Grumdrig的c++ /JavaScript实现对我来说非常有用，所以我提供了我正在使用的Python直接端口。完整的代码在这里。

class Point(object):
  def __init__(self, x, y):
    self.x = float(x)
    self.y = float(y)

def square(x):
  return x * x

def distance_squared(v, w):
  return square(v.x - w.x) + square(v.y - w.y)

def distance_point_segment_squared(p, v, w):
  # Segment length squared, |w-v|^2
  d2 = distance_squared(v, w) 
  if d2 == 0: 
    # v == w, return distance to v
    return distance_squared(p, v)
  # Consider the line extending the segment, parameterized as v + t (w - v).
  # We find projection of point p onto the line.
  # It falls where t = [(p-v) . (w-v)] / |w-v|^2
  t = ((p.x - v.x) * (w.x - v.x) + (p.y - v.y) * (w.y - v.y)) / d2;
  if t < 0:
    # Beyond v end of the segment
    return distance_squared(p, v)
  elif t > 1.0:
    # Beyond w end of the segment
    return distance_squared(p, w)
  else:
    # Projection falls on the segment.
    proj = Point(v.x + t * (w.x - v.x), v.y + t * (w.y - v.y))
    # print proj.x, proj.y
    return distance_squared(p, proj)

下面是HSQLDB的SQL实现:

CREATE FUNCTION dist_to_segment(px double, py double, vx double, vy double, wx double, wy double)
  RETURNS double
BEGIN atomic
   declare l2 double;
   declare t double;
   declare nx double;
   declare ny double;
   set l2 =(vx - wx)*(vx - wx) + (vy - wy)*(vy - wy);
   IF l2 = 0 THEN
     RETURN sqrt((vx - px)*(vx - px) + (vy - py)*(vy - py));
   ELSE
     set t = ((px - vx) * (wx - vx) + (py - vy) * (wy - vy)) / l2;
     set t = GREATEST(0, LEAST(1, t));
     set nx=vx + t * (wx - vx);
     set ny=vy + t * (wy - vy);
     RETURN sqrt((nx - px)*(nx - px) + (ny - py)*(ny - py));
   END IF;
END;

Postgres的实现:

CREATE FUNCTION dist_to_segment(px numeric, py numeric, vx numeric, vy numeric, wx numeric, wy numeric)
  RETURNS numeric
AS $$
   declare l2 numeric;
   declare t numeric;
   declare nx numeric;
   declare ny numeric;
BEGIN 
   l2 := (vx - wx)*(vx - wx) + (vy - wy)*(vy - wy);
   IF l2 = 0 THEN
     RETURN sqrt((vx - px)*(vx - px) + (vy - py)*(vy - py));
   ELSE
     t := ((px - vx) * (wx - vx) + (py - vy) * (wy - vy)) / l2;
     t := GREATEST(0, LEAST(1, t));
     nx := vx + t * (wx - vx);
     ny := vy + t * (wy - vy);
     RETURN sqrt((nx - px)*(nx - px) + (ny - py)*(ny - py));
   END IF;
END;
$$ LANGUAGE plpgsql;

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();
}

使用arctangents的一行解决方案:

思路是将A移动到(0,0)，并顺时针旋转三角形，使C位于X轴上， 当这种情况发生时，By就是距离。

a角= Atan(Cy - Ay, Cx - Ax); b角= Atan(By - Ay, Bx - Ax); AB长度=平方根((Bx - Ax)²+ (By - Ay)²) By = Sin (bAngle - aAngle) * ABLength

C#

public double Distance(Point a, Point b, Point c)
{
    // normalize points
    Point cn = new Point(c.X - a.X, c.Y - a.Y);
    Point bn = new Point(b.X - a.X, b.Y - a.Y);

    double angle = Math.Atan2(bn.Y, bn.X) - Math.Atan2(cn.Y, cn.X);
    double abLength = Math.Sqrt(bn.X*bn.X + bn.Y*bn.Y);

    return Math.Sin(angle)*abLength;
}

一行c#(要转换为SQL)

double distance = Math.Sin(Math.Atan2(b.Y - a.Y, b.X - a.X) - Math.Atan2(c.Y - a.Y, c.X - a.X)) * Math.Sqrt((b.X - a.X) * (b.X - a.X) + (b.Y - a.Y) * (b.Y - a.Y))

I'm assuming you want to find the shortest distance between the point and a line segment; to do this, you need to find the line (lineA) which is perpendicular to your line segment (lineB) which goes through your point, determine the intersection between that line (lineA) and your line which goes through your line segment (lineB); if that point is between the two points of your line segment, then the distance is the distance between your point and the point you just found which is the intersection of lineA and lineB; if the point is not between the two points of your line segment, you need to get the distance between your point and the closer of two ends of the line segment; this can be done easily by taking the square distance (to avoid a square root) between the point and the two points of the line segment; whichever is closer, take the square root of that one.

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))