Find Area under the curve as a non negative

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,:)';

[Caroline Szujewski]

Here is an example of what one of my figures looks like and I have attached the vector data in an excel sheet below. You can see how the waveform is both above and below 0 (figure 1). I want the entire area shaded in red as a positive number, so Area 1 + abs(Area 2) (first figure). However, I cannot just ask for the absolute area of my vector because it flips everything as a mirror over the x axis (figure 2), which inflates the actual area under the curve (shaded in red in figure 3).

Figure 1:

Figure 2:

Figure 3:

Does this answer both of your questions?

[Torsten]

Shift the curve by 0.78 in y-direction (corresponding to the y-value at the minimum at about 52) and make a usual integration.

[Caroline Szujewski]

So this is just one vector in a matrix. How would I write a code to say shift it up by the minimum for each column in the matrix, then take the AUC?

[Torsten]

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]

This works! Thank you!

Answer 1

[Star Strider]

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')

.

[Caroline Szujewski]

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]

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.

.

Answer 2

[John D'Errico]

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.

[Caroline Szujewski]

hello, I left another comment above that I think will clarify what I want. Please let me know if it is still not clear.

[John D'Errico]

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]

I ended up doing that, thnak you for your help!