- 方法1： 用shapely中的geometry包
1）polygon.covers(point)
如果point在多边形polygon上（包括边），返回True，否则False。
2）polygon.contains(point)
如果point在多边形polygon上（不包括边），返回True，否则False。
from shapely import geometry import matplotlib.pyplot as plt pts = [(0, 0), (1, 1), (0, 1), (0, 0)] polygon = geometry.Polygon(pts) pt = 0.1, 0.2 print(*pt) point = geometry.Point(*pt) plt.figure(0) plt.plot([i[0] for i in pts], [i[1] for i in pts]) plt.scatter(*pt) plt.show() print(polygon.covers(point))
方法2： 用blender的内置python api。
将点投影到三角形平面上，并检查其是否在三角形内，如果在，则返回在面上的投影。
没试效果。地址 https://docs.blender.org/api/current/mathutils.geometry.html
mathutils.geometry.intersect_point_tri(pt, tri_p1, tri_p2, tri_p3)
取4个向量：一个是点，下3个定义三角形。将点投影到三角形平面上，并检查其是否在三角形内。
Takes 4 vectors: one is the point and the next 3 define the triangle. Projects the point onto the triangle plane and checks if it is within the triangle.
Parameters
pt (mathutils.Vector) – Point
tri_p1 (mathutils.Vector) – First point of the triangle
tri_p2 (mathutils.Vector) – Second point of the triangle
tri_p3 (mathutils.Vector) – Third point of the triangle
Returns
Point on the triangles plane or None if its outside the triangle
mathutils.Vector or None
from mathutils import Vector
from mathutils.geometry import (intersect_point_tri, distance_point_to_plane)
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import axes3d
pt1 = Vector((0, 0,1))
pt2 = Vector((1, 1, 0))
pt3 = Vector((1, -1, 0))
plane_co = pt1
plane_no = (pt3 - pt1).cross(pt2 - pt1).normalized()
pt = Vector((0.5, 0, 1))
# find distance to plane
d = distance_point_to_plane(pt, plane_co, plane_no)
ptm = pt - d * plane_no
print("move", pt, d, "units to", ptm)
p = intersect_point_tri(pt, pt1, pt2, pt3)
if p:
print("point on triangle", p)
else:
print("point outside triangle")
fig = plt.figure()
ax = plt.axes(projection="3d")
ax.set_xlabel("x axis")
ax.set_ylabel("y axis")
ax.set_zlabel("z axis")
x0 = [pt1[0], pt2[0], pt3[0], pt1[0]]
y0 = [pt1[1], pt2[1], pt3[1], pt1[1]]
z0 = [pt1[2], pt2[2], pt3[2], pt1[2]]
ax.plot3D(x0, y0, z0)
ax.scatter(x0, y0, z0, color='b')
for x, y, z in zip(x0, y0, z0):
text = "({},{},{})".format(x, y, z)
ax.text(x, y, z, text, zdir=(1, 1, 1))
ax.scatter(pt[0], pt[1], pt[2], color='r')
ax.scatter(ptm[0], ptm[1], ptm[2], color='g')
plt.show()
- 方法3： 还是blender的内置python api
mathutils.geometry.intersect_point_tri_2d(pt, tri_p1, tri_p2, tri_p3)
- 取4个向量（仅使用x和y坐标）：一个是点，接下来的3个定义三角形。如果点位于三角形内，则返回1，否则返回0。
- Takes 4 vectors (using only the x and y coordinates): one is the point and the next 3 define the triangle. Returns 1 if the point is within the triangle, otherwise 0.
Parameters
pt (mathutils.Vector) – Point
tri_p1 (mathutils.Vector) – First point of the triangle
tri_p2 (mathutils.Vector) – Second point of the triangle
tri_p3 (mathutils.Vector) – Third point of the triangle
Return type
int
