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
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)。
其他回答
公认的答案行不通 (例如,0,0和(-10,2,10,2)之间的距离应为2)。
下面是工作代码:
def dist2line2(x,y,line):
x1,y1,x2,y2=line
vx = x1 - x
vy = y1 - y
ux = x2-x1
uy = y2-y1
length = ux * ux + uy * uy
det = (-vx * ux) + (-vy * uy) #//if this is < 0 or > length then its outside the line segment
if det < 0:
return (x1 - x)**2 + (y1 - y)**2
if det > length:
return (x2 - x)**2 + (y2 - y)**2
det = ux * vy - uy * vx
return det**2 / length
def dist2line(x,y,line): return math.sqrt(dist2line2(x,y,line))
Lua解决方案
-- distance from point (px, py) to line segment (x1, y1, x2, y2)
function distPointToLine(px,py,x1,y1,x2,y2) -- point, start and end of the segment
local dx,dy = x2-x1,y2-y1
local length = math.sqrt(dx*dx+dy*dy)
dx,dy = dx/length,dy/length -- normalization
local p = dx*(px-x1)+dy*(py-y1)
if p < 0 then
dx,dy = px-x1,py-y1
return math.sqrt(dx*dx+dy*dy), x1, y1 -- distance, nearest point
elseif p > length then
dx,dy = px-x2,py-y2
return math.sqrt(dx*dx+dy*dy), x2, y2 -- distance, nearest point
end
return math.abs(dy*(px-x1)-dx*(py-y1)), x1+dx*p, y1+dy*p -- distance, nearest point
end
对于折线(有两条以上线段的线):
-- if the (poly-)line has several segments, just iterate through all of them:
function nearest_sector_in_line (x, y, line)
local x1, y1, x2, y2, min_dist
local ax,ay = line[1], line[2]
for j = 3, #line-1, 2 do
local bx,by = line[j], line[j+1]
local dist = distPointToLine(x,y,ax,ay,bx,by)
if not min_dist or dist < min_dist then
min_dist = dist
x1, y1, x2, y2 = ax,ay,bx,by
end
ax, ay = bx, by
end
return x1, y1, x2, y2
end
例子:
-- call it:
local x1, y1, x2, y2 = nearest_sector_in_line (7, 4, {0,0, 10,0, 10,10, 0,10})
这个答案是基于公认答案的JavaScript解决方案。 它主要只是格式更好,函数名更长,当然函数语法更短,因为它是在ES6 + CoffeeScript中。
JavaScript版本(ES6)
distanceSquared = (v, w)=> Math.pow(v.x - w.x, 2) + Math.pow(v.y - w.y, 2);
distance = (v, w)=> Math.sqrt(distanceSquared(v, w));
distanceToLineSegmentSquared = (p, v, w)=> {
l2 = distanceSquared(v, w);
if (l2 === 0) {
return distanceSquared(p, v);
}
t = ((p.x - v.x) * (w.x - v.x) + (p.y - v.y) * (w.y - v.y)) / l2;
t = Math.max(0, Math.min(1, t));
return distanceSquared(p, {
x: v.x + t * (w.x - v.x),
y: v.y + t * (w.y - v.y)
});
}
distanceToLineSegment = (p, v, w)=> {
return Math.sqrt(distanceToLineSegmentSquared(p, v));
}
CoffeeScript版本
distanceSquared = (v, w)-> (v.x - w.x) ** 2 + (v.y - w.y) ** 2
distance = (v, w)-> Math.sqrt(distanceSquared(v, w))
distanceToLineSegmentSquared = (p, v, w)->
l2 = distanceSquared(v, w)
return distanceSquared(p, v) if l2 is 0
t = ((p.x - v.x) * (w.x - v.x) + (p.y - v.y) * (w.y - v.y)) / l2
t = Math.max(0, Math.min(1, t))
distanceSquared(p, {
x: v.x + t * (w.x - v.x)
y: v.y + t * (w.y - v.y)
})
distanceToLineSegment = (p, v, w)->
Math.sqrt(distanceToLineSegmentSquared(p, v, w))
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
您可以尝试PHP geo-math-php的库
composer require rkondratuk/geo-math-php:^1
例子:
<?php
use PhpGeoMath\Model\GeoSegment;
use PhpGeoMath\Model\Polar3dPoint;
$polarPoint1 = new Polar3dPoint(
40.758742779050706, -73.97855507715238, Polar3dPoint::EARTH_RADIUS_IN_METERS
);
$polarPoint2 = new Polar3dPoint(
40.74843388072615, -73.98566565776102, Polar3dPoint::EARTH_RADIUS_IN_METERS
);
$polarPoint3 = new Polar3dPoint(
40.74919365249446, -73.98133456388013, Polar3dPoint::EARTH_RADIUS_IN_METERS
);
$arcSegment = new GeoSegment($polarPoint1, $polarPoint2);
$nearestPolarPoint = $arcSegment->calcNearestPoint($polarPoint3);
// Shortest distance from point-3 to segment(point-1, point-2)
$geoDistance = $nearestPolarPoint->calcGeoDistanceToPoint($polarPoint3);