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

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。

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。

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:

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。

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。