For loop inside for loop

Hi, I have the following function that I am using with a single rho(1x654545) -attached. The result gives me the single peaks(1x1023). I would like to modify the script so that I can use as input rho(4x654545) and end up with the result peaks(4x1023).

Can I get some help? Thanks

x = diff(sign(rho)); 
indx = find(x>0); 
for jj=2:length(indx);
   tmp = rho(indx(jj-1):(indx(jj))); 
   test = dt*(indx(jj)-indx(jj-1));
   peaks(jj) = max(tmp);
end

[KL]

Yes, read this .

[Adam]

There doesn't seem to be any guarantee that you will have 1023 values > 0 for each of your 4 rho arrays.

[Isma_gp]

Yes, that is true. The number of values >0 could vary.

[Adam]

In that case you'd have to use a cell array or a 2d numeric array with enough rows for the case with most peaks and leave NaNs in the extra rows for the other columns.

Answer 1

[Arjun]

Hi @Isma_gp,

I see that currently you are processing a row vector to extract a row vector containing peaks and want to extend this functionality for a matrix.

Since each row of the matrix can have different number of peaks and hence returning a matrix of peaks will not be suitable. You can return a “cell array” instead which will contain same number of rows as in the input matrix and a single column containing a vector of peaks.

Kindly refer to the code below for better understanding:

num_rows = size(rho, 1); 
% Initialize a cell array 
peaks = cell(num_rows, 1);  
for i = 1:num_rows 
    x = diff(sign(rho(i, :)));  
    indx = find(x > 0); 
    % Temporary array to collect peaks for the current row 
    row_peaks = [];  
    for jj = 2:length(indx) 
        tmp = rho(i, indx(jj-1):indx(jj)); 
        % Append the peak to the temporary array 
        row_peaks(end+1) = max(tmp);  
    end 
    % Store the peaks for the current row in the cell array 
    peaks{i} = row_peaks;  
end 

Kindly refer to the attached documentation link understand more about the “cell array”: https://www.mathworks.com/help/releases/R2021a/matlab/ref/cell.html

I hope this will help!