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)。

这里它使用Swift

    /* Distance from a point (p1) to line l1 l2 */
func distanceFromPoint(p: CGPoint, toLineSegment l1: CGPoint, and l2: CGPoint) -> CGFloat {
    let A = p.x - l1.x
    let B = p.y - l1.y
    let C = l2.x - l1.x
    let D = l2.y - l1.y

    let dot = A * C + B * D
    let len_sq = C * C + D * D
    let param = dot / len_sq

    var xx, yy: CGFloat

    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
    }

    let dx = p.x - xx
    let dy = p.y - yy

    return sqrt(dx * dx + dy * dy)
}

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})

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, :])

in R

     #distance beetween segment ab and point c in 2D space
getDistance_ort_2 <- function(a, b, c){
  #go to complex numbers
  A<-c(a[1]+1i*a[2],b[1]+1i*b[2])
  q=c[1]+1i*c[2]
  
  #function to get coefficients of line (ab)
  getAlphaBeta <- function(A)
  { a<-Re(A[2])-Re(A[1])
    b<-Im(A[2])-Im(A[1])
    ab<-as.numeric()
    ab[1] <- -Re(A[1])*b/a+Im(A[1])
    ab[2] <-b/a
    if(Im(A[1])==Im(A[2])) ab<- c(Im(A[1]),0)
    if(Re(A[1])==Re(A[2])) ab <- NA
    return(ab)
  }
  
  #function to get coefficients of line ortogonal to line (ab) which goes through point q
  getAlphaBeta_ort<-function(A,q)
  { ab <- getAlphaBeta(A) 
  coef<-c(Re(q)/ab[2]+Im(q),-1/ab[2])
  if(Re(A[1])==Re(A[2])) coef<-c(Im(q),0)
  return(coef)
  }
  
  #function to get coordinates of interception point 
  #between line (ab) and its ortogonal which goes through point q
  getIntersection_ort <- function(A, q){
    A.ab <- getAlphaBeta(A)
    q.ab <- getAlphaBeta_ort(A,q)
    if (!is.na(A.ab[1])&A.ab[2]==0) {
      x<-Re(q)
      y<-Im(A[1])}
    if (is.na(A.ab[1])) {
      x<-Re(A[1])
      y<-Im(q)
    } 
    if (!is.na(A.ab[1])&A.ab[2]!=0) {
      x <- (q.ab[1] - A.ab[1])/(A.ab[2] - q.ab[2])
      y <- q.ab[1] + q.ab[2]*x}
    xy <- x + 1i*y  
    return(xy)
  }
  
  intersect<-getIntersection_ort(A,q)
  if ((Mod(A[1]-intersect)+Mod(A[2]-intersect))>Mod(A[1]-A[2])) {dist<-min(Mod(A[1]-q),Mod(A[2]-q))
  } else dist<-Mod(q-intersect)
  return(dist)
}



 

只是遇到了这个，我想我应该添加一个Lua实现。它假设点以表{x=xVal, y=yVal}给出，直线或线段由包含两个点的表给出(见下面的例子):

function distance( P1, P2 )
    return math.sqrt((P1.x-P2.x)^2 + (P1.y-P2.y)^2)
end

-- Returns false if the point lies beyond the reaches of the segment
function distPointToSegment( line, P )
    if line[1].x == line[2].x and line[1].y == line[2].y then
        print("Error: Not a line!")
        return false
    end

    local d = distance( line[1], line[2] )

    local t = ((P.x - line[1].x)*(line[2].x - line[1].x) + (P.y - line[1].y)*(line[2].y - line[1].y))/(d^2)

    local projection = {}
    projection.x = line[1].x + t*(line[2].x-line[1].x)
    projection.y = line[1].y + t*(line[2].y-line[1].y)

    if t >= 0 and t <= 1 then   -- within line segment?
        return distance( projection, {x=P.x, y=P.y} )
    else
        return false
    end
end

-- Returns value even if point is further down the line (outside segment)
function distPointToLine( line, P )
    if line[1].x == line[2].x and line[1].y == line[2].y then
        print("Error: Not a line!")
        return false
    end

    local d = distance( line[1], line[2] )

    local t = ((P.x - line[1].x)*(line[2].x - line[1].x) + (P.y - line[1].y)*(line[2].y - line[1].y))/(d^2)

    local projection = {}
    projection.x = line[1].x + t*(line[2].x-line[1].x)
    projection.y = line[1].y + t*(line[2].y-line[1].y)

    return distance( projection, {x=P.x, y=P.y} )
end

使用示例:

local P1 = {x = 0, y = 0}
local P2 = {x = 10, y = 10}
local line = { P1, P2 }
local P3 = {x = 7, y = 15}
print(distPointToLine( line, P3 ))  -- prints 5.6568542494924
print(distPointToSegment( line, P3 )) -- prints false