我需要一个基本函数来求点到线段的最短距离。你可以随意用任何你想要的语言来编写解决方案;我可以把它翻译成什么我正在使用(Javascript)。
编辑:线段由两个端点定义。线段AB由两点A (x1,y1)和B (x2,y2)定义。我要求的是这条线段到点C (x3,y3)的距离。我的几何技能生疏了,所以我看到的例子让我很困惑,我很遗憾地承认。
我需要一个基本函数来求点到线段的最短距离。你可以随意用任何你想要的语言来编写解决方案;我可以把它翻译成什么我正在使用(Javascript)。
编辑:线段由两个端点定义。线段AB由两点A (x1,y1)和B (x2,y2)定义。我要求的是这条线段到点C (x3,y3)的距离。我的几何技能生疏了,所以我看到的例子让我很困惑,我很遗憾地承认。
当前回答
我制作了一个交互式Desmos图来演示如何实现这一点:
https://www.desmos.com/calculator/kswrm8ddum
红点是A点,绿点是B点,C点是蓝色点。 您可以拖动图形中的点来查看值的变化。 左边的值“s”是线段的参数(即s = 0表示点A, s = 1表示点B)。 值“d”是第三点到经过A和B的直线的距离。
编辑:
有趣的小见解:坐标(s, d)是坐标系中第三点C的坐标,AB是单位x轴,单位y轴垂直于AB。
其他回答
只是遇到了这个,我想我应该添加一个Lua实现。它假设点以表{x=xVal, y=yVal}给出,直线或线段由包含两个点的表给出(见下面的例子):
function distance( P1, P2 )
return math.sqrt((P1.x-P2.x)^2 + (P1.y-P2.y)^2)
end
-- Returns false if the point lies beyond the reaches of the segment
function distPointToSegment( line, P )
if line[1].x == line[2].x and line[1].y == line[2].y then
print("Error: Not a line!")
return false
end
local d = distance( line[1], line[2] )
local t = ((P.x - line[1].x)*(line[2].x - line[1].x) + (P.y - line[1].y)*(line[2].y - line[1].y))/(d^2)
local projection = {}
projection.x = line[1].x + t*(line[2].x-line[1].x)
projection.y = line[1].y + t*(line[2].y-line[1].y)
if t >= 0 and t <= 1 then -- within line segment?
return distance( projection, {x=P.x, y=P.y} )
else
return false
end
end
-- Returns value even if point is further down the line (outside segment)
function distPointToLine( line, P )
if line[1].x == line[2].x and line[1].y == line[2].y then
print("Error: Not a line!")
return false
end
local d = distance( line[1], line[2] )
local t = ((P.x - line[1].x)*(line[2].x - line[1].x) + (P.y - line[1].y)*(line[2].y - line[1].y))/(d^2)
local projection = {}
projection.x = line[1].x + t*(line[2].x-line[1].x)
projection.y = line[1].y + t*(line[2].y-line[1].y)
return distance( projection, {x=P.x, y=P.y} )
end
使用示例:
local P1 = {x = 0, y = 0}
local P2 = {x = 10, y = 10}
local line = { P1, P2 }
local P3 = {x = 7, y = 15}
print(distPointToLine( line, P3 )) -- prints 5.6568542494924
print(distPointToSegment( line, P3 )) -- prints false
一个2D和3D的解决方案
考虑基底的变化,使得线段变成(0,0,0)-(d, 0,0)和点(u, v, 0)。在这个平面上,最短的距离由
u ≤ 0 -> d(A, C)
0 ≤ u ≤ d -> |v|
d ≤ u -> d(B, C)
(到其中一个端点或到支撑线的距离,取决于到该线的投影。等距轨迹由两个半圆和两条线段组成。)
式中,d为AB线段的长度,u、v分别为AB/d (AB方向的单位矢量)与AC的标量积和外积的模量。
AB.AC ≤ 0 -> |AC|
0 ≤ AB.AC ≤ AB² -> |ABxAC|/|AB|
AB² ≤ AB.AC -> |BC|
用t-sql编码
点为(@px, @py),线段从(@ax, @ay)到(@bx, @by)
create function fn_sqr (@NumberToSquare decimal(18,10))
returns decimal(18,10)
as
begin
declare @Result decimal(18,10)
set @Result = @NumberToSquare * @NumberToSquare
return @Result
end
go
create function fn_Distance(@ax decimal (18,10) , @ay decimal (18,10), @bx decimal(18,10), @by decimal(18,10))
returns decimal(18,10)
as
begin
