回答问题

我发现了两种主要的方法来查看一个点是否属于多边形。一种是在这里使用使用的光线追踪方法,这是最推荐的答案,另一种是使用 matplotlibpath.contains_points(这对我来说似乎有点晦涩)。我将不得不连续检查很多点。有谁知道这两个中的任何一个是否比另一个更值得推荐,或者是否有更好的第三种选择?

更新:

我检查了这两种方法,matplotlib 看起来要快得多。

from time import time
import numpy as np
import matplotlib.path as mpltPath

# regular polygon for testing
lenpoly = 100
polygon = [[np.sin(x)+0.5,np.cos(x)+0.5] for x in np.linspace(0,2*np.pi,lenpoly)[:-1]]

# random points set of points to test 
N = 10000
points = np.random.rand(N,2)


# Ray tracing
def ray_tracing_method(x,y,poly):

    n = len(poly)
    inside = False

    p1x,p1y = poly[0]
    for i in range(n+1):
        p2x,p2y = poly[i % n]
        if y > min(p1y,p2y):
            if y <= max(p1y,p2y):
                if x <= max(p1x,p2x):
                    if p1y != p2y:
                        xints = (y-p1y)*(p2x-p1x)/(p2y-p1y)+p1x
                    if p1x == p2x or x <= xints:
                        inside = not inside
        p1x,p1y = p2x,p2y

    return inside

start_time = time()
inside1 = [ray_tracing_method(point[0], point[1], polygon) for point in points]
print("Ray Tracing Elapsed time: " + str(time()-start_time))

# Matplotlib mplPath
start_time = time()
path = mpltPath.Path(polygon)
inside2 = path.contains_points(points)
print("Matplotlib contains_points Elapsed time: " + str(time()-start_time))

这使,

Ray Tracing Elapsed time: 0.441395998001
Matplotlib contains_points Elapsed time: 0.00994491577148

使用三角形而不是 100 边多边形获得了相同的相对差异。我也会检查身材匀称,因为它看起来是一个专门解决这类问题的包

Answers

你可以考虑shapely:

from shapely.geometry import Point
from shapely.geometry.polygon import Polygon

point = Point(0.5, 0.5)
polygon = Polygon([(0, 0), (0, 1), (1, 1), (1, 0)])
print(polygon.contains(point))

从您提到的方法中,我只使用了第二种方法,path.contains_points,它工作正常。在任何情况下,根据测试所需的精度,我建议创建一个 numpy bool 网格,多边形内的所有节点都为 True(如果不是,则为 False)。如果您要对很多点进行测试,这可能会更快(尽管注意这依赖于您在“像素”容差内进行测试):

from matplotlib import path
import matplotlib.pyplot as plt
import numpy as np

first = -3
size  = (3-first)/100
xv,yv = np.meshgrid(np.linspace(-3,3,100),np.linspace(-3,3,100))
p = path.Path([(0,0), (0, 1), (1, 1), (1, 0)])  # square with legs length 1 and bottom left corner at the origin
flags = p.contains_points(np.hstack((xv.flatten()[:,np.newaxis],yv.flatten()[:,np.newaxis])))
grid = np.zeros((101,101),dtype='bool')
grid[((xv.flatten()-first)/size).astype('int'),((yv.flatten()-first)/size).astype('int')] = flags

xi,yi = np.random.randint(-300,300,100)/100,np.random.randint(-300,300,100)/100
vflag = grid[((xi-first)/size).astype('int'),((yi-first)/size).astype('int')]
plt.imshow(grid.T,origin='lower',interpolation='nearest',cmap='binary')
plt.scatter(((xi-first)/size).astype('int'),((yi-first)/size).astype('int'),c=vflag,cmap='Greens',s=90)
plt.show()

,结果是这样的: