Hi, everyone!

I need assistance, actually I have a matrix b=2500x21, which is imaginary part of a FRF signal. I want to find peaks for every column, each column represents accelerometer response at different location of a beam. The following code i have written.

for i=1:21
   pks(:,i)=findpeaks(b(:,i))
end
pks(:,i)=pks;

Problem is that, the matrix pks just stores last iteration value. I want that all the peak values after every iteration will be stored in a single matrix.

Can any body having suggestions?

[madhan ravi]

Just format the code using the code button. So that it’s easy for others to read your code.

[Walter Roberson]

After your for loop, i will be left at its last value, 21. Then your final statement there would be equivalent to

pks(:,21) = pks;

which tries to store the entire pks array into a single location in the pks array. I am having difficulty coming up with any circumstance under which that could work without giving an error message.

I suspect that the for loop will generate an error, as most of the time the different columns of signal will lead to a different number of peaks being found, leading to errors when you tried to store the different vector lengths into the same array.

[Stephen23]

What is the purpose of this?:

pks(:,i)=pks;

[jonas]

I'm surprised you can even run your code. I'd expect a crash unless you find the same number of peaks in each iteration.

Store output in a preallocated cell array instead.

[Ayisha Nayyar]

Actually its working, it generates a matrix of length 5x21, the problem is that matrix contains peak values in 1st and last column, rest of the columns are zero. I want that matrix containing peak values at 21 different locations.......these peak values are appeared in the command window, but i am unable to store them in matrix properly.

[Ayisha Nayyar]

Shortly, i should say that every next iteration value overrides the previous one....

[jonas]

Can you attach the data? No point in guessing. I find it hard to believe no error is returned.

[Stephen23]

Ayisha Nayyar's "Answer" moved here and formatted correctly:

Hi jonas, I have attached the file, now follow the code and see the results please.

for i=1:21
   pks(:,i)=findpeaks(b(:,i))
end
pks(:,i)=pks;

Answer 1

[jonas]

As expected, your code returned the error

Unable to perform assignment because the size of the left side is 3-by-1 and the size of the right side is 6-by-1.

...because of reasons already listed multiple times in the comments.

You can solve this by storing the data in a cell array.

T=readtable('b.xlsx');
b=table2array(T);
pks=cell(21,1);
for i=1:21
   pks{i}=findpeaks(b(:,i))
end

pks =

21×1 cell array

    {6×1 double}
    {6×1 double}
    {6×1 double}
    {6×1 double}
    {7×1 double}
    {6×1 double}
    {5×1 double}
    {7×1 double}
    ...

[Ayisha Nayyar]

Thanks Jonas, i am partially satisfied by your answer, but as i know that for every column of my matrix 'b', I could have at least five peaks. The values arranged in this format is providing only one peak value of each column of 'b'.

[jonas]

No. You can access the peak values in the first column by:

pks{1}

ans =

   1.0e-04 *

    0.2180
    0.0216
    0.0040
    0.0008
    0.0001
   -0.0000

As you can see, there are 6 values. Some peak values have extremely small prominence. You can adjust the 'MinPeakProminence' argument of findpeaks if you wish to exclude those.

Further reading on accessing data in cell arrays ( link ).

[Ayisha Nayyar]

thanks again