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
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
其他回答
这是Javascript中最简单的完整代码。
(X, y)是目标点(x1, y)到(x2, y)是线段。
更新:修复了评论中0长度的行问题。
function pDistance(x, y, x1, y1, x2, y2) {
var A = x - x1;
var B = y - y1;
var C = x2 - x1;
var D = y2 - y1;
var dot = A * C + B * D;
var len_sq = C * C + D * D;
var param = -1;
if (len_sq != 0) //in case of 0 length line
param = dot / len_sq;
var xx, yy;
if (param < 0) {
xx = x1;
yy = y1;
}
else if (param > 1) {
xx = x2;
yy = y2;
}
else {
xx = x1 + param * C;
yy = y1 + param * D;
}
var dx = x - xx;
var dy = y - yy;
return Math.sqrt(dx * dx + dy * dy);
}
更新:Kotlin版本
fun getDistance(x: Double, y: Double, x1: Double, y1: Double, x2: Double, y2: Double): Double {
val a = x - x1
val b = y - y1
val c = x2 - x1
val d = y2 - y1
val lenSq = c * c + d * d
val param = if (lenSq != .0) { //in case of 0 length line
val dot = a * c + b * d
dot / lenSq
} else {
-1.0
}
val (xx, yy) = when {
param < 0 -> x1 to y1
param > 1 -> x2 to y2
else -> x1 + param * c to y1 + param * d
}
val dx = x - xx
val dy = y - yy
return hypot(dx, dy)
}
以下是Grumdrig解决方案的一个更完整的说明。这个版本还返回最近的点本身。
#include "stdio.h"
#include "math.h"
class Vec2
{
public:
float _x;
float _y;
Vec2()
{
_x = 0;
_y = 0;
}
Vec2( const float x, const float y )
{
_x = x;
_y = y;
}
Vec2 operator+( const Vec2 &v ) const
{
return Vec2( this->_x + v._x, this->_y + v._y );
}
Vec2 operator-( const Vec2 &v ) const
{
return Vec2( this->_x - v._x, this->_y - v._y );
}
Vec2 operator*( const float f ) const
{
return Vec2( this->_x * f, this->_y * f );
}
float DistanceToSquared( const Vec2 p ) const
{
const float dX = p._x - this->_x;
const float dY = p._y - this->_y;
return dX * dX + dY * dY;
}
float DistanceTo( const Vec2 p ) const
{
return sqrt( this->DistanceToSquared( p ) );
}
float DotProduct( const Vec2 p ) const
{
return this->_x * p._x + this->_y * p._y;
}
};
// return minimum distance between line segment vw and point p, and the closest point on the line segment, q
float DistanceFromLineSegmentToPoint( const Vec2 v, const Vec2 w, const Vec2 p, Vec2 * const q )
{
const float distSq = v.DistanceToSquared( w ); // i.e. |w-v|^2 ... avoid a sqrt
if ( distSq == 0.0 )
{
// v == w case
(*q) = v;
return v.DistanceTo( p );
}
// consider the line extending the segment, parameterized as v + t (w - v)
// we find projection of point p onto the line
// it falls where t = [(p-v) . (w-v)] / |w-v|^2
const float t = ( p - v ).DotProduct( w - v ) / distSq;
if ( t < 0.0 )
{
// beyond the v end of the segment
(*q) = v;
return v.DistanceTo( p );
}
else if ( t > 1.0 )
{
// beyond the w end of the segment
(*q) = w;
return w.DistanceTo( p );
}
// projection falls on the segment
const Vec2 projection = v + ( ( w - v ) * t );
(*q) = projection;
return p.DistanceTo( projection );
}
float DistanceFromLineSegmentToPoint( float segmentX1, float segmentY1, float segmentX2, float segmentY2, float pX, float pY, float *qX, float *qY )
{
Vec2 q;
float distance = DistanceFromLineSegmentToPoint( Vec2( segmentX1, segmentY1 ), Vec2( segmentX2, segmentY2 ), Vec2( pX, pY ), &q );
(*qX) = q._x;
(*qY) = q._y;
return distance;
}
void TestDistanceFromLineSegmentToPoint( float segmentX1, float segmentY1, float segmentX2, float segmentY2, float pX, float pY )
{
float qX;
float qY;
float d = DistanceFromLineSegmentToPoint( segmentX1, segmentY1, segmentX2, segmentY2, pX, pY, &qX, &qY );
printf( "line segment = ( ( %f, %f ), ( %f, %f ) ), p = ( %f, %f ), distance = %f, q = ( %f, %f )\n",
segmentX1, segmentY1, segmentX2, segmentY2, pX, pY, d, qX, qY );
}
void TestDistanceFromLineSegmentToPoint()
{
TestDistanceFromLineSegmentToPoint( 0, 0, 1, 1, 1, 0 );
TestDistanceFromLineSegmentToPoint( 0, 0, 20, 10, 5, 4 );
TestDistanceFromLineSegmentToPoint( 0, 0, 20, 10, 30, 15 );
TestDistanceFromLineSegmentToPoint( 0, 0, 20, 10, -30, 15 );
TestDistanceFromLineSegmentToPoint( 0, 0, 10, 0, 5, 1 );
TestDistanceFromLineSegmentToPoint( 0, 0, 0, 10, 1, 5 );
}
c#版本
public static FP DistanceToLineSegment(FPVector3 a, FPVector3 b, FPVector3 point)
{
var d = b - a;
var s = d.SqrMagnitude;
var ds = d / s;
var lambda = FPVector3.Dot(point - a, ds);
var p = FPMath.Clamp01(lambda) * d;
return (a + p - point).Magnitude;
}
用Matlab直接实现Grumdrig
function ans=distP2S(px,py,vx,vy,wx,wy)
% [px py vx vy wx wy]
t=( (px-vx)*(wx-vx)+(py-vy)*(wy-vy) )/idist(vx,wx,vy,wy)^2;
[idist(px,vx,py,vy) idist(px,vx+t*(wx-vx),py,vy+t*(wy-vy)) idist(px,wx,py,wy) ];
ans(1+(t>0)+(t>1)); % <0 0<=t<=1 t>1
end
function d=idist(a,b,c,d)
d=abs(a-b+1i*(c-d));
end
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
