找到点和线段之间的最短距离(不是线)

Question

我已经设置线段(未线),(A1, B1),(A2, B2),(A3, B3),其中A,B被结束的线段的点.每个A并B有(x,y)坐标.

问: 我想知道之间的最短距离point O,并line segments如所示所示图中的代码行实施.我真正理解的代码是伪代码或Python.

代码:我尝试使用此代码解决问题,遗憾的是,它无法正常工作.

def dist(A, B, O):
    A_ = complex(*A)
    B_ = complex(*B)
    O_= complex(*O)
    OA = O_ - A_
    OB = O_ - B_
    return min(OA, OB)
# coordinates are given
A1, B1 = [1, 8], [6,4]
A2, B2 = [3,1], [5,2]
A3, B3 = [2,3], [2, 1]
O = [2, 5]
A = [A1, A2, A3]
B = [B1, B2, B3]
print [ dist(i, j, O)  for i, j in zip(A, B)]

提前致谢.

Answer 1

您可以将这些操作向量化并获得更好的性能，而不是使用 for 循环。这是我的解决方案，它允许您使用矢量化计算来计算从单个点到多个线段的距离。

def lineseg_dists(p, a, b):
    """Cartesian distance from point to line segment

    Edited to support arguments as series, from:
    /sf/answers/3810979301/

    Args:
        - p: np.array of single point, shape (2,) or 2D array, shape (x, 2)
        - a: np.array of shape (x, 2)
        - b: np.array of shape (x, 2)
    """
    # normalized tangent vectors
    d_ba = b - a
    d = np.divide(d_ba, (np.hypot(d_ba[:, 0], d_ba[:, 1])
                           .reshape(-1, 1)))

    # signed parallel distance components
    # rowwise dot products of 2D vectors
    s = np.multiply(a - p, d).sum(axis=1)
    t = np.multiply(p - b, d).sum(axis=1)

    # clamped parallel distance
    h = np.maximum.reduce([s, t, np.zeros(len(s))])

    # perpendicular distance component
    # rowwise cross products of 2D vectors  
    d_pa = p - a
    c = d_pa[:, 0] * d[:, 1] - d_pa[:, 1] * d[:, 0]

    return np.hypot(h, c)

还有一些测试：

p = np.array([0, 0])
a = np.array([[ 1,  1],
              [-1,  0],
              [-1, -1]])
b = np.array([[ 2,  2],
              [ 1,  0],
              [ 1, -1]])

print(lineseg_dists(p, a, b))

p = np.array([[0, 0],
              [1, 1],
              [0, 2]])

print(lineseg_dists(p, a, b))

>>> [1.41421356 0.         1.        ]
    [1.41421356 1.         3.        ]

Answer 2

这是答案。此代码属于Malcolm Kesson，其源代码在此处。我之前只提供了链接本身，但主持人将其删除。我认为这样做的原因是因为没有提供代码（作为答案）。

import math

def dot(v,w):
    x,y,z = v
    X,Y,Z = w
    return x*X + y*Y + z*Z

def length(v):
    x,y,z = v
    return math.sqrt(x*x + y*y + z*z)

def vector(b,e):
    x,y,z = b
    X,Y,Z = e
    return (X-x, Y-y, Z-z)

def unit(v):
    x,y,z = v
    mag = length(v)
    return (x/mag, y/mag, z/mag)

def distance(p0,p1):
    return length(vector(p0,p1))

def scale(v,sc):
    x,y,z = v
    return (x * sc, y * sc, z * sc)

def add(v,w):
    x,y,z = v
    X,Y,Z = w
    return (x+X, y+Y, z+Z)


# Given a line with coordinates 'start' and 'end' and the
# coordinates of a point 'pnt' the proc returns the shortest 
# distance from pnt to the line and the coordinates of the 
# nearest point on the line.
#
# 1  Convert the line segment to a vector ('line_vec').
# 2  Create a vector connecting start to pnt ('pnt_vec').
# 3  Find the length of the line vector ('line_len').
# 4  Convert line_vec to a unit vector ('line_unitvec').
# 5  Scale pnt_vec by line_len ('pnt_vec_scaled').
# 6  Get the dot product of line_unitvec and pnt_vec_scaled ('t').
# 7  Ensure t is in the range 0 to 1.
# 8  Use t to get the nearest location on the line to the end
#    of vector pnt_vec_scaled ('nearest').
# 9  Calculate the distance from nearest to pnt_vec_scaled.
# 10 Translate nearest back to the start/end line. 
# Malcolm Kesson 16 Dec 2012

def pnt2line(pnt, start, end):
    line_vec = vector(start, end)
    pnt_vec = vector(start, pnt)
    line_len = length(line_vec)
    line_unitvec = unit(line_vec)
    pnt_vec_scaled = scale(pnt_vec, 1.0/line_len)
    t = dot(line_unitvec, pnt_vec_scaled)    
    if t < 0.0:
        t = 0.0
    elif t > 1.0:
        t = 1.0
    nearest = scale(line_vec, t)
    dist = distance(nearest, pnt_vec)
    nearest = add(nearest, start)
    return (dist, nearest)