Further processing of contour plot data

Hello,

I would like to analyse a contour plot data.I’m trying to divide the contour plot into 4x3 pieces, calculate the mean for every data “rectangle” (4x3 rectangles=12 mean values).

I gather the contour data with this code below:

clc,close all
x = linspace(-pi,pi);
y = linspace(0,2*pi);
[X,Y] = meshgrid(x,y);
Z = sin(X) + cos(Y);
[C,h] = contourf(X,Y,Z,10)

I'm struggeling to code the divison of the contour data into smaller areas, with subsequent mean calculation.

Maybe someone has a clue to achieve this

Greetings

Steffen

Answer 1

[Mathieu NOE]

hello

maybe this ?

clc,close all
clearvars
N = 100; % <= added 
x = linspace(-pi,pi,N);
y = linspace(0,2*pi,N);
[X,Y] = meshgrid(x,y);
Z = sin(X) + cos(Y);

figure(1)
[C,h] = contourf(X,Y,Z,10);
hold on

% let(s do a 4 x 3 area division and compute the individual cells average
Nx = 4;
Ny = 3;
dx = floor(N/Nx);
dy = floor(N/Ny);

for ci = 1:Nx
    for cj = 1:Ny
        indX = (1:dx)+(ci-1)*dx;
        indY = (1:dy)+(cj-1)*dy;
        % centroid of cell
        xc = mean(x(indX));
        yc = mean(y(indY));
        % mean of submatrix
        Zmean(ci,cj) = mean(Z(indX,indY),'all');
        plot(xc,yc,'*r','Markersize',15);
        t = text(xc+0.15,yc+0.15,num2str(Zmean(ci,cj),3),'FontSize',14,'Color','r');
    end
end


    

[Steffen B.]

Hello Mathieu,

thanks a lot for your help.

Your script is working just fine.

Greetings

Steffen

[Mathieu NOE]

as always, my pleasure !

Answer 2

[Bjorn Gustavsson]

I would not do the average of the contours in a region, because if you have a rather flat region with a single spike in some region you will have very few contours at the flat level, but a large number of contours and contour-points around the spike. If you then take the average of the z-levels of the contour-points it will be significantly skewed away from the flat level. I would make some surface-fit from the conours, either with the spline-fitting functions spap2, spapi, spaps, csaps or tpaps, or use scatteredInterpolant or griddata - because that way I would obtain a sampling that is regular in the x and y-directions.

Regardless of approach you'll need something that extracts the x y and z coordinates from the contours. For this you'll need something like the function extract_contours (attached).

Using this I'd do something like this:

% Mocking up some data:
x = 1:123;
y = 1:123;
z = peaks(123);
nC = 10;
C = contour(x,y,z,nC); % this give us a contour-data-set
[xC,yC,zC] = extract_contours(C); % extracting contour-levels
F = scatteredInterpolant(cell2mat(xC)',cell2mat(yC)',cell2mat(zC)','natural');
xi = 1:2:123;
[xi,yi] = meshgrid(xi);
zi = F(xi,yi); % calculating the interpolated values at [xi,yi]

xlims = [0, 40.5 92.3 124];
ylims = [0 82.3 124];
for i1 = (numel(ylims)-1):-1:1
  for i2 = (numel(xlims)-1):-1:1
    zAVG(i1,i2) = mean(zi(ylims(i1)<yi(:,1)&yi(:,1)<ylims(i1+1),xlims(i1)<xi(1,:)&xi(1,:)<xlims(i2+1)),'all','omitnan');
  end
end

If you want to average over the contours you could proceed something like this after the contour-extraction:

xAll = cell2mat(xC)';
yAll = cell2mat(yC)';
zAll = cell2mat(zC)';
xlims = [0, 40.5 92.3 124];
ylims = [0 82.3 124];
for i1 = (numel(ylims)-1):-1:1
  for i2 = (numel(xlims)-1):-1:1
    zAVG(i1,i2) = mean(zAll(ylims(i1)<yAll & ...
                                      yAll<ylims(i1+1) & ...
                            xlims(i1)<xAll & ...
                                      xAll < xlims(i2+1)),'all','omitnan');
  end
end

I think that should work.

HTH

[Steffen B.]

Hello Björn,

thanks a lot for your help.

Your script is working just fine.

Unfortunately it is not possible to accept two answers at the same time.

Mathieu was faster, therefore he got the point.

Sorry for this, your answer deserved it too.

Greetings

Steffen

[Bjorn Gustavsson]

Well, our answers addressed two completely different problems. I interpreted your problem as if you had only the contours and wanted the averages in different regions estimated from that, and that was obviously not the problem you had. But if you happen to run into that type of problem you already have one solution here.

Cheers.