问题
I have set of line segments (not lines), (A1, B1), (A2, B2), (A3, B3), where A,B are ending points of the line segment. Each A and B has (x,y) coordinates.
QUESTION:
I need to know the shortest distance between point O and line segments as shown in the shown figure implemented in line of codes. The code I can really understand is either pseudo-code or Python.
CODE: I tried to solve the problem with this code, unfortunately, it does not work properly.
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)]
Thanks in advance.
回答1:
Here is the answer. This code belongs to Malcolm Kesson, the source is here. I provided it before with just link itself and it was deleted by the moderator. I assume that the reason for that is because of not providing the code (as an answer).
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)
回答2:
Rather than using a for loop, you can vectorize these operations and get much better performance. Here is my solution that allows you to compute the distance from a single point to multiple line segments with vectorized computation.
def lineseg_dists(p, a, b):
"""Cartesian distance from point to line segment
Edited to support arguments as series, from:
https://stackoverflow.com/a/54442561/11208892
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)
And some tests:
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. ]
回答3:
The explanation is in the docstring of this function:
def point_to_line_dist(point, line):
"""Calculate the distance between a point and a line segment.
To calculate the closest distance to a line segment, we first need to check
if the point projects onto the line segment. If it does, then we calculate
the orthogonal distance from the point to the line.
If the point does not project to the line segment, we calculate the
distance to both endpoints and take the shortest distance.
:param point: Numpy array of form [x,y], describing the point.
:type point: numpy.core.multiarray.ndarray
:param line: list of endpoint arrays of form [P1, P2]
:type line: list of numpy.core.multiarray.ndarray
:return: The minimum distance to a point.
:rtype: float
"""
# unit vector
unit_line = line[1] - line[0]
norm_unit_line = unit_line / np.linalg.norm(unit_line)
# compute the perpendicular distance to the theoretical infinite line
segment_dist = (
np.linalg.norm(np.cross(line[1] - line[0], line[0] - point)) /
np.linalg.norm(unit_line)
)
diff = (
(norm_unit_line[0] * (point[0] - line[0][0])) +
(norm_unit_line[1] * (point[1] - line[0][1]))
)
x_seg = (norm_unit_line[0] * diff) + line[0][0]
y_seg = (norm_unit_line[1] * diff) + line[0][1]
endpoint_dist = min(
np.linalg.norm(line[0] - point),
np.linalg.norm(line[1] - point)
)
# decide if the intersection point falls on the line segment
lp1_x = line[0][0] # line point 1 x
lp1_y = line[0][1] # line point 1 y
lp2_x = line[1][0] # line point 2 x
lp2_y = line[1][1] # line point 2 y
is_betw_x = lp1_x <= x_seg <= lp2_x or lp2_x <= x_seg <= lp1_x
is_betw_y = lp1_y <= y_seg <= lp2_y or lp2_y <= y_seg <= lp1_y
if is_betw_x and is_betw_y:
return segment_dist
else:
# if not, then return the minimum distance to the segment endpoints
return endpoint_dist
回答4:
Basic algorithm: pretend that you have lines, so oriented that A lies to the left of B when O lies above the line (mentally rotate the picture to match as needed).
Find closest point as normal. If the point is between A and B, you're done. If it's to the left of A, the closest point is A. If the point is to the right of B, the closest point is B.
The case when A, B, and O all lie on the same line may or may not need special attention. Be sure to include a few tests of this position.
回答5:
On my side, I've found that the two other answers were broken, especially when the line is purely vertical or horizontal. Here is what I did to properly solve the problem.
