Calculating area under curve with cutoff

Hey Community,

I got a n-by-1 row vector obtained by measured data with n likely beeing about 100.000. The values can be either positive or negative and are derived as the difference from 2 other functions:

I calculated the area with cumtrapz / cumsum (red line):

The value of the cumulated area under the curve should display a state which can not be negative. Therefore, I want to lift up the curve in the upper graph in the consecutive negativ intervals, so it looks like the blue line. To get there i used this code:

dv=diff(v);
for a=1:length(v)-2
    if v(a)<0
        if dv(a)>0 && dv(a+1)<=0
            v(a:end)=v(a:end)-v(a);
        end
        v(a)=0;
    end
end

This code works for me, but it is not very economic. I would like to avoid the for loop and use a faster vectorized solution because of speed. The function is going to be called about thousand times or even more often with vectors about 100k. Could you help me to find a more convenient way for calculation?

A possible solution could seek for consecutive local minima (tn). Detect the first point following after the local minimum which is equal (pn) and then create a stepwise mixed function which looks similar to this one:

[darova]

I don't understand how to get the blue line. Can you please explain more?

[Robin Schäfer]

If you mean the last plot (single blue line): It is the difference between the red and blue line from the upper graph. Therefore, it is somehow the result from my for-loop. If I can create the curve by a vectorized solution it is just a simple addition...

[darova]

Im aksing about this curve

Don't understand how do you want to lift the red one up

[Robin Schäfer]

Okay - The for-loop above is doing what I want:

dv=diff(v);         % difference to determine local minima     
for a=1:length(v)-2
    if v(a)<0                           % for all negative values: set zero
        if dv(a)>0 && dv(a+1)<=0        % detect change of sign for local minimum
            v(a:end)=v(a:end)-v(a);     % lift up the rest of v to the actual value(v(a))
        end
        v(a)=0;                         
    end
end

% After the lift up nothing is going to happen until negative values occur again
% note that actual value v(a) has to be negative, otherwise + abs(v(a)) could be used as well

Again: The formula works like intended, but I would like to prefer a vectorized solution for faster computing =)

Answer 1

[darova]

Don't know about vectorizing. Maybe this part will be simpler and faster

load v.mat
v = cumsum(v);
v1 = v;
dv1 = v*0;
for i = 1:length(v1)-1
    if v1(i+1) < dv1(i)
        dv1(i+1) = v1(i+1);
    else
        dv1(i+1) = dv1(i);
    end
end