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
这个答案是基于公认答案的JavaScript解决方案。 它主要只是格式更好,函数名更长,当然函数语法更短,因为它是在ES6 + CoffeeScript中。
JavaScript版本(ES6)
distanceSquared = (v, w)=> Math.pow(v.x - w.x, 2) + Math.pow(v.y - w.y, 2);
distance = (v, w)=> Math.sqrt(distanceSquared(v, w));
distanceToLineSegmentSquared = (p, v, w)=> {
l2 = distanceSquared(v, w);
if (l2 === 0) {
return distanceSquared(p, v);
}
t = ((p.x - v.x) * (w.x - v.x) + (p.y - v.y) * (w.y - v.y)) / l2;
t = Math.max(0, Math.min(1, t));
return distanceSquared(p, {
x: v.x + t * (w.x - v.x),
y: v.y + t * (w.y - v.y)
});
}
distanceToLineSegment = (p, v, w)=> {
return Math.sqrt(distanceToLineSegmentSquared(p, v));
}
CoffeeScript版本
distanceSquared = (v, w)-> (v.x - w.x) ** 2 + (v.y - w.y) ** 2
distance = (v, w)-> Math.sqrt(distanceSquared(v, w))
distanceToLineSegmentSquared = (p, v, w)->
l2 = distanceSquared(v, w)
return distanceSquared(p, v) if l2 is 0
t = ((p.x - v.x) * (w.x - v.x) + (p.y - v.y) * (w.y - v.y)) / l2
t = Math.max(0, Math.min(1, t))
distanceSquared(p, {
x: v.x + t * (w.x - v.x)
y: v.y + t * (w.y - v.y)
})
distanceToLineSegment = (p, v, w)->
Math.sqrt(distanceToLineSegmentSquared(p, v, w))
其他回答
下面是HSQLDB的SQL实现:
CREATE FUNCTION dist_to_segment(px double, py double, vx double, vy double, wx double, wy double)
RETURNS double
BEGIN atomic
declare l2 double;
declare t double;
declare nx double;
declare ny double;
set l2 =(vx - wx)*(vx - wx) + (vy - wy)*(vy - wy);
IF l2 = 0 THEN
RETURN sqrt((vx - px)*(vx - px) + (vy - py)*(vy - py));
ELSE
set t = ((px - vx) * (wx - vx) + (py - vy) * (wy - vy)) / l2;
set t = GREATEST(0, LEAST(1, t));
set nx=vx + t * (wx - vx);
set ny=vy + t * (wy - vy);
RETURN sqrt((nx - px)*(nx - px) + (ny - py)*(ny - py));
END IF;
END;
Postgres的实现:
CREATE FUNCTION dist_to_segment(px numeric, py numeric, vx numeric, vy numeric, wx numeric, wy numeric)
RETURNS numeric
AS $$
declare l2 numeric;
declare t numeric;
declare nx numeric;
declare ny numeric;
BEGIN
l2 := (vx - wx)*(vx - wx) + (vy - wy)*(vy - wy);
IF l2 = 0 THEN
RETURN sqrt((vx - px)*(vx - px) + (vy - py)*(vy - py));
ELSE
t := ((px - vx) * (wx - vx) + (py - vy) * (wy - vy)) / l2;
t := GREATEST(0, LEAST(1, t));
nx := vx + t * (wx - vx);
ny := vy + t * (wy - vy);
RETURN sqrt((nx - px)*(nx - px) + (ny - py)*(ny - py));
END IF;
END;
$$ LANGUAGE plpgsql;
对于感兴趣的人,这里是Joshua的Javascript代码到Objective-C的简单转换:
- (double)distanceToPoint:(CGPoint)p fromLineSegmentBetween:(CGPoint)l1 and:(CGPoint)l2
{
double A = p.x - l1.x;
double B = p.y - l1.y;
double C = l2.x - l1.x;
double D = l2.y - l1.y;
double dot = A * C + B * D;
double len_sq = C * C + D * D;
double param = dot / len_sq;
double xx, yy;
if (param < 0 || (l1.x == l2.x && l1.y == l2.y)) {
xx = l1.x;
yy = l1.y;
}
else if (param > 1) {
xx = l2.x;
yy = l2.y;
}
else {
xx = l1.x + param * C;
yy = l1.y + param * D;
}
double dx = p.x - xx;
double dy = p.y - yy;
return sqrtf(dx * dx + dy * dy);
}
我需要这个解决方案与MKMapPoint一起工作,所以我将分享它,以防其他人需要它。只是一些小的改变,这将返回米为单位的距离:
- (double)distanceToPoint:(MKMapPoint)p fromLineSegmentBetween:(MKMapPoint)l1 and:(MKMapPoint)l2
{
double A = p.x - l1.x;
double B = p.y - l1.y;
double C = l2.x - l1.x;
double D = l2.y - l1.y;
double dot = A * C + B * D;
double len_sq = C * C + D * D;
double param = dot / len_sq;
double xx, yy;
if (param < 0 || (l1.x == l2.x && l1.y == l2.y)) {
xx = l1.x;
yy = l1.y;
}
else if (param > 1) {
xx = l2.x;
yy = l2.y;
}
else {
xx = l1.x + param * C;
yy = l1.y + param * D;
}
return MKMetersBetweenMapPoints(p, MKMapPointMake(xx, yy));
}
GLSL版:
// line (a -> b ) point p[enter image description here][1]
float distanceToLine(vec2 a, vec2 b, vec2 p) {
float aside = dot((p - a),(b - a));
if(aside< 0.0) return length(p-a);
float bside = dot((p - b),(a - b));
if(bside< 0.0) return length(p-b);
vec2 pointOnLine = (bside*a + aside*b)/pow(length(a-b),2.0);
return length(p - pointOnLine);
}
在f#中,点c到a和b之间的线段的距离为:
let pointToLineSegmentDistance (a: Vector, b: Vector) (c: Vector) =
let d = b - a
let s = d.Length
let lambda = (c - a) * d / s
let p = (lambda |> max 0.0 |> min s) * d / s
(a + p - c).Length
向量d沿着线段从a指向b。d/s与c-a的点积给出了无限直线与点c之间最接近点的参数。使用min和max函数将该参数钳制到范围0..s,使该点位于a和b之间。最后,a+p-c的长度是c到线段上最近点的距离。
使用示例:
pointToLineSegmentDistance (Vector(0.0, 0.0), Vector(1.0, 0.0)) (Vector(-1.0, 1.0))
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 );
}