Findout the peaks in each row of matrix with specific conditions

i would lime to find out the no of peaks in each row with a specific condtion (that is ) peakvalues with (peakvalue-minvalueof each row>1000)

Now i take a single row for easiness.In reality i have a big matrix.

A= [1000 8000 1000 1800 1200 4000 4800 4000 1000 1000]

the min-value of A=1000 The peak values of A= [8000 1800 5000], But the problem here is if i use ordinary threshold limit& findpeak

findpeaks(A,'Threshold',1000)
it will only compare with the values near to it.

I want to find out the total number of peaks which obey my condition (i.e>1000) in detailed way

8000-1000 >1000 (true) 1800-1000<1000(false) 5000-1000>1000(True)

so the aswer i expect peakswithcondition=2

Answer 1

[Star Strider]

Does this do what you want?

A= [1000 8000 1000 1800 1200 4000 4800 4000 1000 1000];
[pks,loc] = findpeaks(A-1000);
pksvct = (pks>1000);
pks = pks(pksvct) + 1000
loc = loc(pksvct)

It first subtracts 1000 from ‘A’, finds the peaks, finds the peaks the meet your criterion, then reports the actual value of the ‘acceptable’ peaks and their locations.

[navan]

Dear Strider, It works well for a single row. how should i modify the program for a m*n matrix? (i would like to get the peaks with specific conditions and no of such peaks in each row of matrix)

[Star Strider]

Since findpeaks takes only vector arguments, you will need to use a for loop to loop through each row. Modify the approach I discussed in my Answer to your other Question: ‘How to use findpeaks for a matrix of size A’ to include the current code:

A = randi(5000, 4, 8);                      % Create Data
for k1 = 1:size(A,1)
    [pks,loc] = findpeaks(A(k1,:)-1000);
    pksvct = (pks>1000);
    pks = pks(pksvct) + 1000;
    loc = loc(pksvct);
    P{k1} = [pks; loc];
end

[navan]

Dear strider, thanks alot. Its a great help again. It works well. These is one more issue. we assumed the minimum value as 1000 before. But now we have 4 different minimum value [1000, 700,900,500], I have tried to apply min(transpose(A)) in the equation insted of 1000. But i could not get my result as before. waiting for your reply and thanks for your great concern.

 A= [1000 8000 1000 1800 1200 4000 4800 4000 1000 1000;
   1000 1200 8000 1800 1200 4000 4800 4000 800 700;
   1000 6000 900 2800 1200 3000 5800 4000 1200 1000;
   1000 1000 1000 1800 1200 1000 1800 1000 500 1000]
a1=min(transpose(A));
%A = randi(5000, 4, 8);                     
for k1 = 1:size(A,1)
   xx=a1(k1,:);
yy=A(k1,:);
    [pks,loc] = findpeaks(A(k1,:)-a1(:,k1));
    pksvct = (pks>1000);
    pks = pks(pksvct) + 1000;
    loc = loc(pksvct);
    P{k1} = [pks; loc];
end
for k1 = 1:size(P,2)
    NrPks(k1) = size(P{k1},2);
end

[Star Strider]

My pleasure.

You have to add back the minimum in this line:

pks = pks(pksvct) + a1(:,k1);

That should work.

You can also use:

a1 = min(A,[],2);

but to use it, you would have to change the references to it to: a1(k1).

I would also delete the ‘xx’ and ‘yy’ assignments, since you’re not doing anything with them.

[navan]

Dear strider great. It resolved my issue thanks a lot again.

[Star Strider]

My pleasure!

Happy we got it sorted.