2D坐标数组的Python Numpy实现:

import numpy as np


def dist2d(p1, p2, coords):
    ''' Distance from points to a finite line btwn p1 -> p2 '''
    assert coords.ndim == 2 and coords.shape[1] == 2, 'coords is not 2 dim'
    dp = p2 - p1
    st = dp[0]**2 + dp[1]**2
    u = ((coords[:, 0] - p1[0]) * dp[0] + (coords[:, 1] - p1[1]) * dp[1]) / st

    u[u > 1.] = 1.
    u[u < 0.] = 0.

    dx = (p1[0] + u * dp[0]) - coords[:, 0]
    dy = (p1[1] + u * dp[1]) - coords[:, 1]

    return np.sqrt(dx**2 + dy**2)


# Usage:
p1 = np.array([0., 0.])
p2 = np.array([0., 10.])

# List of coordinates
coords = np.array(
    [[0., 0.],
     [5., 5.],
     [10., 10.],
     [20., 20.]
     ])

d = dist2d(p1, p2, coords)

# Single coordinate
coord = np.array([25., 25.])
d = dist2d(p1, p2, coord[np.newaxis, :])

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)。

对于感兴趣的人，这里是Joshua的Javascript代码到Objective-C的简单转换:

- (double)distanceToPoint:(CGPoint)p fromLineSegmentBetween:(CGPoint)l1 and:(CGPoint)l2
{
    double A = p.x - l1.x;
    double B = p.y - l1.y;
    double C = l2.x - l1.x;
    double D = l2.y - l1.y;

    double dot = A * C + B * D;
    double len_sq = C * C + D * D;
    double param = dot / len_sq;

    double xx, yy;

    if (param < 0 || (l1.x == l2.x && l1.y == l2.y)) {
        xx = l1.x;
        yy = l1.y;
    }
    else if (param > 1) {
        xx = l2.x;
        yy = l2.y;
    }
    else {
        xx = l1.x + param * C;
        yy = l1.y + param * D;
    }

    double dx = p.x - xx;
    double dy = p.y - yy;

    return sqrtf(dx * dx + dy * dy);
}

我需要这个解决方案与MKMapPoint一起工作，所以我将分享它，以防其他人需要它。只是一些小的改变，这将返回米为单位的距离:

- (double)distanceToPoint:(MKMapPoint)p fromLineSegmentBetween:(MKMapPoint)l1 and:(MKMapPoint)l2
{
    double A = p.x - l1.x;
    double B = p.y - l1.y;
    double C = l2.x - l1.x;
    double D = l2.y - l1.y;

    double dot = A * C + B * D;
    double len_sq = C * C + D * D;
    double param = dot / len_sq;

    double xx, yy;

    if (param < 0 || (l1.x == l2.x && l1.y == l2.y)) {
        xx = l1.x;
        yy = l1.y;
    }
    else if (param > 1) {
        xx = l2.x;
        yy = l2.y;
    }
    else {
        xx = l1.x + param * C;
        yy = l1.y + param * D;
    }

    return MKMetersBetweenMapPoints(p, MKMapPointMake(xx, yy));
}

下面是devnullicus转换为c#的c++版本。对于我的实现，我需要知道交叉点，并找到他的解决方案。

public static bool PointSegmentDistanceSquared(PointF point, PointF lineStart, PointF lineEnd, out double distance, out PointF intersectPoint)
{
    const double kMinSegmentLenSquared = 0.00000001; // adjust to suit.  If you use float, you'll probably want something like 0.000001f
    const double kEpsilon = 1.0E-14; // adjust to suit.  If you use floats, you'll probably want something like 1E-7f
    double dX = lineEnd.X - lineStart.X;
    double dY = lineEnd.Y - lineStart.Y;
    double dp1X = point.X - lineStart.X;
    double dp1Y = point.Y - lineStart.Y;
    double segLenSquared = (dX * dX) + (dY * dY);
    double t = 0.0;

