with three values (radius,thickness,height), (r,t,h).

[Star Strider]

In that event:

x(ri, ti, hi) = [r(ri),t(ti),h(hi)];

and:

hoop_max(ri, ti, hi) = (gamma_molasses*(h(hi)*12)*(r(ri)*12))/(t(ti));

I still do not follow what you are doing. The important aspect appears to be to get the array referencing correct.

Rearrange the indices in the order you want them to appear in your matrices.

[Ivan Arkhipov]

I am getting this error "Assignment has more non-singleton rhs dimensions than non-singleton subscripts" After the

x(ri, ti, hi) = [r(ri),t(ti),h(hi)];

[Rik]

If gamma_molasses is indeed a scalar, why not avoid the loops and use my solution?

[Star Strider]

Add a dimension:

x(ri, ti, hi, :) = [r(ri),t(ti),h(hi)];

That should eliminate the error, at the expense of creating a matrix with an additional dimension.

[Ivan Arkhipov]

This still gives me the same error.

[Rik]

For this syntax to work, you will need to pre-allocate x. (this should do the trick: x = zeros(1,1,1,3);)

Also note that this will give you a different result from my solution.