Re: Finding peaks/max in a graph

[Thomas Shores]

Hmmm.  Worked fine for me. And if you want a row/column independent
version, see below.


octave:67> a = [1 2 3 4 5 4 3 2 1 2 3 2 1]';
octave:68> peaks = find([a(2:n,1)-a(1:n-1,1) < 0;1] &
[1;a(1:n-1,1)-a(2:n,1)<0])
peaks =

   5
  11
octave:69> find(diff(a(1:n-1))>0 & diff(a(2:n))<0)+1
ans =

   5
  11



On Mon, 2004-04-05 at 10:22, edA-qa mort-ora-y wrote:
> David Bateman wrote:
> > A for-loop is not the way to go about this. Consider the code fragment
> > n = 9
> > a = [1 2 3 4 5 4 3 2 1]';
> > peaks = find([a(2:n,1) - a(1:n-1,1) < 0; 1] & [1; a(1:n-1,1) - a(2:n,1) < 
> > 0]);
> > that will find all indexes of the peaks in the data in a single statement
> 
> If you set
>    a = [1 2 3 4 5 4 3 2 1 2 3 2 1]';
> Peaks will only find 5, not the 3 as well.  I will see if somehow, 
> however, I can modify my code not to use a for-loop, rather to use a 
> similar notation to above.



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