如何确定一个点是否在二维三角形中?

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

我在JavaScript中改编的高性能代码(文章如下):

function pointInTriangle (p, p0, p1, p2) {
  return (((p1.y - p0.y) * (p.x - p0.x) - (p1.x - p0.x) * (p.y - p0.y)) | ((p2.y - p1.y) * (p.x - p1.x) - (p2.x - p1.x) * (p.y - p1.y)) | ((p0.y - p2.y) * (p.x - p2.x) - (p0.x - p2.x) * (p.y - p2.y))) >= 0;
}

pointInTriangle(p, p0, p1, p2) -用于逆时针方向的三角形 pointInTriangle(p, p0, p1, p2) -用于顺时针三角形

在jsFiddle(包括性能测试)中，在一个单独的函数中也有缠绕检查。或按下面的“运行代码片段”

受此启发: http://www.phatcode.net/articles.php?id=459

2016-05-13 23:34:09

其他回答

python中的其他函数，比Developer的方法更快(至少对我来说)，并受到Cédric Dufour解决方案的启发:

def ptInTriang(p_test, p0, p1, p2):       
     dX = p_test[0] - p0[0]
     dY = p_test[1] - p0[1]
     dX20 = p2[0] - p0[0]
     dY20 = p2[1] - p0[1]
     dX10 = p1[0] - p0[0]
     dY10 = p1[1] - p0[1]

     s_p = (dY20*dX) - (dX20*dY)
     t_p = (dX10*dY) - (dY10*dX)
     D = (dX10*dY20) - (dY10*dX20)

     if D > 0:
         return (  (s_p >= 0) and (t_p >= 0) and (s_p + t_p) <= D  )
     else:
         return (  (s_p <= 0) and (t_p <= 0) and (s_p + t_p) >= D  )

你可以用:

X_size = 64
Y_size = 64
ax_x = np.arange(X_size).astype(np.float32)
ax_y = np.arange(Y_size).astype(np.float32)
coords=np.meshgrid(ax_x,ax_y)
points_unif = (coords[0].reshape(X_size*Y_size,),coords[1].reshape(X_size*Y_size,))
p_test = np.array([0 , 0])
p0 = np.array([22 , 8]) 
p1 = np.array([12 , 55]) 
p2 = np.array([7 , 19]) 
fig = plt.figure(dpi=300)
for i in range(0,X_size*Y_size):
    p_test[0] = points_unif[0][i]
    p_test[1] = points_unif[1][i]
    if ptInTriang(p_test, p0, p1, p2):
        plt.plot(p_test[0], p_test[1], '.g')
    else:
        plt.plot(p_test[0], p_test[1], '.r')

绘制网格需要花费很多时间，但是该网格在0.0195319652557秒内测试，而开发人员代码为0.0844349861145秒。

最后是代码注释:

# Using barycentric coordintes, any point inside can be described as:
# X = p0.x * r + p1.x * s + p2.x * t
# Y = p0.y * r + p1.y * s + p2.y * t
# with:
# r + s + t = 1  and 0 < r,s,t < 1
# then: r = 1 - s - t
# and then:
# X = p0.x * (1 - s - t) + p1.x * s + p2.x * t
# Y = p0.y * (1 - s - t) + p1.y * s + p2.y * t
#
# X = p0.x + (p1.x-p0.x) * s + (p2.x-p0.x) * t
# Y = p0.y + (p1.y-p0.y) * s + (p2.y-p0.y) * t
#
# X - p0.x = (p1.x-p0.x) * s + (p2.x-p0.x) * t
# Y - p0.y = (p1.y-p0.y) * s + (p2.y-p0.y) * t
#
# we have to solve:
#
# [ X - p0.x ] = [(p1.x-p0.x)   (p2.x-p0.x)] * [ s ]
# [ Y - p0.Y ]   [(p1.y-p0.y)   (p2.y-p0.y)]   [ t ]
#
# ---> b = A*x ; ---> x = A^-1 * b
# 
# [ s ] =   A^-1  * [ X - p0.x ]
# [ t ]             [ Y - p0.Y ]
#
# A^-1 = 1/D * adj(A)
#
# The adjugate of A:
#
# adj(A)   =   [(p2.y-p0.y)   -(p2.x-p0.x)]
#              [-(p1.y-p0.y)   (p1.x-p0.x)]
#
# The determinant of A:
#
# D = (p1.x-p0.x)*(p2.y-p0.y) - (p1.y-p0.y)*(p2.x-p0.x)
#
# Then:
#
# s_p = { (p2.y-p0.y)*(X - p0.x) - (p2.x-p0.x)*(Y - p0.Y) }
# t_p = { (p1.x-p0.x)*(Y - p0.Y) - (p1.y-p0.y)*(X - p0.x) }
#
# s = s_p / D
# t = t_p / D
#
# Recovering r:
#
# r = 1 - (s_p + t_p)/D
#
# Since we only want to know if it is insidem not the barycentric coordinate:
#
# 0 < 1 - (s_p + t_p)/D < 1
# 0 < (s_p + t_p)/D < 1
# 0 < (s_p + t_p) < D
#
# The condition is:
# if D > 0:
#     s_p > 0 and t_p > 0 and (s_p + t_p) < D
# else:
#     s_p < 0 and t_p < 0 and (s_p + t_p) > D
#
# s_p = { dY20*dX - dX20*dY }
# t_p = { dX10*dY - dY10*dX }
# D = dX10*dY20 - dY10*dX20

