Help me determine why the Isolocalmax doesn't work after applying MinProminence.

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Hanna Armstrong
Hanna Armstrong 2024-3-31
编辑： Star Strider 2024-3-31
% 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!

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回答（1 个）

Star Strider
Star Strider 2024-3-31
编辑：Star Strider 2024-3-31
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.
.

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