polygon

问题 I have a class describing a Point (has 2 coordinates x and y) and a class describing a Polygon which has a list of Points which correspond to corners (self.corners) I need to check if a Point is in a Polygon Here is the function that is supposed to check if the Point in in the Polygon. I am using the Ray Casting Method def in_me(self, point): result = False n = len(self.corners) p1x = int(self.corners[0].x) p1y = int(self.corners[0].y) for i in range(n+1): p2x = int(self.corners[i % n].x) p2y

问题 I've created a class that extends the awt.Polygon class. I'm trying to write a method that given the PathIterator of the polygon and a Point representing a vertex, adds the point at the appropriate location in the path. For Example: A Polygon thats points are (0,0) (0,10) (10,10) (10,0) (A square), given the point (1,5) would make the polygon (0,0) (1,5) (0,10) (10,10) (10,0) Thanks in advance 回答1: Expanding on @normalocity's idea, this appears to be a possible approach. Addendum: For

问题 I've created a class that extends the awt.Polygon class. I'm trying to write a method that given the PathIterator of the polygon and a Point representing a vertex, adds the point at the appropriate location in the path. For Example: A Polygon thats points are (0,0) (0,10) (10,10) (10,0) (A square), given the point (1,5) would make the polygon (0,0) (1,5) (0,10) (10,10) (10,0) Thanks in advance 回答1: Expanding on @normalocity's idea, this appears to be a possible approach. Addendum: For

问题 Yesterday I was looking to check if a point was inside a polygon and found this great script: https://github.com/tparkin/Google-Maps-Point-in-Polygon But today at work I was told that our client needs to check if one polygon is inside another polygon. I am wondering if is there a formula where I can take, let's say, two coordinates (instead of one to check a point), and from those two coordinates generate a rectangle and check if that rectangle is inside a polygon. I don't know if I'm asking

问题 Closed. This question is off-topic. It is not currently accepting answers. Want to improve this question? Update the question so it's on-topic for Stack Overflow. Closed last year . I saw on StackOverflow a "point in polygon" raytracing algorithm that I implemented in my PHP Code. Most of the time, it works well, but in some complicated cases, with complex polygons and vicious points, it fails and it says that point in not in polygon when it is. For example: You will find here my Polygon and

问题 I have a 4 side convex Polygon defined by 4 points in 2D, and I want to be able to generate random points inside it. If it really simplifies the problem, I can limit the polygon to a parallelogram, but a more general answer is preferred. Generating random points until one is inside the polygon wouldn't work because it's really unpredictable the time it takes. 回答1: A. If you can restrict your input to parallelogram, this is really simple: Take two random numbers between 0 and 1. We'll call

问题 i have a typical question with the Geometric datatype of mysql, polygon. I have the polygon data, in the form of an array of latitudes and longitudes, ex: [["x":37.628134, "y":-77.458334], ["x":37.629867, "y":-77.449021], ["x":37.62324, "y":-77.445416], ["x":37.622424, "y":-77.457819]] And i have a point (Vertex) with coordinates of latitude and longitude, ex: $location = new vertex($_GET["longitude"], $_GET["latitude"]); Now i want to find whether this vertex (point) is inside the polygon.

问题 I have the created the table below CREATE TABLE geom (g GEOMETRY); and have inserted many rows, example below: INSERT INTO geom (g) VALUES(PolygonFromText('POLYGON(( 9.190586853 45.464518970, 9.190602686 45.463993916, 9.191572471 45.464001929, 9.191613325 45.463884676, 9.192136130 45.463880767, 9.192111509 45.464095594, 9.192427961 45.464117804, 9.192417811 45.464112862, 9.192509035 45.464225851, 9.192493139 45.464371079, 9.192448471 45.464439002, 9.192387444 45.464477861, 9.192051402 45

问题 It doesn't need to be 100% correct, it can be the center of the bounding rectangle. 回答1: Algorithm: Run through all the points in the polygon. For all the points find; x1 , the lowest x coordinate y1 , the lowest y coordinate x2 , the highest x coordinate y2 , the highest y coordinate You now have the bounding rectangle, and can work out the center using: center.x = x1 + ((x2 - x1) / 2); center.y = y1 + ((y2 - y1) / 2); 回答2: Matthew's answer is a good solution. However, when using the Google

问题 From the man page for XFillPolygon: If shape is Complex , the path may self-intersect. Note that contiguous coincident points in the path are not treated as self-intersection. If shape is Convex , for every pair of points inside the polygon, the line segment connecting them does not intersect the path. If known by the client, specifying Convex can improve performance. If you specify Convex for a path that is not convex, the graphics results are undefined. If shape is Nonconvex , the path does