2017-09-25 16:13:20

我同意Andreas Brinck的观点，重心坐标对于这项任务来说非常方便。注意，不需要每次都求解一个方程组:只需计算解析解。使用Andreas的符号，解是:

s = 1/(2*Area)*(p0y*p2x - p0x*p2y + (p2y - p0y)*px + (p0x - p2x)*py);
t = 1/(2*Area)*(p0x*p1y - p0y*p1x + (p0y - p1y)*px + (p1x - p0x)*py);

其中Area是三角形的(带符号的)面积:

Area = 0.5 *(-p1y*p2x + p0y*(-p1x + p2x) + p0x*(p1y - p2y) + p1x*p2y);

只计算st和1-s-t。点p在三角形内当且仅当它们都是正的。

编辑:请注意，上面的区域表达式假设三角形节点编号是逆时针方向的。如果编号是顺时针的，这个表达式将返回一个负的面积(但大小正确)。然而，测试本身(s>0 && t>0 && 1-s-t>0)并不依赖于编号的方向，因为如果三角形节点的方向改变，上面乘以1/(2*Area)的表达式也会改变符号。

编辑2:为了获得更好的计算效率，请参阅下面的coproc注释(其中指出，如果三角形节点的方向(顺时针或逆时针)事先已知，则可以避免在s和t的表达式中除以2*Area)。在Andreas Brinck的回答下面的评论中也可以看到Perro Azul的jsfiddle-code。

2013-01-17 15:42:17

由andreasdr和Perro Azul发布的重心方法的c#版本。我添加了一个检查，当s和t有相反的符号(而且都不为零)时，放弃面积计算，因为潜在地避免三分之一的乘法成本似乎是合理的。

public static bool PointInTriangle(Point p, Point p0, Point p1, Point p2)
{
    var s = (p0.X - p2.X) * (p.Y - p2.Y) - (p0.Y - p2.Y) * (p.X - p2.X);
    var t = (p1.X - p0.X) * (p.Y - p0.Y) - (p1.Y - p0.Y) * (p.X - p0.X);

    if ((s < 0) != (t < 0) && s != 0 && t != 0)
        return false;

    var d = (p2.X - p1.X) * (p.Y - p1.Y) - (p2.Y - p1.Y) * (p.X - p1.X);
    return d == 0 || (d < 0) == (s + t <= 0);
}

2021年更新:这个版本正确处理任意一个缠绕方向(顺时针和逆时针)指定的三角形。请注意，对于恰好位于三角形边缘上的点，本页上的一些其他答案会给出不一致的结果，这取决于三角形三个点的排列顺序。这些点被认为是“在”三角形中，这段代码正确地返回true，而不管缠绕方向如何。

2013-12-31 17:34:51

If you know the co-ordinates of the three vertices and the co-ordinates of the specific point, then you can get the area of the complete triangle. Afterwards, calculate the area of the three triangle segments (one point being the point given and the other two being any two vertices of the triangle). Thus, you will get the area of the three triangle segments. If the sum of these areas are equal to the total area (that you got previously), then, the point should be inside the triangle. Otherwise, the point is not inside the triangle. This should work. If there are any issues, let me know. Thank you.

2014-08-03 18:56:04

一个简单的方法是:

找出连接分别指向三角形的三个点顶点和夹角之和这些向量。如果它们的和角度是2*那么点是在三角形里面。

两个解释替代方案的好网站是:

黑卒和沃尔夫勒姆

2010-01-12 14:31:40

如何确定一个点是否在二维三角形中?

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