Loop output not saving to a cell array. Only last solution saving to cell

This question has been asked and I have read the accepted answers. But I just don't get it. Its been several hours working on this now and I'm hoping for some very clear help.

I have a loop with only 6 iterations. Within that loop is a function. Each time the loop runs, the function solves and a solution array of size (20000, 1) is generated. I want to save each solution array for all 6 iterations. But my attempts have resulted in only the last iteration's solution being stored for all cnew{1,1}, cnew{2,1} ... cnew{6,1}. They're all equal in the cnew cell, but I know with 100% certainty that each solution array z is not equal with changing parameter values.

Please help

Below is my code:

for iter=1:6
if iter==1, y=0; end       
if iter==2, y=0.15; end
if iter==3, y=0.30; end
if iter==4, y=0.45; end
if iter==5, y=0.60; end
if iter==6, y=0.75; end

      z = function( _various inputs that change with y_);

       cnew = cell(6,1);
       for loopiter = 1:6
         cnew{iter}=z;
       end

end

Answer 1

[Stephen23]

You only need one loop:

vec = 0:0.15:0.75;
cnew = cell(numel(vec),1);
for iter = 1:numel(vec)
    y = vec(iter);
    cnew{iter} = function( _various inputs that change with y_);
end

[Stephen23]

The fix to your code is to get rid of the inner loop:

cnew = cell(6,1);
for iter=1:6
    if iter==1, y=0; end       
    if iter==2, y=0.15; end
    if iter==3, y=0.30; end
    if iter==4, y=0.45; end
    if iter==5, y=0.60; end
    if iter==6, y=0.75; end
    cnew{iter} = function( _various inputs that change with y_);
end

[EM]

Unfortunately I tried that and z{iter} is empty[] except for z{6,1}. Which is odd, because I put a breakpoint in and noticed that after one iteration, z{1,1} was not empty.

[Stephen23]

EM's "Answer" moved here:

Hi Stephen, thank you for your answer but I fear it is too complicated and I'm not sure what exactly is wrong with my code, that is causing only the last solution array z, to be saved.

And you tossed a vec(k) in there. What is k?

[Stephen23]

@EM: See my comment where I fixed your code.

"I fear it is too complicated"

My answer uses only things that you already use: vectors and indexing, plus the colon operator to define a vector of all y values. These are all very basic MATLAB concepts that are covered in the introductory tutorials:

https://www.mathworks.com/help/matlab/getting-started-with-matlab.html

Compare your code carefully with the code I gave in my answer: you will see that my answer does much the same thing, just without using a hard-coded number of loop iterations, making it simpler and more flexible at the same time.

"And you tossed a vec(k) in there. What is k?"

My mistake: I often use k for the loop iterator. I changed this to iter.

"Unfortunately I tried that and z{iter} is empty[] except for z{6,1}. Which is odd, because I put a breakpoint in and noticed that after one iteration, z{1,1} was not empty"

What you tested with a breakpoint was not my code, it was the code you showed in your question.

[EM]

Hi Stephen, yes I understand that vec(k) is meant to represent the various values of y. But that does not seem to work.

I just get the error

y = vec(k);

Undefined function or variable 'k'.

[Walter Roberson]

change vec(k) to vec(iter)

[EM]

Thanks Walter, I appreciate that.

[EM]

@Stephen

Unfortunately I tried that and z{iter} is empty[] except for z{6,1}. Which is odd, because I put a breakpoint in and noticed that after one iteration, z{1,1} was not empty"

What you tested with a breakpoint was not my code, it was the code you showed in your question.

That is correct. I tested with a breakpoint on my code. Unfortunately there is not really a way to share with you that I know of. This is my great frustration. Why would only the last solution be saved after all iterations have completed? Perhaps if you know of an explanation, I can troubleshoot.

[Stephen23]

"Perhaps if you know of an explanation, I can troubleshoot."

I wrote that in the very first sentence of my answer. Read my answer again.

Question: how many loops have I used in my code?

Question: how many loops have you used in your code?

[EM]

I only need one loop? Is that your solution?

[Stephen23]

"I only need one loop?"

Yes. Which is why I wrote that in my answer one hour ago.

[EM]

So I just tried the following and still only the last iteration is saved.

vec=0:0.15:0.75;
z = cell(numel(vec),1);
for iter=1:numel(vec)

y   = vec(iter);

      [z{iter}] = function( _various inputs that change with y_)

      cnew=z{iter};
      disp(iter)
  end

[Stephen23]

This should work:

vec = 0:0.15:0.75;
out = cell(numel(vec),1);
for iter = 1:numel(vec)
    y = vec(iter);
    out{iter} = function( _various inputs that change with y_)
end

If this does not work then please upload the exact code that you are using by clicking the paperclip button. It would also be helpful to upload the function too.

[EM]

Stephen that did work. Thank you. I would encourage you to use more explanation when you provide solutions because people come here to LEARN. For example, why does it matter to have only one loop?

[Stephen23]

@EM: The problem with your original code was the two loops:

for iter = 1:6
    if iter==1, y=0; end       
    if iter==2, y=0.15; end
    if iter==3, y=0.30; end
    if iter==4, y=0.45; end
    if iter==5, y=0.60; end
    if iter==6, y=0.75; end
    z = function( _various inputs that change with y_);
    cnew = cell(6,1);
    for loopiter = 1:6
        cnew{iter}=z;
    end
end

Note that the on every iteration of the outper loop you completely redefine cnew, so anything stored in cnew on iterations 1:5 is discarded. That leaves only the final iteration, where y=0.75: on this final iteration you calculate z, redefine cnew again (discarding any existing data), and then use the inner loop to put the same z value into each cell of cnew. Thus cnew only contains the z value from the final outer loop iteration.

Using a breakpoint was a very good idea: step through until the end and you will notice that you redefine cnew on each outer loop iteration, discarding all of the existing data. You will also notice that the inner loop allocates exactly the same z value to all cnew cells.

My answer has one loop, iterating as many times as there are elements in vec. For each element it gets one y value from the vector, calculates the function, and saves the output value in cnew.

[EM]

Great, that makes sense. Thank you for the explanation. Please save this explanation for any future visitors