Timeline for How to check if a 2D point is in a polygon?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 13, 2017 at 12:55 | history | edited | CommunityBot | replaced http://mathematica.stackexchange.com/ with https://mathematica.stackexchange.com/ | |
| Oct 11, 2015 at 16:22 | comment | added | Daniel Lichtblau | @maththeorem It is very difficult to future-proof code of this sort. It appears that CountryData has changed since this thread was in its infancy. | |
| Oct 11, 2015 at 7:03 | comment | added | matheorem | @DanielLichtblau p should be p = CountryData["Canada", "Polygon"][[1, 1,25]];. The main part of Canada is 25th list. | |
| Aug 20, 2012 at 7:25 | comment | added | Mac | Worth pointing out too that the "Canada" polygon example assumes implicitly that each vertex defining the polygon can be connected by a straight line. This has to be a good assumption for such a densely sampled contour. However, for large geographical polygons with few points the connecting edges should be modelled as "great circles". | |
| Aug 16, 2012 at 14:34 | history | edited | Daniel Lichtblau | CC BY-SA 3.0 | minor corrections |
| Aug 16, 2012 at 14:33 | comment | added | Daniel Lichtblau | Thanks. I changed from my point generator to yours and failed to catch the effect of the partitioning. Will correct response accordingly.. | |
| Aug 16, 2012 at 9:27 | comment | added | Mac | I don't think we will find a better solution in terms of performance unless perhaps a GPU-solution can be implemented (way above my programming skills in any case). One minor point - you're actually testing 10000 pairs of coordinates (points) not 20000. | |
| Aug 16, 2012 at 9:17 | comment | added | Mac | For once Canada is mentioned in another context than exporting cold air down south...(Canadians will know what I mean). | |
| Aug 15, 2012 at 22:46 | history | answered | Daniel Lichtblau | CC BY-SA 3.0 |