Azzera filtri
Azzera filtri

trying to find peaks on a time series

7 visualizzazioni (ultimi 30 giorni)
DuckDuck
DuckDuck il 11 Giu 2012
i'm trying to find peaks on some time series data but when i write
[pks,locs]=findpeaks(x)
i get the position of peak on locs. but i want to get the locs the same length i have x files or better if i have locs represented as 0 and 1 i can get the x values without a hassle. is there an elegant and simple way of dooing it, like the find function??
i also am trying this but obviously it takes time because of the loops, is there any simple way of dooing this in matlab.
newx=zeros(length(xn),1);
for p=1:length(xn);
for m=1:length(locs);
if locs(m)==p
newx(p)=1;
else
end;
end;
end;
end;
  2 Commenti
Ryan
Ryan il 11 Giu 2012
I am confused, if you have the locations of the peaks already, what is the issue? What type of location information is stored in locs?
DuckDuck
DuckDuck il 11 Giu 2012
look at my code, i want to know if there is a better way of writing that code, because loops take a long time in matlab

Accedi per commentare.

Risposta accettata

Wayne King
Wayne King il 11 Giu 2012
I'm assuming that xn in your loop above is the signal. You don't show us where xn comes from and it's not used in your call to findpeaks(). In your call to findpeaks, you use x. So If that is the case, then you don't need a loop.
newx = zeros(size(x));
newx(locs) = 1;
Does the same thing as your loop.
  1 Commento
DuckDuck
DuckDuck il 11 Giu 2012
yeah you are right, basically xn is s x.
no it does not, i tried and it did not.

Accedi per commentare.

Più risposte (1)

Wayne King
Wayne King il 12 Giu 2012
Then please compare:
x = [2 12 4 6 9 4 3 1 19 7];
[pks,locs] = findpeaks(x);
% here is your code
newx=zeros(length(x),1);
for p=1:length(x);
for m=1:length(locs);
if locs(m)==p
newx(p)=1;
else
end
end
end
Now what I answered:
newx2 = zeros(length(x),1);
newx2(locs) = 1;
Now
isequal(newx,newx2)
That returns a 1 for me.

Prodotti

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by