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trying to find peaks on a time series

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DuckDuck
DuckDuck 2012 年 6 月 11 日
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 件のコメント
Ryan
Ryan 2012 年 6 月 11 日
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 2012 年 6 月 11 日
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

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採用された回答

Wayne King
Wayne King 2012 年 6 月 11 日
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 件のコメント
DuckDuck
DuckDuck 2012 年 6 月 11 日
yeah you are right, basically xn is s x.
no it does not, i tried and it did not.

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その他の回答 (1 件)

Wayne King
Wayne King 2012 年 6 月 12 日
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.

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