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
这是我最后写的代码。这段代码假设一个点以{x:5, y:7}的形式定义。注意,这不是绝对最有效的方法,但它是我能想到的最简单、最容易理解的代码。
// a, b, and c in the code below are all points
function distance(a, b)
{
var dx = a.x - b.x;
var dy = a.y - b.y;
return Math.sqrt(dx*dx + dy*dy);
}
function Segment(a, b)
{
var ab = {
x: b.x - a.x,
y: b.y - a.y
};
var length = distance(a, b);
function cross(c) {
return ab.x * (c.y-a.y) - ab.y * (c.x-a.x);
};
this.distanceFrom = function(c) {
return Math.min(distance(a,c),
distance(b,c),
Math.abs(cross(c) / length));
};
}
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)
}
Matlab代码,内置“自检”,如果他们调用函数没有参数:
function r = distPointToLineSegment( xy0, xy1, xyP )
% r = distPointToLineSegment( xy0, xy1, xyP )
if( nargin < 3 )
selfTest();
r=0;
else
vx = xy0(1)-xyP(1);
vy = xy0(2)-xyP(2);
ux = xy1(1)-xy0(1);
uy = xy1(2)-xy0(2);
lenSqr= (ux*ux+uy*uy);
detP= -vx*ux + -vy*uy;
if( detP < 0 )
r = norm(xy0-xyP,2);
elseif( detP > lenSqr )
r = norm(xy1-xyP,2);
else
r = abs(ux*vy-uy*vx)/sqrt(lenSqr);
end
end
function selfTest()
%#ok<*NASGU>
disp(['invalid args, distPointToLineSegment running (recursive) self-test...']);
ptA = [1;1]; ptB = [-1;-1];
ptC = [1/2;1/2]; % on the line
ptD = [-2;-1.5]; % too far from line segment
ptE = [1/2;0]; % should be same as perpendicular distance to line
ptF = [1.5;1.5]; % along the A-B but outside of the segment
distCtoAB = distPointToLineSegment(ptA,ptB,ptC)
distDtoAB = distPointToLineSegment(ptA,ptB,ptD)
distEtoAB = distPointToLineSegment(ptA,ptB,ptE)
distFtoAB = distPointToLineSegment(ptA,ptB,ptF)
figure(1); clf;
circle = @(x, y, r, c) rectangle('Position', [x-r, y-r, 2*r, 2*r], ...
'Curvature', [1 1], 'EdgeColor', c);
plot([ptA(1) ptB(1)],[ptA(2) ptB(2)],'r-x'); hold on;
plot(ptC(1),ptC(2),'b+'); circle(ptC(1),ptC(2), 0.5e-1, 'b');
plot(ptD(1),ptD(2),'g+'); circle(ptD(1),ptD(2), distDtoAB, 'g');
plot(ptE(1),ptE(2),'k+'); circle(ptE(1),ptE(2), distEtoAB, 'k');
plot(ptF(1),ptF(2),'m+'); circle(ptF(1),ptF(2), distFtoAB, 'm');
hold off;
axis([-3 3 -3 3]); axis equal;
end
end
在javascript中使用几何:
var a = { x:20, y:20};//start segment
var b = { x:40, y:30};//end segment
var c = { x:37, y:14};//point
// magnitude from a to c
var ac = Math.sqrt( Math.pow( ( a.x - c.x ), 2 ) + Math.pow( ( a.y - c.y ), 2) );
// magnitude from b to c
var bc = Math.sqrt( Math.pow( ( b.x - c.x ), 2 ) + Math.pow( ( b.y - c.y ), 2 ) );
// magnitude from a to b (base)
var ab = Math.sqrt( Math.pow( ( a.x - b.x ), 2 ) + Math.pow( ( a.y - b.y ), 2 ) );
// perimeter of triangle
var p = ac + bc + ab;
// area of the triangle
var area = Math.sqrt( p/2 * ( p/2 - ac) * ( p/2 - bc ) * ( p/2 - ab ) );
// height of the triangle = distance
var h = ( area * 2 ) / ab;
console.log ("height: " + h);
在我自己的问题线程如何计算在C, c# / .NET 2.0或Java的所有情况下一个点和线段之间的最短2D距离?当我找到一个c#的答案时,我被要求把它放在这里:所以它是从http://www.topcoder.com/tc?d1=tutorials&d2=geometry1&module=Static修改的:
//Compute the dot product AB . BC
private double DotProduct(double[] pointA, double[] pointB, double[] pointC)
{
double[] AB = new double[2];
double[] BC = new double[2];
AB[0] = pointB[0] - pointA[0];
AB[1] = pointB[1] - pointA[1];
BC[0] = pointC[0] - pointB[0];
BC[1] = pointC[1] - pointB[1];
double dot = AB[0] * BC[0] + AB[1] * BC[1];
return dot;
}
//Compute the cross product AB x AC
private double CrossProduct(double[] pointA, double[] pointB, double[] pointC)
{
double[] AB = new double[2];
double[] AC = new double[2];
AB[0] = pointB[0] - pointA[0];
AB[1] = pointB[1] - pointA[1];
AC[0] = pointC[0] - pointA[0];
AC[1] = pointC[1] - pointA[1];
double cross = AB[0] * AC[1] - AB[1] * AC[0];
return cross;
}
//Compute the distance from A to B
double Distance(double[] pointA, double[] pointB)
{
double d1 = pointA[0] - pointB[0];
double d2 = pointA[1] - pointB[1];
return Math.Sqrt(d1 * d1 + d2 * d2);
}
//Compute the distance from AB to C
//if isSegment is true, AB is a segment, not a line.
double LineToPointDistance2D(double[] pointA, double[] pointB, double[] pointC,
bool isSegment)
{
double dist = CrossProduct(pointA, pointB, pointC) / Distance(pointA, pointB);
if (isSegment)
{
double dot1 = DotProduct(pointA, pointB, pointC);
if (dot1 > 0)
return Distance(pointB, pointC);
double dot2 = DotProduct(pointB, pointA, pointC);
if (dot2 > 0)
return Distance(pointA, pointC);
}
return Math.Abs(dist);
}
我不是要回答问题,而是要问问题,所以我希望我不会因为某些原因而得到数百万张反对票,而是批评。我只是想(并被鼓励)分享其他人的想法,因为这个帖子中的解决方案要么是用一些奇异的语言(Fortran, Mathematica),要么被某人标记为错误。对我来说唯一有用的(由Grumdrig编写)是用c++编写的,没有人标记它有错误。但是它缺少被调用的方法(dot等)。
WPF版本:
public class LineSegment
{
private readonly Vector _offset;
private readonly Vector _vector;
public LineSegment(Point start, Point end)
{
_offset = (Vector)start;
_vector = (Vector)(end - _offset);
}
public double DistanceTo(Point pt)
{
var v = (Vector)pt - _offset;
// first, find a projection point on the segment in parametric form (0..1)
var p = (v * _vector) / _vector.LengthSquared;
// and limit it so it lays inside the segment
p = Math.Min(Math.Max(p, 0), 1);
// now, find the distance from that point to our point
return (_vector * p - v).Length;
}
}