How to calculate the area dissimilarity between two curves?

I have two smoothed curves 'f' and 'g'.

     f = [4.66356058704069;4.76003678995220;4.85195856216057;4.93937125386862;5.02241091568476;5.10134964882580;5.17664095531976;5.24896508820865;5.31927440175128;5.38883870162610;5.45908891231901;5.53146074451118;5.60723836246689;5.68739805142138;5.77245188496874;5.86229139244967;5.95603122633945;6.05185282963574;6.14684810324653;6.23686307337801;6.31713266653654;6.38291548453061;6.43012857947280;6.45598222878181;6.45961471018444;6.44272707671764;6.41021793173056;6.37081820388660;6.33483553209117;6.31189865041951;6.30870177304451;6.32932410300064;6.37554934094787;6.44718519393563;6.54238288416678;6.65795665776154;6.78970329352158;6.93272161169404;7.08173198273553;7.23139583607635;7.37663516888456;7.51295205483015;7.63674815284922;7.74522475322546;7.83628282367168;7.90842305541129;7.96064590925989;7.99235166170680;8.00324045099670;7.99321232321114]

and

g= [3.42595434180783;3.52889117917543;3.63122045210296;3.73294216049285;3.83405630405245;3.93456288219636;4.03446189394889;4.13375333784643;4.23243721183987;4.33051351319701;4.42798223853624;4.52484338386025;4.62109694458974;4.71674291559714;4.81178129124035;4.90621206539645;5.00003523149543;5.09325078264601;5.18585871176133;5.27785901168476;5.36925167531561;5.46003669573492;5.55021406633123;5.63978378092629;5.72874583375865;5.81710021946721;5.90484693307479;5.9919859699717;6.07851732589953;6.16444099693433;6.24975697947062;6.33446527020481;6.41856586611880;6.50205876447018;6.58494396278248;6.66722145883533;6.74889125065473;6.82995333650325;6.91040771487023;6.99025438446208;7.06949334419241;7.14812459317236;7.22614813069081;7.30356395619462;7.38037206926889;7.45657246961725;7.53216515704210;7.60715013142487;7.68152739270630;7.75529694086660]

'f' is produced by smoothing

x_f  =[1.36; 1.26; 1.15; 1.07; 1.04; 0.975; 0.919; 0.902; 0.85]
y_f  =[8.166216269;7.843848638;7.365180126;7.170119543;7.192934221;6.956545443;6.459904454;6.257667588;5.379897354]

And 'g' is produced by smoothing

x_g  = [1.43;1.33;1.25;1.18;1.11;1.05;1;0.952;0.909;0.87;0.833]
y_g  =[9.193091461;8.14221752;7.560356469;7.320262202;7.209229304;7.139303413;7.064348687;6.618999297;6.065270152;5.49321781;4.925745302]

I want to find the area dissimilarity between them. I am using this Equation

where D is the intersection between domains of two functions. The curves(f and g) needs to be rescaled before computing these dissimilarity. Dividing 'f' and 'g' by the maximum value of f. So the maximum value is 1.

Can you please help me, how can i implement this in MATLAB?

[KSSV]

What you want ? Is D the area of intersection of curves?

[Mr. 206]

I want to implement this dissimilarity term in MATLAB.

I don't understand this term (D), But that's the only info given in a paper. Someone said it's an area to normalize the dissimilarity term.

[Mr. 206]

normalized_f = f/max(f);
normalized_g = g/max(f);
Euclidean_distance = norm(normalized_f-normalized_g);

This is so far i can proceed then don't understand this Term 'D'

[Torsten]

d^0_L^2(f,g) is the integral distance between two functions:

d^0_L^2(f,g) = 1/(b-a) * sqrt[integral_{a}^{b} (f(x)-g(x))^2]

So, D = b-a, a normalization by the length of the interval of integration.

Best wishes

Torsten.

[Mr. 206]

Following your lead this is the code...

             a_f=x_f(1,1);
             b_f=x_f(end,1);
             D = b_f.-a_f;
             v = (f.-g); 
             d0_L2_pre = norm(v)/abs(D);

Though your explanation seems legit. However it is not giving desired results.

Answer 1

[KSSV]

If D is the area you can calculate it using :

x_f  =[1.36; 1.26; 1.15; 1.07; 1.04; 0.975; 0.919; 0.902; 0.85] ;
y_f  =[8.166216269;7.843848638;7.365180126;7.170119543;7.192934221;6.956545443;6.459904454;6.257667588;5.379897354]  ;

x_g  = [1.43;1.33;1.25;1.18;1.11;1.05;1;0.952;0.909;0.87;0.833] ;
y_g  =[9.193091461;8.14221752;7.560356469;7.320262202;7.209229304;7.139303413;7.064348687;6.618999297;6.065270152;5.49321781;4.925745302]  ;
% Smoothe the curves 
N = 1000 ;
xi_f = linspace(min(x_f),max(x_f),N)'; 
yi_f = interp1(x_f,y_f,xi_f) ;
f = [xi_f yi_f] ;

xi_g = linspace(min(x_g),max(x_g),N)'; 
yi_g = interp1(x_g,y_g,xi_g) ;
g = [xi_g yi_g] ;

P = InterX(f',g') ;
P(:,1) = [] ;

%%Get interscetion area
x0 = min(P(1,:)) ; x1 = max(P(1,:)) ;
f0 = f ; g0 = g ;
f(f(:,1)<x0,:) = [] ;f(f(:,1)>x1,:) = [] ;
g(g(:,1)<x0,:) = [] ;g(g(:,1)>x1,:) = [] ;

d = abs(trapz(f(:,1),f(:,2))-trapz(g(:,1),g(:,2))) ;

plot(f(:,1),f(:,2),'r')
hold on
plot(g(:,1),g(:,2),'b')
plot(P(1,:),P(2,:),'*k')

Get the function InterX from the link: https://in.mathworks.com/matlabcentral/fileexchange/22441-curve-intersections

[Mr. 206]

The length of f and g are not same!

[KSSV]

The above code worked fine for me......let them be not same...doesn't matter....did you get any error?

[Mr. 206]

So far it also worked for me, but i need to use the 'D'. And when using it in

           q =  f/max(f)
           r =  g/max(f)
           v = norm(q-r) 
           d0_L2_pre = norm(v)/d

Then it is saying "Matrix dimensions must agree."