有没有一种简单的方法来确定一个点是否在三角形内?是2D的,不是3D的。
当前回答
这是确定一个点是在三角形的内、外还是在三角形的臂上的最简单的概念。
用行列式确定三角形内的点:
最简单的工作代码:
#-*- coding: utf-8 -*-
import numpy as np
tri_points = [(1,1),(2,3),(3,1)]
def pisinTri(point,tri_points):
Dx , Dy = point
A,B,C = tri_points
Ax, Ay = A
Bx, By = B
Cx, Cy = C
M1 = np.array([ [Dx - Bx, Dy - By, 0],
[Ax - Bx, Ay - By, 0],
[1 , 1 , 1]
])
M2 = np.array([ [Dx - Ax, Dy - Ay, 0],
[Cx - Ax, Cy - Ay, 0],
[1 , 1 , 1]
])
M3 = np.array([ [Dx - Cx, Dy - Cy, 0],
[Bx - Cx, By - Cy, 0],
[1 , 1 , 1]
])
M1 = np.linalg.det(M1)
M2 = np.linalg.det(M2)
M3 = np.linalg.det(M3)
print(M1,M2,M3)
if(M1 == 0 or M2 == 0 or M3 ==0):
print("Point: ",point," lies on the arms of Triangle")
elif((M1 > 0 and M2 > 0 and M3 > 0)or(M1 < 0 and M2 < 0 and M3 < 0)):
#if products is non 0 check if all of their sign is same
print("Point: ",point," lies inside the Triangle")
else:
print("Point: ",point," lies outside the Triangle")
print("Vertices of Triangle: ",tri_points)
points = [(0,0),(1,1),(2,3),(3,1),(2,2),(4,4),(1,0),(0,4)]
for c in points:
pisinTri(c,tri_points)
其他回答
求解如下方程组:
p = p0 + (p1 - p0) * s + (p2 - p0) * t
当0 <= s <= 1和0 <= t <= 1以及s + t <= 1时,点p在三角形内。
S,t和1 - S - t称为点p的重心坐标。
重心法Java版:
class Triangle {
Triangle(double x1, double y1, double x2, double y2, double x3,
double y3) {
this.x3 = x3;
this.y3 = y3;
y23 = y2 - y3;
x32 = x3 - x2;
y31 = y3 - y1;
x13 = x1 - x3;
det = y23 * x13 - x32 * y31;
minD = Math.min(det, 0);
maxD = Math.max(det, 0);
}
boolean contains(double x, double y) {
double dx = x - x3;
double dy = y - y3;
double a = y23 * dx + x32 * dy;
if (a < minD || a > maxD)
return false;
double b = y31 * dx + x13 * dy;
if (b < minD || b > maxD)
return false;
double c = det - a - b;
if (c < minD || c > maxD)
return false;
return true;
}
private final double x3, y3;
private final double y23, x32, y31, x13;
private final double det, minD, maxD;
}
上面的代码可以准确地处理整数,假设没有溢出。它也适用于顺时针和逆时针三角形。它不适用于共线三角形(但您可以通过测试det==0来检查)。
如果你要用同一个三角形测试不同的点,以重心为中心的版本是最快的。
重心版本在3个三角形点上是不对称的,所以它可能不如Kornel Kisielewicz的边缘半平面版本一致,因为浮点舍入误差。
图片来源:我根据维基百科关于重心坐标的文章制作了上面的代码。
其中一个最简单的方法来检查是否由三角形的顶点组成的面积 (x1,y1) (x2,y2) (x3,y3)是否为正。
面积可由公式计算:
1/2 [x1(y2–y3) + x2(y3–y1) + x3(y1–y2)]
或者python代码可以写成:
def triangleornot(p1,p2,p3):
return (1/ 2) [p1[0](p2[1]–p3[1]) + p2[0] (p3[1]–p1[1]) + p3[0] (p1[0]–p2[0])]
bool point2Dtriangle(double e,double f, double a,double b,double c, double g,double h,double i, double v, double w){
/* inputs: e=point.x, f=point.y
a=triangle.Ax, b=triangle.Bx, c=triangle.Cx
g=triangle.Ay, h=triangle.By, i=triangle.Cy */
v = 1 - (f * (b - c) + h * (c - e) + i * (e - b)) / (g * (b - c) + h * (c - a) + i * (a - b));
w = (f * (a - b) + g * (b - e) + h * (e - a)) / (g * (b - c) + h * (c - a) + i * (a - b));
if (*v > -0.0 && *v < 1.0000001 && *w > -0.0 && *w < *v) return true;//is inside
else return false;//is outside
return 0;
}
从质心转换而来的几乎完美的笛卡尔坐标 在*v (x)和*w (y)双精度内导出。 在每种情况下,两个导出双精度对象前面都应该有一个*字符,可能是*v和*w 代码也可以用于四边形的另一个三角形。 特此签名只写三角形abc从顺时针abcd的四边形。
A---B
|..\\.o|
