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
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;
}
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;
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
下面是devnullicus转换为c#的c++版本。对于我的实现,我需要知道交叉点,并找到他的解决方案。
public static bool PointSegmentDistanceSquared(PointF point, PointF lineStart, PointF lineEnd, out double distance, out PointF intersectPoint)
{
const double kMinSegmentLenSquared = 0.00000001; // adjust to suit. If you use float, you'll probably want something like 0.000001f
const double kEpsilon = 1.0E-14; // adjust to suit. If you use floats, you'll probably want something like 1E-7f
double dX = lineEnd.X - lineStart.X;
double dY = lineEnd.Y - lineStart.Y;
double dp1X = point.X - lineStart.X;
double dp1Y = point.Y - lineStart.Y;
double segLenSquared = (dX * dX) + (dY * dY);
double t = 0.0;
if (segLenSquared >= -kMinSegmentLenSquared && segLenSquared <= kMinSegmentLenSquared)
{
// segment is a point.
intersectPoint = lineStart;
t = 0.0;
distance = ((dp1X * dp1X) + (dp1Y * dp1Y));
}
else
{
// Project a line from p to the segment [p1,p2]. By considering the line
// extending the segment, parameterized as p1 + (t * (p2 - p1)),
// we find projection of point p onto the line.
// It falls where t = [(p - p1) . (p2 - p1)] / |p2 - p1|^2
t = ((dp1X * dX) + (dp1Y * dY)) / segLenSquared;
if (t < kEpsilon)
{
// intersects at or to the "left" of first segment vertex (lineStart.X, lineStart.Y). If t is approximately 0.0, then
// intersection is at p1. If t is less than that, then there is no intersection (i.e. p is not within
// the 'bounds' of the segment)
if (t > -kEpsilon)
{
// intersects at 1st segment vertex
t = 0.0;
}
// set our 'intersection' point to p1.
intersectPoint = lineStart;
// Note: If you wanted the ACTUAL intersection point of where the projected lines would intersect if
// we were doing PointLineDistanceSquared, then intersectPoint.X would be (lineStart.X + (t * dx)) and intersectPoint.Y would be (lineStart.Y + (t * dy)).
}
else if (t > (1.0 - kEpsilon))
{
// intersects at or to the "right" of second segment vertex (lineEnd.X, lineEnd.Y). If t is approximately 1.0, then
// intersection is at p2. If t is greater than that, then there is no intersection (i.e. p is not within
// the 'bounds' of the segment)
if (t < (1.0 + kEpsilon))
{
// intersects at 2nd segment vertex
t = 1.0;
}
// set our 'intersection' point to p2.
intersectPoint = lineEnd;
// Note: If you wanted the ACTUAL intersection point of where the projected lines would intersect if
// we were doing PointLineDistanceSquared, then intersectPoint.X would be (lineStart.X + (t * dx)) and intersectPoint.Y would be (lineStart.Y + (t * dy)).
}
else
{
// The projection of the point to the point on the segment that is perpendicular succeeded and the point
// is 'within' the bounds of the segment. Set the intersection point as that projected point.
intersectPoint = new PointF((float)(lineStart.X + (t * dX)), (float)(lineStart.Y + (t * dY)));
}
// return the squared distance from p to the intersection point. Note that we return the squared distance
// as an optimization because many times you just need to compare relative distances and the squared values
// works fine for that. If you want the ACTUAL distance, just take the square root of this value.
double dpqX = point.X - intersectPoint.X;
double dpqY = point.Y - intersectPoint.Y;
distance = ((dpqX * dpqX) + (dpqY * dpqY));
}
return true;
}
这是一个为有限线段而做的实现,而不是像这里的大多数其他函数那样的无限线(这就是为什么我做这个)。
Paul Bourke的理论实施。
Python:
def dist(x1, y1, x2, y2, x3, y3): # x3,y3 is the point
px = x2-x1
py = y2-y1
norm = px*px + py*py
u = ((x3 - x1) * px + (y3 - y1) * py) / float(norm)
if u > 1:
u = 1
elif u < 0:
u = 0
x = x1 + u * px
y = y1 + u * py
dx = x - x3
dy = y - y3
# Note: If the actual distance does not matter,
# if you only want to compare what this function
# returns to other results of this function, you
# can just return the squared distance instead
# (i.e. remove the sqrt) to gain a little performance
dist = (dx*dx + dy*dy)**.5
return dist
AS3:
public static function segmentDistToPoint(segA:Point, segB:Point, p:Point):Number
{
var p2:Point = new Point(segB.x - segA.x, segB.y - segA.y);
var something:Number = p2.x*p2.x + p2.y*p2.y;
var u:Number = ((p.x - segA.x) * p2.x + (p.y - segA.y) * p2.y) / something;
if (u > 1)
u = 1;
else if (u < 0)
u = 0;
var x:Number = segA.x + u * p2.x;
var y:Number = segA.y + u * p2.y;
var dx:Number = x - p.x;
var dy:Number = y - p.y;
var dist:Number = Math.sqrt(dx*dx + dy*dy);
return dist;
}
Java
private double shortestDistance(float x1,float y1,float x2,float y2,float x3,float y3)
{
float px=x2-x1;
float py=y2-y1;
float temp=(px*px)+(py*py);
float u=((x3 - x1) * px + (y3 - y1) * py) / (temp);
if(u>1){
u=1;
}
else if(u<0){
u=0;
}
float x = x1 + u * px;
float y = y1 + u * py;
float dx = x - x3;
float dy = y - y3;
double dist = Math.sqrt(dx*dx + dy*dy);
return dist;
}