Hi, I need some advice/help with cell accessing and operation in a for loop.

So I have a "100x20001" matrix variable " i_w " which are the data points of 100 sine waves i.e. each row can map a sine wave. I have found the locations of the peaks (3 peaks would just be enough) and store them inorder to find the distance between 2 consecuitive peaks. i have used a cell in a for loop as below:

i=cell(100,2);
for n=1:100
    % I have use findpeaks here to get the respective 
    % peaks and its respective location
    [i{n,:},LOCS]=findpeaks(i_w(n,:),'NPeaks',3);  
end

now this gave me a cell array such as

the locations of the first 3 peaks are stored in the second column.

now all i need to do is to access and find the difference between the 3rd location and 2nd location of each row until the 100th row and store them all.

eg:

751-418 = 333 and so

i have tried this, but seem to get an error which i cannot seem to understand:

dist=zeros(100,1);
for n=1:100
    dist{n,:}=i{n,2}(3)-i{n,2}(2)
end


Index exceeds matrix dimensions.

Error in matrices (line 113)
    dist{n,:}=i{n,2}(3)-i{n,2}(2)

so I know I have made a mistake but it would be a great help to hear some advice. Thanks in advance.

I am sorry the file "i_w" is too big to attach i will present the code instead:

   
t=0:0.001:20; % time
w=[0 3 6 9 12 15 18 21 24 0];    %frequency
a=[4 8 12 16 20 24 28 32 36 40];  % Amplitude

val=[sin(2*pi*w(1)*t);sin(2*pi*w(2)*t);sin(2*pi*w(3)*t);...
     sin(2*pi*w(4)*t);sin(2*pi*w(5)*t);sin(2*pi*w(6)*t);...
     sin(2*pi*w(7)*t);sin(2*pi*w(8)*t);sin(2*pi*w(9)*t);sin(2*pi*w(10)*t)];
 
 out01_10=val*a(1);
 out11_20=val*a(2);
 out21_30=val*a(3);
 out31_40=val*a(4);
 out41_50=val*a(5);
 out51_60=val*a(6);
 out61_70=val*a(7);
 out71_80=val*a(8);
 out81_90=val*a(9);
 out91_100=val*a(10);
 
 i_w=[out01_10;out11_20;out21_30;out31_40;out41_50;...
      out51_60;out61_70;out71_80;out81_90;out91_100];

[Image Analyst]

And how can we run your code if you didn't attach i_w in a .mat file???

[Kirankumar Thangadurai]

sorry file was too long, hence i have present the code instead

Answer 1

[Star Strider]

The error is likely because of the first result being empty.

Try this:

for n=1:100
    % I have use findpeaks here to get the respective 
    % peaks and its respective location
    [i{n,1},LOCS]=findpeaks(i_w(n,:));  
    if ~isempty(LOCS)
        peridx(:,n) = median(diff(LOCS,[],2));          % Use Either ‘median’ Or ‘mean’
    end
end

Here ‘peridx’ is the median (or mean if you prefer) of the differences between the ‘LOCS’ index values for each row.

[Kirankumar Thangadurai]

Thank you. This works and yes the first result was empty and hence the error.

[Star Strider]

As always, my pleasure.

Answer 2

[madhan ravi]

Since dist is a column vector use single index but you used double index which caused the problem :

dist{n}=i{n,1}-i{n,2} %column 1 subtracted by column 2

[Kirankumar Thangadurai]

Unfortunately the error remains the same

[madhan ravi]

see the edited answer

[Image Analyst]

If you'd just attach your data, a smaller, cropped version of it if it's too big, then we can avoid this kind of back and forth and get it right the first time.

[madhan ravi]

exactly I second sir Image Analyst

[Kirankumar Thangadurai]

a shorter version, reducted the sampling time