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
嘿,我昨天才写的。它在Actionscript 3.0中,基本上是Javascript,尽管你可能没有相同的Point类。
//st = start of line segment
//b = the line segment (as in: st + b = end of line segment)
//pt = point to test
//Returns distance from point to line segment.
//Note: nearest point on the segment to the test point is right there if we ever need it
public static function linePointDist( st:Point, b:Point, pt:Point ):Number
{
var nearestPt:Point; //closest point on seqment to pt
var keyDot:Number = dot( b, pt.subtract( st ) ); //key dot product
var bLenSq:Number = dot( b, b ); //Segment length squared
if( keyDot <= 0 ) //pt is "behind" st, use st
{
nearestPt = st
}
else if( keyDot >= bLenSq ) //pt is "past" end of segment, use end (notice we are saving twin sqrts here cuz)
{
nearestPt = st.add(b);
}
else //pt is inside segment, reuse keyDot and bLenSq to get percent of seqment to move in to find closest point
{
var keyDotToPctOfB:Number = keyDot/bLenSq; //REM dot product comes squared
var partOfB:Point = new Point( b.x * keyDotToPctOfB, b.y * keyDotToPctOfB );
nearestPt = st.add(partOfB);
}
var dist:Number = (pt.subtract(nearestPt)).length;
return dist;
}
此外,这里有一个关于这个问题的相当完整和可读的讨论:notejot.com
忍不住用python来编码:)
from math import sqrt, fabs
def pdis(a, b, c):
t = b[0]-a[0], b[1]-a[1] # Vector ab
dd = sqrt(t[0]**2+t[1]**2) # Length of ab
t = t[0]/dd, t[1]/dd # unit vector of ab
n = -t[1], t[0] # normal unit vector to ab
ac = c[0]-a[0], c[1]-a[1] # vector ac
return fabs(ac[0]*n[0]+ac[1]*n[1]) # Projection of ac to n (the minimum distance)
print pdis((1,1), (2,2), (2,0)) # Example (answer is 1.414)
fortran也是一样:)
real function pdis(a, b, c)
real, dimension(0:1), intent(in) :: a, b, c
real, dimension(0:1) :: t, n, ac
real :: dd
t = b - a ! Vector ab
dd = sqrt(t(0)**2+t(1)**2) ! Length of ab
t = t/dd ! unit vector of ab
n = (/-t(1), t(0)/) ! normal unit vector to ab
ac = c - a ! vector ac
pdis = abs(ac(0)*n(0)+ac(1)*n(1)) ! Projection of ac to n (the minimum distance)
end function pdis
program test
print *, pdis((/1.0,1.0/), (/2.0,2.0/), (/2.0,0.0/)) ! Example (answer is 1.414)
end program test
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;
}
%Matlab solution by Tim from Cody
function ans=distP2S(x0,y0,x1,y1,x2,y2)
% Point is x0,y0
z=complex(x0-x1,y0-y1);
complex(x2-x1,y2-y1);
abs(z-ans*min(1,max(0,real(z/ans))));
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})