Find Area under the curve as a non negative

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Caroline Szujewski
Caroline Szujewski el 1 de Mzo. de 2024
Comentada: Caroline Szujewski el 1 de Mzo. de 2024
The vector I am plotting is partially negative because the y axis is a zscore of my fiber photometry signal. I am trying to calculate the area under the curve, but it is returning negative numbers because my y-axis is partially 0. Using the absolue value inflates my actual area under the curve (my signal is a waveform, so making everything positive makes waht was outside of the ewaveform inside of the waveform when flipped bove x). I tried fliplr, but the resulting graph looks the exact same. How do I just get an absolute area under the curve without inflating the AUC?
Current code: raw_dff is a matrix where each column is a waveform from a previoujs pek detection I performed. I get the area under the curve for each column in that matrix with my AUC_counter, but againn, this is sometimes negative.
auc = cumtrapz(raw_dff);
AUC_counter = auc(81,:)';
  5 comentarios
Torsten
Torsten el 1 de Mzo. de 2024
Editada: Torsten el 1 de Mzo. de 2024
for i = 1:size(raw_dff,2)
vector = raw_dff(:,i);
m = min(vector);
if m < 0
vector = vector - m;
end
auc(:,i) = cumtrapz(vector);
end
Caroline Szujewski
Caroline Szujewski el 1 de Mzo. de 2024
This works! Thank you!

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Respuesta aceptada

Star Strider
Star Strider el 1 de Mzo. de 2024
It would help to have the data.
Perhaps selecting everything with both x and y coordinates greater than zero, and then integrating that result will do what you want.
Try this —
x = linspace(-1, 2, 500);
y = sin(2*pi*x);
Lv = x>0 & y>0;
inty = cumtrapz(x(Lv), y(Lv));
figure
plot(x, y, 'DisplayName','Original Data')
hold on
plot(x(Lv), y(Lv), '-r', 'DisplayName','Selected Data')
plot(x(Lv), inty, '-g', 'DisplayName','Integral')
hold off
legend('Location','best')
.
  2 comentarios
Caroline Szujewski
Caroline Szujewski el 1 de Mzo. de 2024
hello, I left another comment above that I think will clarify what I want. Please let me know if it is still not clear.
Star Strider
Star Strider el 1 de Mzo. de 2024
Editada: Star Strider el 1 de Mzo. de 2024
Perhaps this —
y = readmatrix('vector_ex_data.xlsx');
L = numel(y);
x = linspace(0, L-1, L); % Corresponds To Indices Into 'y', Use A Different Vector As Desired (Used In 'cumtrapz' Call)
miny = min(y)
miny = -0.7757
ys = y-miny; % 'ys' —> y-shifted
figure
plot(x, y)
grid
xlabel('X')
ylabel('Y')
title('Original')
AUC = cumtrapz(x, ys);
figure
yyaxis left
plot(x, ys, 'DisplayName','y-shifted')
hold on
patch([x flip(x)], [zeros(size(ys)) flip(ys)], 'r', 'DisplayName','AUC')
hold off
text(45, 1.05, sprintf('Total Area = %.4f', AUC(end)))
Ax1 = gca;
ylabel('Y')
yyaxis right
plot(x, AUC, '-g', 'DisplayName','Cumulative Area', 'LineWidth',2)
Ax2 = gca;
grid
xlabel('X')
ylabel('Y')
title('Y-Shifted')
legend('Location','best')
EDIT — Corrected typographical errors.
.

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Más respuestas (1)

John D'Errico
John D'Errico el 1 de Mzo. de 2024
You don't show your data, so I need to make something up, and I'm guessing what your problem is, because your question is confusing. But I think this is your question:
What is the area between my curve and the x axis? So negative regions of the curve, they will still have positive area. For example:
x = linspace(0,2*pi,8);
y = sin(x);
plot(x,y,'-o')
yline(0)
If we use trapz to integrate this curve, we will get zero.
trapz(x,y)
ans = 0
The positive and negative parts compensate exactly. So I think you want to find the are between the curve and the x axis. BUT, if you do this:
trapz(x,abs(y))
ans = 3.9326
it mistakes the area, because of the region near x==pi.
plot(x,abs(y),'o-')
The problem is, that part of the curve wants to hit zero between the sampled points, so right at x==pi.
Instead, you need to interpolate the data, then use integral on the absolute value of the interpolation function. For example...
spl = spline(x,y);
fplot(@(x) abs(fnval(spl,x)),[0,2*pi])
You should see it now does touch down to zero at x==pi. And integral gives a better estimate of the true area.
integral(@(x) abs(fnval(spl,x)),0,2*pi)
ans = 4.0094
Note that trapz will under-estimate that area because it does so on an always concave-down function, and it over-estimates the area because of the problem near x==pi.
  3 comentarios
John D'Errico
John D'Errico el 1 de Mzo. de 2024
If you just shift the entire curve upwards, it does not do what you want. It just adds a linear bias to the cumulative area.
I explained how to do, what I still think you are asking.
Are you looking for the cumulative area between the curve and the x-axis, or the total area?
Caroline Szujewski
Caroline Szujewski el 1 de Mzo. de 2024
I ended up doing that, thnak you for your help!

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