Determining 1, 2 and 3 sigma radii for a bivariate Gaussian histogram

8 vues (au cours des 30 derniers jours)
Tim Fulcher
Tim Fulcher le 18 Déc 2021
Commenté : Star Strider le 20 Déc 2021
Hi all,
I've a 2d scatter plot which I've converted into a 2d histogram using scatter_kde (credit: Nils). I'd like to know whether it's possible to determine the 1, 2 and 3 sigma radii for the 2d histogram? It would also be useful if I could quantify the "roundness" of those circles.
Thanks and compliments of the season.
Tim

Réponse acceptée

Star Strider
Star Strider le 18 Déc 2021
If the point positions are in matrix format (or can be interpolated to matrix format using griddata or a similar function), the contour function would likely be appropriate. Set the contours to be whatever values are desired.
Example —
xv = linspace(-30, 30);
yv = linspace(-40, 40);
[Xm,Ym] = ndgrid(xv, yv);
zfcn = @(x,y) exp(-(x/15).^2) + exp(-(y/20).^2);
Zm = zfcn(Xm,Ym) + randn(size(Xm))*0.25;
Xm(Zm < 1) = NaN;
Ym(Zm < 1) = NaN;
Zm(Zm < 1) = NaN;
Zmm = mean(Zm(:),'omitnan');
sgma1 = std(Zm,1,1,'omitnan');
sgma1m = mean(sgma1);
sgma2 = std(Zm,1,2,'omitnan');
sgma2m = mean(sgma2);
NoNaN = ~isnan(Zm);
Xmn = Xm(NoNaN);
Ymn = Ym(NoNaN);
Zmn = Zm(NoNaN);
xvi = linspace(-30, 30, 25);
yvi = linspace(-40, 40, 25);
[Xi,Yi] = ndgrid(xvi, yvi);
Zi = griddata(Xmn(:), Ymn(:), Zmn(:), Xi, Yi);
figure
scatter3(Xm(:), Ym(:), Zm(:), [], Zm(:), 'Marker','.')
xlabel('X')
ylabel('Y')
hold on
surf(Xi, Yi, Zi, 'FaceAlpha',0.75, 'EdgeColor','k')
contour3(Xi, Yi, Zi, Zmm+[0.15 0.30 0.45], '-r', 'LineWidth',5)
hold off
grid on
figure
scatter3(Xm(:), Ym(:), Zm(:), [], Zm(:), 'Marker','.')
xlabel('X')
ylabel('Y')
view(0,90)
hold on
contour3(Xi, Yi, Zi, Zmm+[0.15 0.30 0.45], '-r', 'LineWidth',2.5)
hold off
grid on
This is definitely not a perfect illustration of the data, however it illustrates the approach.
.
  2 commentaires
Tim Fulcher
Tim Fulcher le 20 Déc 2021
Thanks Star Strider. I've not finished implementation yet but so far...so good.
Regards and compliments of the season.
Star Strider
Star Strider le 20 Déc 2021
As always, my pleasure!
Thank you! And to you, too!
.

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