Python code:
def sq_shortest_dist_to_point(self, other_point):
dx = self.b.x - self.a.x
dy = self.b.y - self.a.y
dr2 = float(dx ** 2 + dy ** 2)
lerp = ((other_point.x - self.a.x) * dx + (other_point.y - self.a.y) * dy) / dr2
if lerp < 0:
lerp = 0
elif lerp > 1:
lerp = 1
x = lerp * dx + self.a.x
y = lerp * dy + self.a.y
_dx = x - other_point.x
_dy = y - other_point.y
square_dist = _dx ** 2 + _dy ** 2
return square_dist
def shortest_dist_to_point(self, other_point):
return math.sqrt(self.sq_shortest_dist_to_point(other_point))
A test case:
def test_distance_to_other_point(self):
# Parametrize test with multiple cases:
segments_and_point_and_answer = [
[Segment(Point(1.0, 1.0), Point(1.0, 3.0)), Point(2.0, 4.0), math.sqrt(2.0)],
[Segment(Point(1.0, 1.0), Point(1.0, 3.0)), Point(2.0, 3.0), 1.0],
[Segment(Point(0.0, 0.0), Point(0.0, 3.0)), Point(1.0, 1.0), 1.0],
[Segment(Point(1.0, 1.0), Point(3.0, 3.0)), Point(2.0, 2.0), 0.0],
[Segment(Point(-1.0, -1.0), Point(3.0, 3.0)), Point(2.0, 2.0), 0.0],
[Segment(Point(1.0, 1.0), Point(1.0, 3.0)), Point(2.0, 3.0), 1.0],
[Segment(Point(1.0, 1.0), Point(1.0, 3.0)), Point(2.0, 4.0), math.sqrt(2.0)],
[Segment(Point(1.0, 1.0), Point(-3.0, -3.0)), Point(-3.0, -4.0), 1],
[Segment(Point(1.0, 1.0), Point(-3.0, -3.0)), Point(-4.0, -3.0), 1],
[Segment(Point(1.0, 1.0), Point(-3.0, -3.0)), Point(1, 2), 1],
[Segment(Point(1.0, 1.0), Point(-3.0, -3.0)), Point(2, 1), 1],
[Segment(Point(1.0, 1.0), Point(-3.0, -3.0)), Point(-3, -1), math.sqrt(2.0)],
[Segment(Point(1.0, 1.0), Point(-3.0, -3.0)), Point(-1, -3), math.sqrt(2.0)],
[Segment(Point(-1.0, -1.0), Point(3.0, 3.0)), Point(3, 1), math.sqrt(2.0)],
[Segment(Point(-1.0, -1.0), Point(3.0, 3.0)), Point(1, 3), math.sqrt(2.0)],
[Segment(Point(1.0, 1.0), Point(3.0, 3.0)), Point(3, 1), math.sqrt(2.0)],
[Segment(Point(1.0, 1.0), Point(3.0, 3.0)), Point(1, 3), math.sqrt(2.0)]
]
for i, (segment, point, answer) in enumerate(segments_and_point_and_answer):
result = segment.shortest_dist_to_point(point)
self.assertAlmostEqual(result, answer, delta=0.001, msg=str((i, segment, point, answer)))
Note: I assume this function is inside a Segment class.
In case your line is infinite, don't limit the lerp from 0 to 1 only, but still at least provide two distinct a and b points.
回答6:
I had to solve this problem as well, so for the sake of availability I'll post my code here. I did some cursory validation but nothing particularly serious. Your question actually helped me identify a bug in mine where a vertical or horizontal line segment would have broken the code and bypassed the intersection point on segment logic.
from math import sqrt
def dist_to_segment(ax, ay, bx, by, cx, cy):
"""
Computes the minimum distance between a point (cx, cy) and a line segment with endpoints (ax, ay) and (bx, by).
:param ax: endpoint 1, x-coordinate
:param ay: endpoint 1, y-coordinate
:param bx: endpoint 2, x-coordinate
:param by: endpoint 2, y-coordinate
:param cx: point, x-coordinate
:param cy: point, x-coordinate
:return: minimum distance between point and line segment
"""
# avoid divide by zero error
a = max(by - ay, 0.00001)
b = max(ax - bx, 0.00001)
# compute the perpendicular distance to the theoretical infinite line
dl = abs(a * cx + b * cy - b * ay - a * ax) / sqrt(a**2 + b**2)
# compute the intersection point
x = ((a / b) * ax + ay + (b / a) * cx - cy) / ((b / a) + (a / b))
y = -1 * (a / b) * (x - ax) + ay
# decide if the intersection point falls on the line segment
if (ax <= x <= bx or bx <= x <= ax) and (ay <= y <= by or by <= y <= ay):
return dl
else:
# if it does not, then return the minimum distance to the segment endpoints
return min(sqrt((ax - cx)**2 + (ay - cy)**2), sqrt((bx - cx)**2 + (by - cy)**2))
来源：https://stackoverflow.com/questions/27161533/find-the-shortest-distance-between-a-point-and-line-segments-not-line