MATLAB Answers

Translated by

このページのコンテンツは英語から自動翻訳されています。自動翻訳をオフにする場合は「<a class="turn_off_mt" href="#">ここ</a>」をクリックしてください。

0

How to normalize curves with multiple peaks such that each peak will be 1 and each valley will be 0?

Haguy Wolfenson さんによって質問されました 2019 年 1 月 20 日
最新アクティビティ dpb
さんによって 編集されました 2019 年 1 月 21 日
Hello,
I am looking for a way to normalize curves with multiple peaks such that each local maximum and minimum will be set to 1 and 0 respectively, and that the values between the maxima and minima will be set accordingly.
Thanks,
Haguy.

  2 件のコメント

Only thing comes to mind would be to find each max/min pair and compute the necessary factor for those positions and then interpolate that scaling factor between each.
Providing an example data set would probably be beneficial.
Thanks. That's what I was thinking. Is there a specific function for that?
This would be an example for a data set:
7.758833
9.813105
11.43802
10.68728
6.67001
0.737689
-4.47406
-6.98998
-6.52002
-4.11176
-1.15441
1.4053
3.232571
4.339376
4.729456
4.145157
2.181519
-1.10699
-4.71235
-7.04357
-6.92214
-4.30292
-0.38969
2.694517
2.948959
0.215726
-2.97796
-2.9788
1.951206
9.78937
15.93909
16.79721
12.67508
7.168603
3.742826
3.560238
6.438025
12.06834
18.94483
23.36877
21.61921
13.23558
1.492308
-8.82432
-13.8726
-12.363

サインイン to comment.

タグ

2 件の回答

回答者: dpb
2019 年 1 月 20 日
編集済み: dpb
2019 年 1 月 21 日
 採用された回答

If you mean 'builtin' already coded to do that, "No, I don't think so!" I've never seen it actually done before; don't think it's terribly common.
I don't have time to write full-blown code, but you should be able to follow along and write a function from a demo of the idea done at command line for the sample data--
y=x+abs(min(x)); % move whole trace to baseline first
findpeaks(y) % just to make the fancy plot
[pks,ipk]=findpeaks(y); % get the peaks, locations
[vlys,ivly]=findpeaks(-y); % and the valleys
vlys=-vlys; % back to correct sign of signal
hold on
plot(ivly,vlys-1,'^r','MarkerFaceColor','r');% add markers for the valleys just for fun
b=polyfit([ipk(1) ivly(1)],[1/pks(1) 0],1); % linear transformation peak-->1, valley-->0 first section
yhat=polyval(b,ipk(1):ivly(1)); % get the scale factor for the section
yhat=yhat.*y(ipk(1):ivly(1)).'; % and scale the actual data
yyaxis right % to plot the scaled on appropriate scale
plot((ipk(1):ivly(1)).',yhat,'k-'); % and plot that first section...
ylim([-0.2 1.6]) % just to make origins aline left/right axes...
polyfit([ivly(1) ipk(2)],[0 1/pks(2)],1) % now do the next section vly(1)-->0; pk(2)==>1
yhat=polyval(b,[ivly(1):ipk(2)]); % and rinse and repeat above pattern...
yhat=yhat.*y(ivly(1):ipk(2)).';
hold on
plot((ivly(1):ipk(2)).',yhat,'k-');
hLg=legend(hLy(1),'Normalized','Location','northwest');
Above yields inserted image; your mission, should you choose to accept it, is to take the above and turn it into the general function walking through the located peaks and valleys. You'll have to decide just what to do about the end sections where you don't have the alternate peak/valley to use to fully define the linear transformation...it's purely a guess as to what that scaling factor should
be as it is not known what the other extremum or its location would have been.

  1 件のコメント

サインイン to comment.


回答者: Sargondjani
2019 年 1 月 20 日

Yeah, you would have to calculate a scaling factor for each peak/trough. In one dimension you can then use the scaling factor for each interval. In multiple dimensions you would have to interpolate the scaling factor.

  1 件のコメント

That would be discontinuous tho, between the sections excepting in very unusual circumstances.

サインイン to comment.



Translated by