判断点在四边形内（Python实现）

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