In an assignment, number of elements in A and B must be same

Hello

I am reproducing the Bayesian Change Detection code by Adams and MacKay. The following is a code for a time series data tem of size T.

   T = 3600;
   lambda        = 200;
   window_size=100;
   hazard_func  = @(r) constant_hazard(r, lambda); % a separate function
   R = zeros([T+1]);
   R(1,1) = 1;
   muT    = mean(tem(window_size:end));
   stdT=std(tem(window_size:end));
   alphaT = 1;
   mins  = zeros([T+1]);
   for t=1:T
   predprobs = studentpdf(tem(t), muT, stdT, 2 * alphaT); % studentpdf is a separate function
   H = hazard_func([1:t]');
   R(2:t+1,t+1) = R(1:t,t) .* predprobs .* (1-H);
   R(1,t+1) = sum( R(1:t,t) .* predprobs .* H );

   R(:,t+1) = R(:,t+1) ./ sum(R(:,t+1));

   muT=[muT; mean(tem(1:t))]
   stdT=[stdT; std(tem(1:t))];
   alphaT  = [alphaT; alphaT + 0.5];

   mins(t) = find(R(:,t)==min(R(:,t)));
   end

I keep getting the error

    In an assignment  A(:) = B, the number of elements in A and B must be the same.
    Error in bayesian_change (line 84) 
    mins(t) = find(R(:,t)==min(R(:,t)));

I've checked multiple times and the dimensions of both mins and R are the same. I know it is tough to understand the code without prior knowledge, but can anyone help me figure out why I keep getting this error even though the dimensions are the same?

Thanks!

Answer 1

[Star Strider]

This line:

mins(t) = find(R(:,t)==min(R(:,t)));

will return all indices equal to ‘min(R(:,t))’. Apparently, there are more than one. If you want to return all of them, the easiest way is to do that is to use a cell array for ‘mins’:

mins{t} = find(R(:,t)==min(R(:,t)));

If you only want the index of the first minimum in the column, you can also just use the min function:

[~,mins(t)] = min(R(:,t));

NOTE — This is UNTESTED CODE but should work.

[Mridul Garg]

Thank you, that worked!

[Star Strider]

My pleasure!