Timeline for How to check if a 2D point is in a polygon?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 11, 2015 at 6:59 | comment | added | matheorem | @mac p is not right, should be p = CountryData["Canada", "Polygon"][[1, 1,25]];. Because there are 25 parts in Canada. The main part is 25th. Tested in V10 | |
| S Aug 22, 2012 at 2:26 | history | suggested | CommunityBot | CC BY-SA 3.0 | formatting |
| Aug 22, 2012 at 2:25 | review | Suggested edits | |||
| S Aug 22, 2012 at 2:26 | |||||
| Aug 15, 2012 at 21:32 | history | edited | Mac | CC BY-SA 3.0 | Tested all point in polygon routines in terms of speed of execution |
| Aug 14, 2012 at 14:21 | comment | added | user21 | @Mac, you could use a slightly larger polygon (e.g. the polygon from a country from CountryData) and some Random points (with a SeedRandom) and see how that performs ;-) | |
| Aug 14, 2012 at 13:21 | comment | added | J. M.'s missing motivation | I'd replace the For[] with a Do[] myself... | |
| Aug 14, 2012 at 11:48 | comment | added | Mac | Thanks for the suggestion - have updated the code accordingly with i and j now localised. | |
| Aug 14, 2012 at 11:46 | history | edited | Mac | CC BY-SA 3.0 | deleted 13 characters in body |
| Aug 14, 2012 at 11:26 | comment | added | Mr.Wizard | Mac, thanks for contributing. Looks interesting! I recommend that you localize i and j. | |
| Aug 14, 2012 at 11:22 | history | answered | Mac | CC BY-SA 3.0 |