How to calculate the area dissimilarity between two curves?

5 次查看（过去 30 天）
Mr. 206
Mr. 206 2018-10-9
编辑： Mr. 206 2018-10-9
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?
  5 个评论
显示 3更早的评论
Torsten
Torsten 2018-10-9
编辑：Torsten 2018-10-9
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
Mr. 206 2018-10-9
编辑：Mr. 206 2018-10-9
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.

请先登录，再进行评论。

请先登录，再回答此问题。

回答（1 个）

KSSV
KSSV 2018-10-9
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
  3 个评论
显示 1更早的评论
KSSV
KSSV 2018-10-9
The above code worked fine for me......let them be not same...doesn't matter....did you get any error?
Mr. 206
Mr. 206 2018-10-9
编辑：Mr. 206 2018-10-9
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."

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 Correlation and Convolution 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by