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
in R
#distance beetween segment ab and point c in 2D space
getDistance_ort_2 <- function(a, b, c){
#go to complex numbers
A<-c(a[1]+1i*a[2],b[1]+1i*b[2])
q=c[1]+1i*c[2]
#function to get coefficients of line (ab)
getAlphaBeta <- function(A)
{ a<-Re(A[2])-Re(A[1])
b<-Im(A[2])-Im(A[1])
ab<-as.numeric()
ab[1] <- -Re(A[1])*b/a+Im(A[1])
ab[2] <-b/a
if(Im(A[1])==Im(A[2])) ab<- c(Im(A[1]),0)
if(Re(A[1])==Re(A[2])) ab <- NA
return(ab)
}
#function to get coefficients of line ortogonal to line (ab) which goes through point q
getAlphaBeta_ort<-function(A,q)
{ ab <- getAlphaBeta(A)
coef<-c(Re(q)/ab[2]+Im(q),-1/ab[2])
if(Re(A[1])==Re(A[2])) coef<-c(Im(q),0)
return(coef)
}
#function to get coordinates of interception point
#between line (ab) and its ortogonal which goes through point q
getIntersection_ort <- function(A, q){
A.ab <- getAlphaBeta(A)
q.ab <- getAlphaBeta_ort(A,q)
if (!is.na(A.ab[1])&A.ab[2]==0) {
x<-Re(q)
y<-Im(A[1])}
if (is.na(A.ab[1])) {
x<-Re(A[1])
y<-Im(q)
}
if (!is.na(A.ab[1])&A.ab[2]!=0) {
x <- (q.ab[1] - A.ab[1])/(A.ab[2] - q.ab[2])
y <- q.ab[1] + q.ab[2]*x}
xy <- x + 1i*y
return(xy)
}
intersect<-getIntersection_ort(A,q)
if ((Mod(A[1]-intersect)+Mod(A[2]-intersect))>Mod(A[1]-A[2])) {dist<-min(Mod(A[1]-q),Mod(A[2]-q))
} else dist<-Mod(q-intersect)
return(dist)
}
其他回答
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));
};
}
对于懒人来说,以下是我在Objective-C语言中移植@Grumdrig的解决方案:
CGFloat sqr(CGFloat x) { return x*x; }
CGFloat dist2(CGPoint v, CGPoint w) { return sqr(v.x - w.x) + sqr(v.y - w.y); }
CGFloat distanceToSegmentSquared(CGPoint p, CGPoint v, CGPoint w)
{
CGFloat l2 = dist2(v, w);
if (l2 == 0.0f) return dist2(p, v);
CGFloat t = ((p.x - v.x) * (w.x - v.x) + (p.y - v.y) * (w.y - v.y)) / l2;
if (t < 0.0f) return dist2(p, v);
if (t > 1.0f) return dist2(p, w);
return dist2(p, CGPointMake(v.x + t * (w.x - v.x), v.y + t * (w.y - v.y)));
}
CGFloat distanceToSegment(CGPoint point, CGPoint segmentPointV, CGPoint segmentPointW)
{
return sqrtf(distanceToSegmentSquared(point, segmentPointV, segmentPointW));
}
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))
}
使用arctangents的一行解决方案:
思路是将A移动到(0,0),并顺时针旋转三角形,使C位于X轴上, 当这种情况发生时,By就是距离。
a角= Atan(Cy - Ay, Cx - Ax); b角= Atan(By - Ay, Bx - Ax); AB长度=平方根((Bx - Ax)²+ (By - Ay)²) By = Sin (bAngle - aAngle) * ABLength
C#
public double Distance(Point a, Point b, Point c)
{
// normalize points
Point cn = new Point(c.X - a.X, c.Y - a.Y);
Point bn = new Point(b.X - a.X, b.Y - a.Y);
double angle = Math.Atan2(bn.Y, bn.X) - Math.Atan2(cn.Y, cn.X);
double abLength = Math.Sqrt(bn.X*bn.X + bn.Y*bn.Y);
return Math.Sin(angle)*abLength;
}
一行c#(要转换为SQL)
double distance = Math.Sin(Math.Atan2(b.Y - a.Y, b.X - a.X) - Math.Atan2(c.Y - a.Y, c.X - a.X)) * Math.Sqrt((b.X - a.X) * (b.X - a.X) + (b.Y - a.Y) * (b.Y - a.Y))
我需要一个Godot (GDscript)的实现,所以我写了一个基于grumdrig接受的答案:
func minimum_distance(v: Vector2, w: Vector2, p: Vector2):
# Return minimum distance between line segment vw and point p
var l2: float = (v - w).length_squared() # i.e. |w-v|^2 - avoid a sqrt
if l2 == 0.0:
return p.distance_to(v) # v == w case
# 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
# We clamp t from [0,1] to handle points outside the segment vw.
var t: float = max(0, min(1, (p - v).dot(w - v) / l2))
var projection: Vector2 = v + t * (w - v) # Projection falls on the segment
return p.distance_to(projection)