how does the function findpeaks work in matlab?

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yawen kang
yawen kang 2017-5-3
评论： Rik 2017-5-3
I use the function findpeaks to detect peak in a vector,it worked as i expected,but I want to understand how does it work,and what the principle it use.
n=0:1:10;
x = [25 8 15 5 6 10 10 3 1 20 7];
[k,v]=findpeaks(x);
plot(n,x,'b');grid;hold on
plot(n(k),v,'r.');
Can anyone can help me?I will be very appreciate for your kindness.

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回答（2 个）

Rik
Rik 2017-5-3
Excerpt from the documentation:
pks = findpeaks(data) returns a vector with the local maxima (peaks) of the input signal vector, data. A local peak is a data sample that is either larger than its two neighboring samples or is equal to Inf. Non- Inf signal endpoints are excluded. If a peak is flat, the function returns only the point with the lowest index.
So that's how it works.
  2 个评论
yawen kang
yawen kang 2017-5-3
first,thanks for your answer.but I already know how to use it.I want to know what algorithm it use to find the peak.I want to understand what the code in erery line mean,and what principle it use to find the peak.
n=0:1:10; x = [25 8 15 5 6 10 10 3 1 20 7]; nx=length(x); x=x(:); dx=x(2:end)-x(1:end-1); r=find(dx>0); f=find(dx<0); if length(r)>0 & length(f)>0 % we must have at least one rise and one fall dr=r; dr(2:end)=r(2:end)-r(1:end-1); rc=repmat(1,nx,1); rc(r+1)=1-dr; rc(1)=0; rs=cumsum(rc); % = time since the last rise df=f; df(2:end)=f(2:end)-f(1:end-1); fc=repmat(1,nx,1); fc(f+1)=1-df; fc(1)=0; fs=cumsum(fc); % = time since the last fall rp=repmat(-1,nx,1); rp([1; r+1])=[dr-1; nx-r(end)-1]; rq=cumsum(rp); % = time to the next rise fp=repmat(-1,nx,1); fp([1; f+1])=[df-1; nx-f(end)-1]; fq=cumsum(fp); % = time to the next fall k=find((rs<fs) & (fq<rq) & (floor((fq-rs)/2)==0)); % the final term centres peaks within a plateau v=x(k); else k=[]; v=[]; end
Jan
Jan 2017-5-3
@yawen: It seems like you noticed, that the code is not readable. Please edit your comment and use the "{} Code" button for a proper formatting. Thanks.

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yawen kang
yawen kang 2017-5-3
first,thanks for your answer.but I already know how to use it.I want to know what algorithm it use to find the peak.I want to understand what the code in erery line mean,and what principle it use to find the peak.
n=0:1:10;
x = [25 8 15 5 6 10 10 3 1 20 7];
nx=length(x);
x=x(:);
dx=x(2:end)-x(1:end-1);
r=find(dx>0);
f=find(dx<0);
if length(r)>0 & length(f)>0 % we must have at least one rise and one fall
dr=r;
dr(2:end)=r(2:end)-r(1:end-1);
rc=repmat(1,nx,1);
rc(r+1)=1-dr;
rc(1)=0;
rs=cumsum(rc); % = time since the last rise
df=f;
df(2:end)=f(2:end)-f(1:end-1);
fc=repmat(1,nx,1);
fc(f+1)=1-df;
fc(1)=0;
fs=cumsum(fc); % = time since the last fall
rp=repmat(-1,nx,1);
rp([1; r+1])=[dr-1; nx-r(end)-1];
rq=cumsum(rp); % = time to the next rise
fp=repmat(-1,nx,1);
fp([1; f+1])=[df-1; nx-f(end)-1];
fq=cumsum(fp); % = time to the next fall
k=find((rs<fs) & (fq<rq) & (floor((fq-rs)/2)==0)); % the final term centres peaks within a plateau
v=x(k);
else
k=[];
v=[];
end
  3 个评论
显示 1更早的评论
Adam
Adam 2017-5-3
What is the relevance of that code anyway? It doesn't even include a call to findpeaks!
Rik
Rik 2017-5-3
Skimming this code, it looks like it aims to replicate findpeaks, although this version doesn't return the first index of a plateau, but the middle.

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