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
上面的函数在垂直线上不起作用。这是一个工作正常的函数! 与点p1 p2相交。CheckPoint为p;
public float DistanceOfPointToLine2(PointF p1, PointF p2, PointF p)
{
// (y1-y2)x + (x2-x1)y + (x1y2-x2y1)
//d(P,L) = --------------------------------
// sqrt( (x2-x1)pow2 + (y2-y1)pow2 )
double ch = (p1.Y - p2.Y) * p.X + (p2.X - p1.X) * p.Y + (p1.X * p2.Y - p2.X * p1.Y);
double del = Math.Sqrt(Math.Pow(p2.X - p1.X, 2) + Math.Pow(p2.Y - p1.Y, 2));
double d = ch / del;
return (float)d;
}
其他回答
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.
嘿,我昨天才写的。它在Actionscript 3.0中,基本上是Javascript,尽管你可能没有相同的Point类。
//st = start of line segment
//b = the line segment (as in: st + b = end of line segment)
//pt = point to test
//Returns distance from point to line segment.
//Note: nearest point on the segment to the test point is right there if we ever need it
public static function linePointDist( st:Point, b:Point, pt:Point ):Number
{
var nearestPt:Point; //closest point on seqment to pt
var keyDot:Number = dot( b, pt.subtract( st ) ); //key dot product
var bLenSq:Number = dot( b, b ); //Segment length squared
if( keyDot <= 0 ) //pt is "behind" st, use st
{
nearestPt = st
}
else if( keyDot >= bLenSq ) //pt is "past" end of segment, use end (notice we are saving twin sqrts here cuz)
{
nearestPt = st.add(b);
}
else //pt is inside segment, reuse keyDot and bLenSq to get percent of seqment to move in to find closest point
{
var keyDotToPctOfB:Number = keyDot/bLenSq; //REM dot product comes squared
var partOfB:Point = new Point( b.x * keyDotToPctOfB, b.y * keyDotToPctOfB );
nearestPt = st.add(partOfB);
}
var dist:Number = (pt.subtract(nearestPt)).length;
return dist;
}
此外,这里有一个关于这个问题的相当完整和可读的讨论:notejot.com
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})
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));
}
请参见以下网站中的Matlab几何工具箱: http://people.sc.fsu.edu/~jburkardt/m_src/geometry/geometry.html
按Ctrl +f,输入“segment”,查找线段相关函数。函数“segment_point_dist_2d.”和segment_point_dist_3d。M "是你需要的。
几何代码有C版本、c++版本、FORTRAN77版本、FORTRAN90版本和MATLAB版本。