Error: Assignment has more non-singleton rhs dimensions than non-singleton subscripts

Hello all,

I tried to run a particle tracking model on matlab. When i try to run the code below:

clear I J u v u1 u2 v1 v2
    I = find(xq<x(it,ip),1);
    J = find(yq<y(it,ip),1);
    u1 = MASK(J,I).*U(J,I);
    v1 = MASK(J,I).*V(J,I);

      % advection:
      ddx = u1*1800;%m.DT;
      ddy = v1*1800;%m.DT;

an error message appears saying:

Assignment has more non-singleton rhs dimensions than non-singleton subscripts

Error in PTM_northMenai - Copy (line 141)
    x(it+1,ip) = x(it,ip)+ ddx;

However if i change the sign < for > it works properly!! I would like to know if someone could explain it to me and how could I fix it?

Thank you very much all

[Jonathan Demmer]

   clear I J u v u1 u2 v1 v2
    I = find(xq<x(it,ip),1);
    J = find(yq<y(it,ip),1);
    u1 = MASK(J,I).*U(J,I);
    v1 = MASK(J,I).*V(J,I);
    % advection:
    ddx = u1*1800;%m.DT;
    ddy = v1*1800;%m.DT;

Answer 1

[Walter Roberson]

If the find() does not find anything then it returns empty, which results in empty when used as an index, which results in ddx and ddy being empty. When you add the empty ddx to a value you get empty. You then try to store the empty results in a location with size 1, but that doesn't fit.

[Jonathan Demmer]

Thank you very much for your answer. But I paid attention to place the particle in the middle of the grid to make sure find() can find something and the thing i don't understand is that find() can find a value when i use the sign >... Why it does not work when i use the sign <?

[Walter Roberson]

Are xq and yq vectors?

It is difficult to analyze without your code and data.

[Walter Roberson]

Instead of looping over it and ip, you can use something like

[~, xbin] = histc(x, qx);
[~, ybin] = histc(y, qy);
ind = sub2ind(size(MASK), ybin, xbin);
u1 = MASK(ind) .* U(ind);
v1 = MASK(ind) .* V(ind);
ddx = u1*1800;%m.DT;
ddy = v1*1800;%m.DT;

Watch out for the boundary conditions: x values outside the range xmin to xmax is obvious, but the behaviour at xmax can be a problem. histc() gives the last entry a bin all of its own.

The newer

[~, ~, xbin] = histcounts(x, qx);

would treat the upper bound exactly as part of the previous bin.

[Jonathan Demmer]

Thank you all for your answer!! Thje problem is solved!

Answer 2

[Jonathan Demmer]

I know sorry but the code is really long that is why i did not copy it...

xq and yq are define as: dx = 50; dy = 50; xq = xmin : dx : xmax; yq = ymin : dy : ymax;

Where xmin, xmax, ymin and ymax are UTM coordinates value