Better “centerpoint” than centroid of a concave polygon

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David Franco
David Franco 2020 年 4 月 18 日
コメント済み: Chad Greene 2022 年 3 月 1 日
I'm using the centroid of polygons to attach a marker in a map application. This works definitely fine for convex polygons and quite good for many concave polygons.
However, some polygons (banana, donut) obviously don't produce the desired result: The centroid is in these cases outside the polygons area.
Does anybody know a better approach to find a suitable point within any polygons area (which may contain holes!) to attach a marker?
Thank you!
  10 件のコメント
Image Analyst
Image Analyst 2020 年 4 月 21 日
Can you trim it down to a subset of shapes?
David Franco
David Franco 2020 年 4 月 21 日
Done!

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採用された回答

darova
darova 2020 年 4 月 20 日
Here is the simplest solution for this task
x = rand(4,1);
y = rand(4,1);
x = [x; x(1)];
y = [y; y(1)];
ii = 2; % corner to place point
n = 3; % how far inside from corner
x0 = (x(ii-1)+2*n*x(ii)+x(ii+1))/(n+1)/2;
y0 = (y(ii-1)+2*n*y(ii)+y(ii+1))/(n+1)/2;
if ~inpolygon(x0,y0,x,y)
x0 = x(ii) - (x0-x(ii));
y0 = y(ii) - (y0-y(ii));
end
plot([x(ii) x0],[y(ii) y0],'o-r')
line(x,y)
axis equal

その他の回答 (3 件)

Chad Greene
Chad Greene 2021 年 10 月 5 日
I just ran into this problem when trying to place a text label in the middle of a crescent-shaped ice shelf. The centroid or the mean or median of the coordinates of the ice shelf polygon are all outside the bounds of the ice shelf. Here's the best solution I could come up with:
% Convert the outline to a polyshape:
P = polyshape(x,y);
% And get the delaunay triangulation of the polygon:
T = triangulation(P);
% Now find the center points of all the triangles:
[C,r] = circumcenter(T);
% Get the index of the centerpoint that has the largest radius from the boundary:
[~,ind] = max(r);
% These center coordinates are in the center of the fattest part of the polygon:
xc = C(ind,1);
yc = C(ind,2);

Image Analyst
Image Analyst 2020 年 4 月 18 日
Well if you compute the distance transform of the shapes you'll get a "spine" that runs along the midline of the shape. I don't know if any point along there is better than any other point. Maybe you can just pick the point half way from one end to the other, if there even ARE endpoints. The only way I know how to do it is with digital images, not analytically with a set of (x,y) vertex points. And it requires the Image Processing Toolbox so you can use bwskel() or bwdist().
  1 件のコメント
David Franco
David Franco 2020 年 4 月 19 日
Thanks, but I am working with shapefiles and I need to find the skeleton of a polygon...

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Chad Greene
Chad Greene 2021 年 10 月 8 日
I added a polycenter function to do exactly what you need in the Climate Data Toolbox for Matlab.
If you have a shapefile, the sytax is:
S = shaperead('myshapefile.shp');
[xc,yc] = polycenter(S);
Then xc,yc are the centerpoints of any entries in the shapefile structure S.
  3 件のコメント
Sim
Sim 2022 年 3 月 1 日
Hi @Chad Greene, cool tool!
However, I got the following error:
>> [xc,yc] = polycenter(S);
Reference to non-existent field 'Lon'.
Error in polycenter (line 116)
x = S(k).Lon;
Any idea on how to solve it ?
Chad Greene
Chad Greene 2022 年 3 月 1 日
@Sim Check the field names of the shapefile and then use x,y coordinates as inputs, like this
[xc,yc] = polycenter(x,y);
which would be the same as
[lonc,latc] = polycenter(lon,lat);
if your input coordinates are geo.

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