Find peak values and locations in two matrices containing zeros

Dear All Community Members,

I have two matrices, 'A' and 'B'. Matrix 'A' represents the time history of velocities, while matrix 'B' represents the corresponding times. I want to extract all the peak velocity values from matrix 'A' that are above a threshold value of 3.

Matrix A is as follows:

A=[ 1 3 -1 1 5 -1 0;1 6 3 -2 0 0 0;2 3 9 -2 1 11 -1;4 1 -2 8 5 0 0;2 -2 6 -1 0 0 0;-1 13 -2 0 0 0 0];

Matrix B is as follows:

B=[1 1.5 2.5 3 3.5 4 0;1.5 3 5 6 0 0 0;0.5 2 2.5 2.75 3 4 5;2 4 4.25 4.5 6 0 0;1 2 3 4 0 0 0;0.5 3 4 0 0 0 0];

Specifically, I am looking to extract the value 5 from the first row, the value 6 from the second row, the values 9 and 11 from the third row, and then values 8,6,13 from the from the fourth and to the last row in matrix A, as well as their locations.

I.e. Peak values: 5,6,9,11,8,6,13 at locations: 3.5, 3, 2.5, 4, 4.5, 3, 3.

I have tried the findpeaks function, but I am getting errors because of the ending zeros in matrix B.

Stephen23, gave me a brilliant tips in order to find the peaks using cells.

F = @(a)findpeaks(a,'Threshold',3);

C = cellfun(F,num2cell(A,2),'uni',0)

Is there a way I can do this without using cells? For example, using matrices and extending the rows with zeros.

I.e. Peak values, C=[5 0;6 0;9 11;8 0;6 0;13 0];

Locations, D=[3.5 0;3 0; 2.5 4;4.5 0; 3 0;3 0];

Thank you in advance.

Answer 1

[Hassaan]

A = [1 3 -1 1 5 -1 0; 1 6 3 -2 0 0 0; 2 3 9 -2 1 11 -1; 4 1 -2 8 5 0 0; 2 -2 6 -1 0 0 0; -1 13 -2 0 0 0 0];
B = [1 1.5 2.5 3 3.5 4 0; 1.5 3 5 6 0 0 0; 0.5 2 2.5 2.75 3 4 5; 2 4 4.25 4.5 6 0 0; 1 2 3 4 0 0 0; 0.5 3 4 0 0 0 0];

threshold = 3;
C = []; % Matrix for peak values
D = []; % Matrix for locations

for i = 1:size(A, 1)
    peaks = []; % Temp array for peaks in current row
    times = []; % Temp array for times in current row

    for j = 1:size(A, 2)
        if A(i, j) > threshold && (j == 1 || A(i, j) > A(i, j-1)) && (j == size(A, 2) || A(i, j) > A(i, j+1))
            peaks = [peaks, A(i, j)];
            times = [times, B(i, j)];
        end
    end

    % Padding with zeros to match the longest row
    maxLen = max(size(C, 2), length(peaks));
    C(i, 1:maxLen) = [peaks, zeros(1, maxLen - length(peaks))];
    D(i, 1:maxLen) = [times, zeros(1, maxLen - length(times))];
end

% Display the results
disp('Peak Values:');

Peak Values:

disp(C);

     5     0
     6     0
     9    11
     4     8
     6     0
    13     0

disp('Locations:');

Locations:

disp(D);

    3.5000         0
    3.0000         0
    2.5000    4.0000
    2.0000    4.5000
    3.0000         0
    3.0000         0

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