0 个投票

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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madhan ravi
madhan ravi 2018-8-31
编辑：madhan ravi 2018-8-31
Just format the code using the code button. So that it’s easy for others to read your code.
Walter Roberson
Walter Roberson 2018-8-31
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
Stephen23 2018-8-31
What is the purpose of this?:
pks(:,i)=pks;
jonas
jonas 2018-8-31
编辑：jonas 2018-8-31
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
Ayisha Nayyar 2018-8-31
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
Ayisha Nayyar 2018-8-31
Shortly, i should say that every next iteration value overrides the previous one....
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

0 个投票

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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Ayisha Nayyar
Ayisha Nayyar 2018-8-31
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
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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