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
和这个答案一样,只是用的是Visual Basic。使其可作为Microsoft Excel和VBA/宏中的用户定义函数使用。
函数返回点(x,y)到由(x1,y1)和(x2,y2)定义的线段的最近距离。
Function DistanceToSegment(x As Double, y As Double, x1 As Double, y1 As Double, x2 As Double, y2 As Double)
Dim A As Double
A = x - x1
Dim B As Double
B = y - y1
Dim C As Double
C = x2 - x1
Dim D As Double
D = y2 - y1
Dim dot As Double
dot = A * C + B * D
Dim len_sq As Double
len_sq = C * C + D * D
Dim param As Double
param = -1
If (len_sq <> 0) Then
param = dot / len_sq
End If
Dim xx As Double
Dim yy As Double
If (param < 0) Then
xx = x1
yy = y1
ElseIf (param > 1) Then
xx = x2
yy = y2
Else
xx = x1 + param * C
yy = y1 + param * D
End If
Dim dx As Double
dx = x - xx
Dim dy As Double
dy = y - yy
DistanceToSegment = Math.Sqr(dx * dx + dy * dy)
End Function
其他回答
用t-sql编码
点为(@px, @py),线段从(@ax, @ay)到(@bx, @by)
create function fn_sqr (@NumberToSquare decimal(18,10))
returns decimal(18,10)
as
begin
declare @Result decimal(18,10)
set @Result = @NumberToSquare * @NumberToSquare
return @Result
end
go
create function fn_Distance(@ax decimal (18,10) , @ay decimal (18,10), @bx decimal(18,10), @by decimal(18,10))
returns decimal(18,10)
as
begin
declare @Result decimal(18,10)
set @Result = (select dbo.fn_sqr(@ax - @bx) + dbo.fn_sqr(@ay - @by) )
return @Result
end
go
create function fn_DistanceToSegmentSquared(@px decimal(18,10), @py decimal(18,10), @ax decimal(18,10), @ay decimal(18,10), @bx decimal(18,10), @by decimal(18,10))
returns decimal(18,10)
as
begin
declare @l2 decimal(18,10)
set @l2 = (select dbo.fn_Distance(@ax, @ay, @bx, @by))
if @l2 = 0
return dbo.fn_Distance(@px, @py, @ax, @ay)
declare @t decimal(18,10)
set @t = ((@px - @ax) * (@bx - @ax) + (@py - @ay) * (@by - @ay)) / @l2
if (@t < 0)
return dbo.fn_Distance(@px, @py, @ax, @ay);
if (@t > 1)
return dbo.fn_Distance(@px, @py, @bx, @by);
return dbo.fn_Distance(@px, @py, @ax + @t * (@bx - @ax), @ay + @t * (@by - @ay))
end
go
create function fn_DistanceToSegment(@px decimal(18,10), @py decimal(18,10), @ax decimal(18,10), @ay decimal(18,10), @bx decimal(18,10), @by decimal(18,10))
returns decimal(18,10)
as
begin
return sqrt(dbo.fn_DistanceToSegmentSquared(@px, @py , @ax , @ay , @bx , @by ))
end
go
--example execution for distance from a point at (6,1) to line segment that runs from (4,2) to (2,1)
select dbo.fn_DistanceToSegment(6, 1, 4, 2, 2, 1)
--result = 2.2360679775
--example execution for distance from a point at (-3,-2) to line segment that runs from (0,-2) to (-2,1)
select dbo.fn_DistanceToSegment(-3, -2, 0, -2, -2, 1)
--result = 2.4961508830
--example execution for distance from a point at (0,-2) to line segment that runs from (0,-2) to (-2,1)
select dbo.fn_DistanceToSegment(0,-2, 0, -2, -2, 1)
--result = 0.0000000000
该算法基于求出指定直线与包含指定点的正交直线的交点,并计算其距离。在线段的情况下,我们必须检查交点是否在线段的点之间,如果不是这样,则最小距离是指定点与线段的一个端点之间的距离。这是一个c#实现。
Double Distance(Point a, Point b)
{
double xdiff = a.X - b.X, ydiff = a.Y - b.Y;
return Math.Sqrt((long)xdiff * xdiff + (long)ydiff * ydiff);
}
Boolean IsBetween(double x, double a, double b)
{
return ((a <= b && x >= a && x <= b) || (a > b && x <= a && x >= b));
}
Double GetDistance(Point pt, Point pt1, Point pt2, out Point intersection)
{
Double a, x, y, R;
if (pt1.X != pt2.X) {
a = (double)(pt2.Y - pt1.Y) / (pt2.X - pt1.X);
x = (a * (pt.Y - pt1.Y) + a * a * pt1.X + pt.X) / (a * a + 1);
y = a * x + pt1.Y - a * pt1.X; }
else { x = pt1.X; y = pt.Y; }
if (IsBetween(x, pt1.X, pt2.X) && IsBetween(y, pt1.Y, pt2.Y)) {
intersection = new Point((int)x, (int)y);
R = Distance(intersection, pt); }
else {
double d1 = Distance(pt, pt1), d2 = Distance(pt, pt2);
if (d1 < d2) { intersection = pt1; R = d1; }
else { intersection = pt2; R = d2; }}
return R;
}
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();
}
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
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: