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
快速实现http://paulbourke.net/geometry/pointlineplane/source.c
static func magnitude(p1: CGPoint, p2: CGPoint) -> CGFloat {
let vector = CGPoint(x: p2.x - p1.x, y: p2.y - p1.y)
return sqrt(pow(vector.x, 2) + pow(vector.y, 2))
}
/// http://paulbourke.net/geometry/pointlineplane/
/// http://paulbourke.net/geometry/pointlineplane/source.c
static func pointDistanceToLine(point: CGPoint, lineStart: CGPoint, lineEnd: CGPoint) -> CGFloat? {
let lineMag = magnitude(p1: lineEnd, p2: lineStart)
let u = (((point.x - lineStart.x) * (lineEnd.x - lineStart.x)) +
((point.y - lineStart.y) * (lineEnd.y - lineStart.y))) /
(lineMag * lineMag)
if u < 0 || u > 1 {
// closest point does not fall within the line segment
return nil
}
let intersectionX = lineStart.x + u * (lineEnd.x - lineStart.x)
let intersectionY = lineStart.y + u * (lineEnd.y - lineStart.y)
return magnitude(p1: point, p2: CGPoint(x: intersectionX, y: intersectionY))
}
其他回答
JavaScript中一个基于这个公式的更简洁的解决方案:
distToSegment: function (point, linePointA, linePointB){
var x0 = point.X;
var y0 = point.Y;
var x1 = linePointA.X;
var y1 = linePointA.Y;
var x2 = linePointB.X;
var y2 = linePointB.Y;
var Dx = (x2 - x1);
var Dy = (y2 - y1);
var numerator = Math.abs(Dy*x0 - Dx*y0 - x1*y2 + x2*y1);
var denominator = Math.sqrt(Dx*Dx + Dy*Dy);
if (denominator == 0) {
return this.dist2(point, linePointA);
}
return numerator/denominator;
}
现在我的解决方案...... (Javascript)
这是非常快的,因为我试图避免任何数学。战俘的功能。
如你所见,在函数的最后,我得到了直线的距离。
代码来自lib http://www.draw2d.org/graphiti/jsdoc/#!/例子
/**
* Static util function to determine is a point(px,py) on the line(x1,y1,x2,y2)
* A simple hit test.
*
* @return {boolean}
* @static
* @private
* @param {Number} coronaWidth the accepted corona for the hit test
* @param {Number} X1 x coordinate of the start point of the line
* @param {Number} Y1 y coordinate of the start point of the line
* @param {Number} X2 x coordinate of the end point of the line
* @param {Number} Y2 y coordinate of the end point of the line
* @param {Number} px x coordinate of the point to test
* @param {Number} py y coordinate of the point to test
**/
graphiti.shape.basic.Line.hit= function( coronaWidth, X1, Y1, X2, Y2, px, py)
{
// Adjust vectors relative to X1,Y1
// X2,Y2 becomes relative vector from X1,Y1 to end of segment
X2 -= X1;
Y2 -= Y1;
// px,py becomes relative vector from X1,Y1 to test point
px -= X1;
py -= Y1;
var dotprod = px * X2 + py * Y2;
var projlenSq;
if (dotprod <= 0.0) {
// px,py is on the side of X1,Y1 away from X2,Y2
// distance to segment is length of px,py vector
// "length of its (clipped) projection" is now 0.0
projlenSq = 0.0;
} else {
// switch to backwards vectors relative to X2,Y2
// X2,Y2 are already the negative of X1,Y1=>X2,Y2
// to get px,py to be the negative of px,py=>X2,Y2
// the dot product of two negated vectors is the same
// as the dot product of the two normal vectors
px = X2 - px;
py = Y2 - py;
dotprod = px * X2 + py * Y2;
if (dotprod <= 0.0) {
// px,py is on the side of X2,Y2 away from X1,Y1
// distance to segment is length of (backwards) px,py vector
// "length of its (clipped) projection" is now 0.0
projlenSq = 0.0;
} else {
// px,py is between X1,Y1 and X2,Y2
// dotprod is the length of the px,py vector
// projected on the X2,Y2=>X1,Y1 vector times the
// length of the X2,Y2=>X1,Y1 vector
projlenSq = dotprod * dotprod / (X2 * X2 + Y2 * Y2);
}
}
// Distance to line is now the length of the relative point
// vector minus the length of its projection onto the line
// (which is zero if the projection falls outside the range
// of the line segment).
var lenSq = px * px + py * py - projlenSq;
if (lenSq < 0) {
lenSq = 0;
}
return Math.sqrt(lenSq)<coronaWidth;
};
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;
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)
}
在数学
它使用线段的参数描述,并将点投影到线段定义的直线中。当参数在线段内从0到1时,如果投影在这个范围之外,我们计算到相应端点的距离,而不是法线到线段的直线。
Clear["Global`*"];
distance[{start_, end_}, pt_] :=
Module[{param},
param = ((pt - start).(end - start))/Norm[end - start]^2; (*parameter. the "."
here means vector product*)
Which[
param < 0, EuclideanDistance[start, pt], (*If outside bounds*)
param > 1, EuclideanDistance[end, pt],
True, EuclideanDistance[pt, start + param (end - start)] (*Normal distance*)
]
];
策划的结果:
Plot3D[distance[{{0, 0}, {1, 0}}, {xp, yp}], {xp, -1, 2}, {yp, -1, 2}]
画出比截断距离更近的点:
等高线图: