判断点在四边形内的算法，在网上流传的方法大多是适合矩形，例如链接，并不适合所有的凸四边形。
先说对于矩形常用的方法，也是网上流传比较广的一种方法，是判断点与四边形的边的夹角是锐角，也即点积>0。实现如下：
def point_in_box_rect2(point, corners):
"""check if a point lies in a rectangle defined by corners. un-support quadrangle
idea: check projection
Args:
point (2,): coordinate of point
corners (4, 2): coordinate of corners
Returns:
True if point in box
"""
assert corners.shape == (4, 2)
a = corners[0, :]
b = corners[1, :]
c = corners[2, :]
d = corners[3, :]
ab = b - a
am = point - a
bc = c - b
bm = point - b
cd = d - c
cm = point - c
da = a - d
dm = point - d
p_ab = np.dot(ab, am)
p_bc = np.dot(bc, bm)
p_cd = np.dot(cd, cm)
p_da = np.dot(da, dm)
cond1 = p_ab > 0 and p_bc > 0 and p_cd > 0 and p_da > 0
cond2 = p_ab < 0 and p_bc < 0 and p_cd < 0 and p_da < 0
return cond1 or cond2以上方法不适用平行四边形的情况，例如：
?对于更一般的情况，可以将四边形考虑成2个三角形，判断点是不是在三角形其中之一内即可，具体参考链接。实现如下：
def sign(p1, p2, p3):
return (p1[0] - p3[0]) * (p2[1] - p3[1]) - \
(p2[0] - p3[0]) * (p1[1] - p3[1])
def point_in_triangle(point, corners):
d1 = sign(point, corners[0, :], corners[1, :])
d2 = sign(point, corners[1, :], corners[2, :])
d3 = sign(point, corners[2, :], corners[0, :])
has_neg = (d1 < 0) or (d2 < 0) or (d3 < 0)
has_pos = (d1 > 0) or (d2 > 0) or (d3 > 0)
return not(has_neg and has_pos)
def point_in_box(point, corners):
triangle1 = corners[:3, :]
ind = [2,3,0]
triangle2 = corners[ind, :]
is_in1 = point_in_triangle(point, triangle1)
is_in2 = point_in_triangle(point, triangle2)
return is_in1 or is_in2