This page: http://www.ecse.rpi.edu/Homepages/wrf/Research/Short_Notes/pnpoly.html shows how to do this for any polygon. I have a Java implementation of this, but it is too big to post here in its entirety. However, you should be able to work it out: class Boundary { private final Point[] points; // Points making up the boundary … /** * Return … Read more
for (i = 0; i < n; i++) { printf(“%f %f\n”,r * Math.cos(2 * Math.PI * i / n), r * Math.sin(2 * Math.PI * i / n)); } where r is the radius of the circumsribing circle. Sorry for the wrong language No Habla C#. Basically the angle between any two vertices is 2 … Read more
One algorithm to solve this is ray-casting. See an explanation here. And you can find code implementing this for the Google Maps JS API V3 here. HTH.
Using matplotlib.pyplot import matplotlib.pyplot as plt coord = [[1,1], [2,1], [2,2], [1,2], [0.5,1.5]] coord.append(coord[0]) #repeat the first point to create a ‘closed loop’ xs, ys = zip(*coord) #create lists of x and y values plt.figure() plt.plot(xs,ys) plt.show() # if you need…
I would suggest using the Path class from matplotlib import matplotlib.path as mplPath import numpy as np poly = [190, 50, 500, 310] bbPath = mplPath.Path(np.array([[poly[0], poly[1]], [poly[1], poly[2]], [poly[2], poly[3]], [poly[3], poly[0]]])) bbPath.contains_point((200, 100)) (There is also a contains_points function if you want to test for multiple points)
If you specifically want to construct your Polygon from the shapely geometry Points, then call their x, y properties in a list comprehension. In other words: from shapely import geometry poly = geometry.Polygon([[p.x, p.y] for p in pointList]) print(poly.wkt) # prints: ‘POLYGON ((0 0, 1 0, 1 1, 0 1, 0 0))’ Note that shapely … Read more
I took @MitchWheat and @templatetypedef’s idea of sampling points on a circle and took it a bit farther. In my application I need to be able to control how weird the polygons are, ie start with regular polygons and as I crank up the parameters they get increasingly chaotic. The basic idea is as stated … Read more
This type of intersection is easily done by the “min of the maxes” and “max of the mins” idea. To write it out one needs a specific notion for the rectangle, and, just to make things clear I’ll use a namedtuple: from collections import namedtuple Rectangle = namedtuple(‘Rectangle’, ‘xmin ymin xmax ymax’) ra = Rectangle(3., … Read more
The question by the OP is a bit ambiguous so the question I will answer is: How to generate a point from a uniform distribution within an arbitrary quadrilateral, which is actually a generalization of How to generate a point from a uniform distribution within an arbitrary (convex) polygon. The answer is based on the … Read more
The key to solving this problem is first making an observation: the center of the largest circle that will fit inside an arbitrary polygon is the point that is: Inside the polygon; and Furthest from any point on the edges of the polygon. Why? Because every point on the edge of a circle is equidistant … Read more