|....\\.|
D---C
o点在ABC三角形内 对于带有第二个三角形的测试,将此函数称为CDA方向,*v=1-*v后的结果应正确;* w = 1 - * w;为了四合院
下面是一个python解决方案,它是高效的,文档化的,包含三个单元测试。它具有专业级的质量,并且可以以模块的形式放入您的项目中。
import unittest
###############################################################################
def point_in_triangle(point, triangle):
"""Returns True if the point is inside the triangle
and returns False if it falls outside.
- The argument *point* is a tuple with two elements
containing the X,Y coordinates respectively.
- The argument *triangle* is a tuple with three elements each
element consisting of a tuple of X,Y coordinates.
It works like this:
Walk clockwise or counterclockwise around the triangle
and project the point onto the segment we are crossing
by using the dot product.
Finally, check that the vector created is on the same side
for each of the triangle's segments.
"""
# Unpack arguments
x, y = point
ax, ay = triangle[0]
bx, by = triangle[1]
cx, cy = triangle[2]
# Segment A to B
side_1 = (x - bx) * (ay - by) - (ax - bx) * (y - by)
# Segment B to C
side_2 = (x - cx) * (by - cy) - (bx - cx) * (y - cy)
# Segment C to A
side_3 = (x - ax) * (cy - ay) - (cx - ax) * (y - ay)
# All the signs must be positive or all negative
return (side_1 < 0.0) == (side_2 < 0.0) == (side_3 < 0.0)
###############################################################################
class TestPointInTriangle(unittest.TestCase):
triangle = ((22 , 8),
(12 , 55),
(7 , 19))
def test_inside(self):
point = (15, 20)
self.assertTrue(point_in_triangle(point, self.triangle))
def test_outside(self):
point = (1, 7)
self.assertFalse(point_in_triangle(point, self.triangle))
def test_border_case(self):
"""If the point is exactly on one of the triangle's edges,
we consider it is inside."""
point = (7, 19)
self.assertTrue(point_in_triangle(point, self.triangle))
###############################################################################
if __name__ == "__main__":
suite = unittest.defaultTestLoader.loadTestsFromTestCase(TestPointInTriangle)
unittest.TextTestRunner().run(suite)
上面的算法有一个额外的可选图形测试,以确认其有效性:
import random
from matplotlib import pyplot
from triangle_test import point_in_triangle
###############################################################################
# The area #
size_x = 64
size_y = 64
# The triangle #
triangle = ((22 , 8),
(12 , 55),
(7 , 19))
# Number of random points #
count_points = 10000
# Prepare the figure #
figure = pyplot.figure()
axes = figure.add_subplot(111, aspect='equal')
axes.set_title("Test the 'point_in_triangle' function")
axes.set_xlim(0, size_x)
axes.set_ylim(0, size_y)
# Plot the triangle #
from matplotlib.patches import Polygon
axes.add_patch(Polygon(triangle, linewidth=1, edgecolor='k', facecolor='none'))
# Plot the points #
for i in range(count_points):
x = random.uniform(0, size_x)
y = random.uniform(0, size_y)
if point_in_triangle((x,y), triangle): pyplot.plot(x, y, '.g')
else: pyplot.plot(x, y, '.b')
# Save it #
figure.savefig("point_in_triangle.pdf")
制作以下图表: