快速实现http://paulbourke.net/geometry/pointlineplane/source.c

    static func magnitude(p1: CGPoint, p2: CGPoint) -> CGFloat {
        let vector = CGPoint(x: p2.x - p1.x, y: p2.y - p1.y)
        return sqrt(pow(vector.x, 2) + pow(vector.y, 2))
    }

    /// http://paulbourke.net/geometry/pointlineplane/
    /// http://paulbourke.net/geometry/pointlineplane/source.c
    static func pointDistanceToLine(point: CGPoint, lineStart: CGPoint, lineEnd: CGPoint) -> CGFloat? {

        let lineMag = magnitude(p1: lineEnd, p2: lineStart)
        let u = (((point.x - lineStart.x) * (lineEnd.x - lineStart.x)) +
                ((point.y - lineStart.y) * (lineEnd.y - lineStart.y))) /
                (lineMag * lineMag)

        if u < 0 || u > 1 {
            // closest point does not fall within the line segment
            return nil
        }

        let intersectionX = lineStart.x + u * (lineEnd.x - lineStart.x)
        let intersectionY = lineStart.y + u * (lineEnd.y - lineStart.y)

        return magnitude(p1: point, p2: CGPoint(x: intersectionX, y: intersectionY))
    }

现在我的解决方案...... (Javascript)

这是非常快的，因为我试图避免任何数学。战俘的功能。

如你所见，在函数的最后，我得到了直线的距离。

代码来自lib http://www.draw2d.org/graphiti/jsdoc/#!/例子

/**
 * Static util function to determine is a point(px,py) on the line(x1,y1,x2,y2)
 * A simple hit test.
 * 
 * @return {boolean}
 * @static
 * @private
 * @param {Number} coronaWidth the accepted corona for the hit test
 * @param {Number} X1 x coordinate of the start point of the line
 * @param {Number} Y1 y coordinate of the start point of the line
 * @param {Number} X2 x coordinate of the end point of the line
 * @param {Number} Y2 y coordinate of the end point of the line
 * @param {Number} px x coordinate of the point to test
 * @param {Number} py y coordinate of the point to test
 **/
graphiti.shape.basic.Line.hit= function( coronaWidth, X1, Y1,  X2,  Y2, px, py)
{
  // Adjust vectors relative to X1,Y1
  // X2,Y2 becomes relative vector from X1,Y1 to end of segment
  X2 -= X1;
  Y2 -= Y1;
  // px,py becomes relative vector from X1,Y1 to test point
  px -= X1;
  py -= Y1;
  var dotprod = px * X2 + py * Y2;
  var projlenSq;
  if (dotprod <= 0.0) {
      // px,py is on the side of X1,Y1 away from X2,Y2
      // distance to segment is length of px,py vector
      // "length of its (clipped) projection" is now 0.0
      projlenSq = 0.0;
  } else {
      // switch to backwards vectors relative to X2,Y2
      // X2,Y2 are already the negative of X1,Y1=>X2,Y2
      // to get px,py to be the negative of px,py=>X2,Y2
      // the dot product of two negated vectors is the same
      // as the dot product of the two normal vectors
      px = X2 - px;
      py = Y2 - py;
      dotprod = px * X2 + py * Y2;
      if (dotprod <= 0.0) {
          // px,py is on the side of X2,Y2 away from X1,Y1
          // distance to segment is length of (backwards) px,py vector
          // "length of its (clipped) projection" is now 0.0
          projlenSq = 0.0;
      } else {
          // px,py is between X1,Y1 and X2,Y2
          // dotprod is the length of the px,py vector
          // projected on the X2,Y2=>X1,Y1 vector times the
          // length of the X2,Y2=>X1,Y1 vector
          projlenSq = dotprod * dotprod / (X2 * X2 + Y2 * Y2);
      }
  }
    // Distance to line is now the length of the relative point
    // vector minus the length of its projection onto the line
    // (which is zero if the projection falls outside the range
    //  of the line segment).
    var lenSq = px * px + py * py - projlenSq;
    if (lenSq < 0) {
        lenSq = 0;
    }
    return Math.sqrt(lenSq)<coronaWidth;
};

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));
}

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, :])

用Matlab直接实现Grumdrig

function ans=distP2S(px,py,vx,vy,wx,wy)
% [px py vx vy wx wy]
  t=( (px-vx)*(wx-vx)+(py-vy)*(wy-vy) )/idist(vx,wx,vy,wy)^2;
  [idist(px,vx,py,vy) idist(px,vx+t*(wx-vx),py,vy+t*(wy-vy)) idist(px,wx,py,wy) ];
  ans(1+(t>0)+(t>1)); % <0 0<=t<=1 t>1     
 end

function d=idist(a,b,c,d)
 d=abs(a-b+1i*(c-d));
end

这里没有看到Java实现，所以我将Javascript函数从接受的答案转换为Java代码:

static double sqr(double x) {
    return x * x;
}
static double dist2(DoublePoint v, DoublePoint w) {
    return sqr(v.x - w.x) + sqr(v.y - w.y);
}
static double distToSegmentSquared(DoublePoint p, DoublePoint v, DoublePoint w) {
    double l2 = dist2(v, w);
    if (l2 == 0) return dist2(p, v);
    double t = ((p.x - v.x) * (w.x - v.x) + (p.y - v.y) * (w.y - v.y)) / l2;
    if (t < 0) return dist2(p, v);
    if (t > 1) return dist2(p, w);
    return dist2(p, new DoublePoint(
            v.x + t * (w.x - v.x),
            v.y + t * (w.y - v.y)
    ));
}
static double distToSegment(DoublePoint p, DoublePoint v, DoublePoint w) {
    return Math.sqrt(distToSegmentSquared(p, v, w));
}
static class DoublePoint {
    public double x;
    public double y;

    public DoublePoint(double x, double y) {
        this.x = x;
        this.y = y;
    }
}