Question

我试图检测给定点（x，y）是否在n * 2数组的多边形中。但似乎多边形边界上的某些点返回它不包括在内。

def point_inside_polygon(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:
                        xinters = (y-p1y)*(p2x-p1x)/float((p2y-p1y))+p1x
                    if p1x == p2x or x <= xinters:
                        inside = not inside
        p1x,p1y = p2x,p2y

    return inside

Answer 1

这是具有多个选项的简单方法

from shapely.geometry import Point, Polygon

# Point objects(Geo-coordinates)
p1 = Point(24.952242, 60.1696017)
p2 = Point(24.976567, 60.1612500)

# Polygon
coords = [(24.950899, 60.169158), (24.953492, 60.169158), (24.953510, 60.170104), (24.950958, 60.169990)]
poly = Polygon(coords)

方法1：

在功能内使用：

语法：point.within（多边形）

# Check if p1 is within the polygon using the within function
print(p1.within(poly))
# True

# Check if p2 is within the polygon
print(p2.within(poly))
# False

方法2：

使用包含功能：

语法：polygon.contains（点）

# Check if polygon contains p1 
print(poly.contains(p1))
# True

# Check if polygon contains p2
print(poly.contains(p2))
# False

要检查点是否位于多边形的边界上：

使用触摸功能：

语法：polygon.touches（点）

poly1 = Polygon([(0, 0), (1, 0), (1, 1)])
point1 = Point(0, 0)

poly1.touches(point1)
# True

如果您想加快此过程，请使用

import shapely.speedups
shapely.speedups.enable()

和

利用Geopandas

参考：

Answer 2

您可以使用来自contains_point的{{1}}函数，具有较小的负半径和正半径（小技巧）。像这样：

matplotlib.path

结果是

import matplotlib.path as mplPath
import numpy as np

crd = np.array([[0,0], [0,1], [1,1], [1,0]])# poly
bbPath = mplPath.Path(crd)
pnts = [[0.0, 0.0],[1,1],[0.0,0.5],[0.5,0.0]] # points on edges
r = 0.001 # accuracy
isIn = [ bbPath.contains_point(pnt,radius=r) or bbPath.contains_point(pnt,radius=-r) for pnt in pnts]

默认情况下（或[True, True, True, True]）不包括边框上的所有点，结果为

r=0

Answer 3

以下是包含边缘的正确代码：

def point_inside_polygon(x, y, poly, include_edges=True):
    '''
    Test if point (x,y) is inside polygon poly.

    poly is N-vertices polygon defined as 
    [(x1,y1),...,(xN,yN)] or [(x1,y1),...,(xN,yN),(x1,y1)]
    (function works fine in both cases)

    Geometrical idea: point is inside polygon if horisontal beam
    to the right from point crosses polygon even number of times. 
    Works fine for non-convex polygons.
    '''
    n = len(poly)
    inside = False

    p1x, p1y = poly[0]
    for i in range(1, n + 1):
        p2x, p2y = poly[i % n]
        if p1y == p2y:
            if y == p1y:
                if min(p1x, p2x) <= x <= max(p1x, p2x):
                    # point is on horisontal edge
                    inside = include_edges
                    break
                elif x < min(p1x, p2x):  # point is to the left from current edge
                    inside = not inside
        else:  # p1y!= p2y
            if min(p1y, p2y) <= y <= max(p1y, p2y):
                xinters = (y - p1y) * (p2x - p1x) / float(p2y - p1y) + p1x

                if x == xinters:  # point is right on the edge
                    inside = include_edges
                    break

                if x < xinters:  # point is to the left from current edge
                    inside = not inside

        p1x, p1y = p2x, p2y

    return inside

更新：修正了错误

确定Point是否在Polygon python中

3 个答案:

方法1：

方法2：

要检查点是否位于多边形的边界上：