Operations between matrix, different calculation according to conditions.

Question

Isai Fernandez 2018-4-17

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编辑： Isai Fernandez 2018-4-17

I want to multiply two matrices, but depends on conditions I will operate with other expression.

If we name matrices "A" and "B", column matrices, I want to multiply them. I have two operations which I will use according to conditions to proceed.

I tried with this code:

d = [64;
  108;
  454;
  498;
  542;
  586];
As = [852;
  568;
  568;
  568;
  568;
  852];
fs = [-344.4991;
 -168.8422;
  420.0000;
  420.0000;
  420.0000;
  420.0000];
c = 150;
if d<c
    Cs(:,1) = (fs-.85*21).*As/1000;
else 
    Cs(:,1) = (fs).*As/1000;
end

This results in a column matrix "Cs".

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Answer 1

Guillaume 2018-4-17

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if d<c will be true if all elements of d are smaller than c, in which case, all elements of Cs are calculated according to your first formula. Otherwise, the if will be false, and all elements of Cs are calculated according to your second formula. Matlab will not loop for you over the element of d to apply your if element by element.

You could write that loop:

for idx = 1:numel(d)
   if d(idx) < c
      Cs(idx, 1) = ...
   else
      Cs(idx, 1) = ...
   end
end

but that's not the way to do it in matlab. This is much better:

Cs = zeros(size(d));  %preallocate Cs
Cs(d<c) = (420-.85*21).*As(d<c)/1000;   %use logical indexing to assign the required elements
Cs(d>=c) = (420).*As(d>=c)/1000;

Or this is shorter but a bit more obscure:

Cs = (d<c) * ((420-.85*21).*As/1000) + (d>=c) * (420.*As/1000);

For more on logical indexing, see the doc

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Isai Fernandez 2018-4-17

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Thanks Guillaume.

I wrote

for idx = 1:numel(d)
   if d(idx,1) < c
      Cs(idx, 1) = (fs-.85*21).*As/1000;
   else
      Cs(idx, 1) = fs.*As/1000;
   end
end

And I have this, despite of the same sizes of matrices:

Subscripted assignment dimension mismatch.

The same for the second code:

Matrix dimensions must agree.

The idea of the third one works well.

Guillaume 2018-4-17

在 MATLAB Online 中打开

You must have edited your question. I'm sure there was no fs originally, just a constant 420.

With the loop you need to index all matrices simultaneously, thus

for idx = 1:numel(d)
   if d(idx) < c
      Cs(idx) = (fs(idx)-.85*21).*As(idx)/1000;
   else
      Cs(idx) = fs.*As(idx)/1000;
   end
end

With the second code, similarly, you've got to apply the logical indexing to all vectors, so

Cs = zeros(size(d));  %preallocate Cs
Cs(d<c) = (fs(d<c)-.85*21).*As(d<c)/1000;   %use logical indexing to assign the required elements
Cs(d>=c) = fs(d<c).*As(d>=c)/1000;

Or you can use James' answer that applies the filtering to the only part that is not common between the two expression. It is actually simpler.

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