How to save/store/recuparate data from inside a nested loop.

So basically a have a nested loop in the following form:

... %Stuff up here
for X=1:N_aoa 
initial=[0;aoa(X);0;0];
Y=[initial,zeros(4,N)];

    for n=1:N
    ... %Stuff happens here and I calculate several values "Y"
    end

[val,idx]=max(Y, [], 2);
Mat=[aoa(X),val(1),val(2)]
end

Mat %This will only print the last stored value of Mat, which is when X=N_aoa
%Since i'll have lots of values my idea was to do again max but for Mat.

As you can se the problem is that I make 2 evaluations, the inner loop evaluates for a range of 1 to N, while the outer loop is for different values of the intial vector state "initial", basically im varying the second value of [0;aoa(X);0;0], calculating my state in the inner loop and finally I want to calculate the maximum value corresponding to that initial state aoa(X) for the entire range of 1:N, however I dont know how to store since it will be erased once X changes...

Answer 1

[Sulaymon Eshkabilov]

This can be done:

for X=1:N_aoa 
initial=[0;aoa(X);0;0];
Y=[initial,zeros(4,N)];
    for n=1:N
    ... %Stuff happens here and I calculate several values "Y"
    end
[val,idx]=max(Y, [], 2);
Mat(X,:)=[aoa(X),val(1),val(2)]; % In a Matrix form
end

Or in a cell array form:

for X=1:N_aoa 
initial=[0;aoa(X);0;0];
Y=[initial,zeros(4,N)];
    for n=1:N
    ... %Stuff happens here and I calculate several values "Y"
    end
[val,idx]=max(Y, [], 2);
Mat{X}=[aoa(X),val(1),val(2)]; % In a cell array form
end

Answer 2

[Bjorn Gustavsson]

This should be simple enough if you do something like this:

Y_best = zeros(n_something,N_aoa); % Not clear to me what your Y contains
for X=1:N_aoa 
  initial=[0;aoa(X);0;0];
  Y=[initial,zeros(4,N)];
  best_val = -inf; % threshold-value for best "Y"
  for n=1:N
    ... %Stuff happens here and I calculate several values "Y"
    if you_maximizing_variable_val > best_val
      best_val = you_maximizing_variable_val;
      Y_best(1:numel(Y_current),X) = Y_current; % save away the best values from whatever
    end
  end
  [val,idx]=max(Y, [], 2);
  Mat=[aoa(X),val(1),val(2)]
end

Something like that should solve this type of problem.

HTH