declare @Result decimal(18,10)
set @Result = (select dbo.fn_sqr(@ax - @bx) + dbo.fn_sqr(@ay - @by) )
return @Result
end
go
create function fn_DistanceToSegmentSquared(@px decimal(18,10), @py decimal(18,10), @ax decimal(18,10), @ay decimal(18,10), @bx decimal(18,10), @by decimal(18,10))
returns decimal(18,10)
as
begin
declare @l2 decimal(18,10)
set @l2 = (select dbo.fn_Distance(@ax, @ay, @bx, @by))
if @l2 = 0
return dbo.fn_Distance(@px, @py, @ax, @ay)
declare @t decimal(18,10)
set @t = ((@px - @ax) * (@bx - @ax) + (@py - @ay) * (@by - @ay)) / @l2
if (@t < 0)
return dbo.fn_Distance(@px, @py, @ax, @ay);
if (@t > 1)
return dbo.fn_Distance(@px, @py, @bx, @by);
return dbo.fn_Distance(@px, @py, @ax + @t * (@bx - @ax), @ay + @t * (@by - @ay))
end
go
create function fn_DistanceToSegment(@px decimal(18,10), @py decimal(18,10), @ax decimal(18,10), @ay decimal(18,10), @bx decimal(18,10), @by decimal(18,10))
returns decimal(18,10)
as
begin
return sqrt(dbo.fn_DistanceToSegmentSquared(@px, @py , @ax , @ay , @bx , @by ))
end
go
--example execution for distance from a point at (6,1) to line segment that runs from (4,2) to (2,1)
select dbo.fn_DistanceToSegment(6, 1, 4, 2, 2, 1)
--result = 2.2360679775
--example execution for distance from a point at (-3,-2) to line segment that runs from (0,-2) to (-2,1)
select dbo.fn_DistanceToSegment(-3, -2, 0, -2, -2, 1)
--result = 2.4961508830
--example execution for distance from a point at (0,-2) to line segment that runs from (0,-2) to (-2,1)
select dbo.fn_DistanceToSegment(0,-2, 0, -2, -2, 1)
--result = 0.0000000000
这是一个基于向量数学的;这个解决方案也适用于更高的维度,并报告交点(在线段上)。
def dist(x1,y1,x2,y2,px,py):
a = np.array([[x1,y1]]).T
b = np.array([[x2,y2]]).T
x = np.array([[px,py]]).T
tp = (np.dot(x.T, b) - np.dot(a.T, b)) / np.dot(b.T, b)
tp = tp[0][0]
tmp = x - (a + tp*b)
d = np.sqrt(np.dot(tmp.T,tmp)[0][0])
return d, a+tp*b
x1,y1=2.,2.
x2,y2=5.,5.
px,py=4.,1.
d, inters = dist(x1,y1, x2,y2, px,py)
print (d)
print (inters)
结果是
2.1213203435596424
[[2.5]
[2.5]]
这里解释了数学
https://brilliant.org/wiki/distance-between-point-and-line/
2D坐标数组的Python Numpy实现:
import numpy as np
def dist2d(p1, p2, coords):
''' Distance from points to a finite line btwn p1 -> p2 '''
assert coords.ndim == 2 and coords.shape[1] == 2, 'coords is not 2 dim'
dp = p2 - p1
st = dp[0]**2 + dp[1]**2
u = ((coords[:, 0] - p1[0]) * dp[0] + (coords[:, 1] - p1[1]) * dp[1]) / st
u[u > 1.] = 1.
u[u < 0.] = 0.
dx = (p1[0] + u * dp[0]) - coords[:, 0]
dy = (p1[1] + u * dp[1]) - coords[:, 1]
return np.sqrt(dx**2 + dy**2)
# Usage:
p1 = np.array([0., 0.])
p2 = np.array([0., 10.])
# List of coordinates
coords = np.array(
[[0., 0.],
[5., 5.],
[10., 10.],
[20., 20.]
])
d = dist2d(p1, p2, coords)
# Single coordinate
coord = np.array([25., 25.])
d = dist2d(p1, p2, coord[np.newaxis, :])