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
C#
改编自@Grumdrig
public static double MinimumDistanceToLineSegment(this Point p,
Line line)
{
var v = line.StartPoint;
var w = line.EndPoint;
double lengthSquared = DistanceSquared(v, w);
if (lengthSquared == 0.0)
return Distance(p, v);
double t = Math.Max(0, Math.Min(1, DotProduct(p - v, w - v) / lengthSquared));
var projection = v + t * (w - v);
return Distance(p, projection);
}
public static double Distance(Point a, Point b)
{
return Math.Sqrt(DistanceSquared(a, b));
}
public static double DistanceSquared(Point a, Point b)
{
var d = a - b;
return DotProduct(d, d);
}
public static double DotProduct(Point a, Point b)
{
return (a.X * b.X) + (a.Y * b.Y);
}
其他回答
Lua: 查找线段(不是整条线)与点之间的最小距离
function solveLinearEquation(A1,B1,C1,A2,B2,C2)
--it is the implitaion of a method of solving linear equations in x and y
local f1 = B1*C2 -B2*C1
local f2 = A2*C1-A1*C2
local f3 = A1*B2 -A2*B1
return {x= f1/f3, y= f2/f3}
end
function pointLiesOnLine(x,y,x1,y1,x2,y2)
local dx1 = x-x1
local dy1 = y-y1
local dx2 = x-x2
local dy2 = y-y2
local crossProduct = dy1*dx2 -dx1*dy2
if crossProduct ~= 0 then return false
else
if ((x1>=x) and (x>=x2)) or ((x2>=x) and (x>=x1)) then
if ((y1>=y) and (y>=y2)) or ((y2>=y) and (y>=y1)) then
return true
else return false end
else return false end
end
end
function dist(x1,y1,x2,y2)
local dx = x1-x2
local dy = y1-y2
return math.sqrt(dx*dx + dy* dy)
end
function findMinDistBetnPointAndLine(x1,y1,x2,y2,x3,y3)
-- finds the min distance between (x3,y3) and line (x1,y2)--(x2,y2)
local A2,B2,C2,A1,B1,C1
local dx = y2-y1
local dy = x2-x1
if dx == 0 then A2=1 B2=0 C2=-x3 A1=0 B1=1 C1=-y1
elseif dy == 0 then A2=0 B2=1 C2=-y3 A1=1 B1=0 C1=-x1
else
local m1 = dy/dx
local m2 = -1/m1
A2=m2 B2=-1 C2=y3-m2*x3 A1=m1 B1=-1 C1=y1-m1*x1
end
local intsecPoint= solveLinearEquation(A1,B1,C1,A2,B2,C2)
if pointLiesOnLine(intsecPoint.x, intsecPoint.y,x1,y1,x2,y2) then
return dist(intsecPoint.x, intsecPoint.y, x3,y3)
else
return math.min(dist(x3,y3,x1,y1),dist(x3,y3,x2,y2))
end
end
上面的函数在垂直线上不起作用。这是一个工作正常的函数! 与点p1 p2相交。CheckPoint为p;
public float DistanceOfPointToLine2(PointF p1, PointF p2, PointF p)
{
// (y1-y2)x + (x2-x1)y + (x1y2-x2y1)
//d(P,L) = --------------------------------
// sqrt( (x2-x1)pow2 + (y2-y1)pow2 )
double ch = (p1.Y - p2.Y) * p.X + (p2.X - p1.X) * p.Y + (p1.X * p2.Y - p2.X * p1.Y);
double del = Math.Sqrt(Math.Pow(p2.X - p1.X, 2) + Math.Pow(p2.Y - p1.Y, 2));
double d = ch / del;
return (float)d;
}
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)
}
Grumdrig的c++ /JavaScript实现对我来说非常有用,所以我提供了我正在使用的Python直接端口。完整的代码在这里。
class Point(object):
def __init__(self, x, y):
self.x = float(x)
self.y = float(y)
def square(x):
return x * x
def distance_squared(v, w):
return square(v.x - w.x) + square(v.y - w.y)
def distance_point_segment_squared(p, v, w):
# Segment length squared, |w-v|^2
d2 = distance_squared(v, w)
if d2 == 0:
# v == w, return distance to v
return distance_squared(p, v)
# 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
t = ((p.x - v.x) * (w.x - v.x) + (p.y - v.y) * (w.y - v.y)) / d2;
if t < 0:
# Beyond v end of the segment
return distance_squared(p, v)
elif t > 1.0:
# Beyond w end of the segment
return distance_squared(p, w)
else:
# Projection falls on the segment.
proj = Point(v.x + t * (w.x - v.x), v.y + t * (w.y - v.y))
# print proj.x, proj.y
return distance_squared(p, proj)
这里是与c++答案相同的东西,但移植到pascal。点参数的顺序已经改变,以适应我的代码,但还是一样的东西。
function Dot(const p1, p2: PointF): double;
begin
Result := p1.x * p2.x + p1.y * p2.y;
end;
function SubPoint(const p1, p2: PointF): PointF;
begin
result.x := p1.x - p2.x;
result.y := p1.y - p2.y;
end;
function ShortestDistance2(const p,v,w : PointF) : double;
var
l2,t : double;
projection,tt: PointF;
begin
// Return minimum distance between line segment vw and point p
//l2 := length_squared(v, w); // i.e. |w-v|^2 - avoid a sqrt
l2 := Distance(v,w);
l2 := MPower(l2,2);
if (l2 = 0.0) then begin
result:= Distance(p, v); // v == w case
exit;
end;
// 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
t := Dot(SubPoint(p,v),SubPoint(w,v)) / l2;
if (t < 0.0) then begin
result := Distance(p, v); // Beyond the 'v' end of the segment
exit;
end
else if (t > 1.0) then begin
result := Distance(p, w); // Beyond the 'w' end of the segment
exit;
end;
//projection := v + t * (w - v); // Projection falls on the segment
tt.x := v.x + t * (w.x - v.x);
tt.y := v.y + t * (w.y - v.y);
result := Distance(p, tt);
end;