Hi, everyone!

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Ayisha Nayyar
Ayisha Nayyar 2018-8-31
评论： Ayisha Nayyar 2018-8-31
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?
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jonas
jonas 2018-8-31
编辑：jonas 2018-8-31
Can you attach the data? No point in guessing. I find it hard to believe no error is returned.
Stephen23
Stephen23 2018-8-31
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;

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jonas
jonas 2018-8-31
编辑：jonas 2018-8-31
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}
...
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jonas
jonas 2018-8-31
编辑：jonas 2018-8-31
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
Ayisha Nayyar 2018-8-31
thanks again

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