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
特征c++版本的3D线段和点
// Return minimum distance between line segment: head--->tail and point
double MinimumDistance(Eigen::Vector3d head, Eigen::Vector3d tail,Eigen::Vector3d point)
{
double l2 = std::pow((head - tail).norm(),2);
if(l2 ==0.0) return (head - point).norm();// head == tail case
// Consider the line extending the segment, parameterized as head + t (tail - point).
// We find projection of point onto the line.
// It falls where t = [(point-head) . (tail-head)] / |tail-head|^2
// We clamp t from [0,1] to handle points outside the segment head--->tail.
double t = max(0,min(1,(point-head).dot(tail-head)/l2));
Eigen::Vector3d projection = head + t*(tail-head);
return (point - projection).norm();
}
其他回答
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;
}
}
这是一个自成体系的Delphi / Pascal版本的函数,基于上面约书亚的答案。使用TPoint用于VCL屏幕图形,但应该易于根据需要进行调整。
function DistancePtToSegment( pt, pt1, pt2: TPoint): double;
var
a, b, c, d: double;
len_sq: double;
param: double;
xx, yy: double;
dx, dy: double;
begin
a := pt.x - pt1.x;
b := pt.y - pt1.y;
c := pt2.x - pt1.x;
d := pt2.y - pt1.y;
len_sq := (c * c) + (d * d);
param := -1;
if (len_sq <> 0) then
begin
param := ((a * c) + (b * d)) / len_sq;
end;
if param < 0 then
begin
xx := pt1.x;
yy := pt1.y;
end
else if param > 1 then
begin
xx := pt2.x;
yy := pt2.y;
end
else begin
xx := pt1.x + param * c;
yy := pt1.y + param * d;
end;
dx := pt.x - xx;
dy := pt.y - yy;
result := sqrt( (dx * dx) + (dy * dy))
end;
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
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)
}
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))