Help me determine why the Isolocalmax doesn't work after applying MinProminence.
조회 수: 8 (최근 30일)
이전 댓글 표시
% Identify indirect peak values (aka vallies)
dataDirect = dataOut;
x = 1: (length(dataDirect));
[ TFm, Pm] = islocalmax(dataDirect);
min_localmax = (mode(dataDirect(TFm))+ mean(dataDirect(TFm)))/2;
% max2 = max(dataDirect);
% max_minlimit2 = max2*.9990;
[ TFd, Pd] = islocalmax(dataDirect, 'MinProminence', min_localmax);% max_minlimit2); % aveDiff,
Peak_location1 = (x(TFd))';
Peak_values1 = dataDirect(TFd);
nexttile
hold on
yline(min_localmax,'r-')
plot(x, dataDirect,x(TFd),dataDirect(TFd),'r*')
title('butter ')
For some reason, the islocalmax function is not identifying any peaks when I use 'MinProminence.' If anyone could suggest alternative functions that would be great. The 'islocalmax' works well when I am trying to determine by 'MinProminence' value but after that it no longer works and I can't figure out why since I use this function all of the time.
If anyone has the time to help I would really appreciate it!
댓글 수: 0
답변 (1개)
Star Strider
2024년 3월 31일
편집: Star Strider
2024년 3월 31일
You need to check the variable values.
I cannot run this here (the Run feature is currently not working, at least for me, although MATLAB Online works), however running this:
dataOut = randn(500,1);
dataDirect = dataOut;
x = 1: (length(dataDirect));
[ TFm, Pm] = islocalmax(dataDirect);
min_localmax = (mode(dataDirect(TFm))+ mean(dataDirect(TFm)))/2
datamax = max(dataOut) % Added
minPM = min(Pm(Pm ~= 0 ))
produces this result:
min_localmax =
-0.2613
datamax =
3.5784
minPM =
0.0082
So I suspect using an extremely low value for 'MinProminence' (much less than the lowest prominence value returned) is likely the problem.
EDIT — (31 Mar 2024 at 17:00)
Corrected typographical errors.
.
댓글 수: 0
참고 항목
카테고리
Help Center 및 File Exchange에서 AI for Signals에 대해 자세히 알아보기
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!