Flipping y-axis of findpeaks-plot
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Nils Norlander
le 9 Mar 2017
Commenté : Greg Dionne
le 15 Mar 2017
Hi!
To find minima-values of an array I multiplied my array with -1 and used findpeaks. To graphically show the minimas I want to reverse the findpeaks-plot but can't figure out how. Any suggestions?
E_bat3 = E_bat2.*(-1);
findpeaks(E_bat3(winter_hours)) %Graph I want to flip.
Thanks!
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Réponse acceptée
Star Strider
le 9 Mar 2017
You may be making this too difficult.
Example —
a = linspace(0, 4*pi);
y = sin(a);
[pks,pkloc] = findpeaks( y,a); % Maxima
[vls,vlloc] = findpeaks(-y,a); % Minima
figure(1)
plot(a, y, '-b')
hold on
plot(pkloc, pks, '^r', 'MarkerFaceColor','r')
plot(vlloc, -vls, 'vg', 'MarkerFaceColor','g')
hold off
grid
axis([xlim -1.1 1.1])
legend('Data', 'Peaks', 'Valleys')
Note that to plot the valleys (‘vls’) or minima correctly, simply negate them in the plot call. So to find them, use:
[vls,vlloc] = findpeaks(-y,a); % Minima
and to plot them or use them in other calculations correctly, negate them again to restore their correct values:
plot(vlloc, -vls, 'vg', 'MarkerFaceColor','g')
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Plus de réponses (2)
Adam
le 9 Mar 2017
Modifié(e) : Adam
le 9 Mar 2017
Don't you just want to negate the peak amplitude results rather than flip?
2 commentaires
Adam
le 9 Mar 2017
But if they are minima then they will be larger as an absolute value than those around them so how can you display them still as minima if you plot the result as positive only?
I guess if you really want to do that then you need to subtract from whatever the largest absolute value in the results is.
Greg Dionne
le 9 Mar 2017
Modifié(e) : Greg Dionne
le 9 Mar 2017
I think Star Strider's answer is the right approach, but if you prefer the graphical 'look and feel' of the FINDPEAKS plot, then try this:
load mtlb
findpeaks(mtlb,Fs,'MinPeakProminence',1);
figure
findpeaks(-mtlb,Fs,'MinPeakProminence',1);
set(gca,'YDir','reverse');
set(findobj(gca,'Tag','Peak'),'Marker','^');
set(gca,'YTickLabel',cellfun(@(x)num2str(-str2num(x)),get(gca,'YTickLabel'),'UniformOutput',false))
2 commentaires
Greg Dionne
le 15 Mar 2017
Yes, that should work. FINDPEAKS expects vector input for Y. So in your case you could do findpeaks(-y, a) just like you did before, but then do the graphics commands above. I think they'll work no matter what input you have (so long as you actually have peaks). Otherwise, finding the 'peak' tag will fail because no peaks will actually be plotted.
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