    if (segLenSquared >= -kMinSegmentLenSquared && segLenSquared <= kMinSegmentLenSquared)
    {
        // segment is a point.
        intersectPoint = lineStart;
        t = 0.0;
        distance = ((dp1X * dp1X) + (dp1Y * dp1Y));
    }
    else
    {
        // Project a line from p to the segment [p1,p2].  By considering the line
        // extending the segment, parameterized as p1 + (t * (p2 - p1)),
        // we find projection of point p onto the line. 
        // It falls where t = [(p - p1) . (p2 - p1)] / |p2 - p1|^2
        t = ((dp1X * dX) + (dp1Y * dY)) / segLenSquared;
        if (t < kEpsilon)
        {
            // intersects at or to the "left" of first segment vertex (lineStart.X, lineStart.Y).  If t is approximately 0.0, then
            // intersection is at p1.  If t is less than that, then there is no intersection (i.e. p is not within
            // the 'bounds' of the segment)
            if (t > -kEpsilon)
            {
                // intersects at 1st segment vertex
                t = 0.0;
            }
            // set our 'intersection' point to p1.
            intersectPoint = lineStart;
            // Note: If you wanted the ACTUAL intersection point of where the projected lines would intersect if
            // we were doing PointLineDistanceSquared, then intersectPoint.X would be (lineStart.X + (t * dx)) and intersectPoint.Y would be (lineStart.Y + (t * dy)).
        }
        else if (t > (1.0 - kEpsilon))
        {
            // intersects at or to the "right" of second segment vertex (lineEnd.X, lineEnd.Y).  If t is approximately 1.0, then
            // intersection is at p2.  If t is greater than that, then there is no intersection (i.e. p is not within
            // the 'bounds' of the segment)
            if (t < (1.0 + kEpsilon))
            {
                // intersects at 2nd segment vertex
                t = 1.0;
            }
            // set our 'intersection' point to p2.
            intersectPoint = lineEnd;
            // Note: If you wanted the ACTUAL intersection point of where the projected lines would intersect if
            // we were doing PointLineDistanceSquared, then intersectPoint.X would be (lineStart.X + (t * dx)) and intersectPoint.Y would be (lineStart.Y + (t * dy)).
        }
        else
        {
            // The projection of the point to the point on the segment that is perpendicular succeeded and the point
            // is 'within' the bounds of the segment.  Set the intersection point as that projected point.
            intersectPoint = new PointF((float)(lineStart.X + (t * dX)), (float)(lineStart.Y + (t * dY)));
        }
        // return the squared distance from p to the intersection point.  Note that we return the squared distance
        // as an optimization because many times you just need to compare relative distances and the squared values
        // works fine for that.  If you want the ACTUAL distance, just take the square root of this value.
        double dpqX = point.X - intersectPoint.X;
        double dpqY = point.Y - intersectPoint.Y;

        distance = ((dpqX * dpqX) + (dpqY * dpqY));
    }

    return true;
}

在javascript中使用几何:

var a = { x:20, y:20};//start segment    
var b = { x:40, y:30};//end segment   
var c = { x:37, y:14};//point   

// magnitude from a to c    
var ac = Math.sqrt( Math.pow( ( a.x - c.x ), 2 ) + Math.pow( ( a.y - c.y ), 2) );    
// magnitude from b to c   
var bc = Math.sqrt( Math.pow( ( b.x - c.x ), 2 ) + Math.pow( ( b.y - c.y ), 2 ) );    
// magnitude from a to b (base)     
var ab = Math.sqrt( Math.pow( ( a.x - b.x ), 2 ) + Math.pow( ( a.y - b.y ), 2 ) );    
 // perimeter of triangle     
var p = ac + bc + ab;    
 // area of the triangle    
var area = Math.sqrt( p/2 * ( p/2 - ac) * ( p/2 - bc ) * ( p/2 - ab ) );    
 // height of the triangle = distance    
var h = ( area * 2 ) / ab;    
console.log ("height: " + h);

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