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
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))
其他回答
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)
}
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。
该算法基于求出指定直线与包含指定点的正交直线的交点,并计算其距离。在线段的情况下,我们必须检查交点是否在线段的点之间,如果不是这样,则最小距离是指定点与线段的一个端点之间的距离。这是一个c#实现。
Double Distance(Point a, Point b)
{
double xdiff = a.X - b.X, ydiff = a.Y - b.Y;
return Math.Sqrt((long)xdiff * xdiff + (long)ydiff * ydiff);
}
Boolean IsBetween(double x, double a, double b)
{
return ((a <= b && x >= a && x <= b) || (a > b && x <= a && x >= b));
}
Double GetDistance(Point pt, Point pt1, Point pt2, out Point intersection)
{
Double a, x, y, R;
if (pt1.X != pt2.X) {
a = (double)(pt2.Y - pt1.Y) / (pt2.X - pt1.X);
x = (a * (pt.Y - pt1.Y) + a * a * pt1.X + pt.X) / (a * a + 1);
y = a * x + pt1.Y - a * pt1.X; }
else { x = pt1.X; y = pt.Y; }
if (IsBetween(x, pt1.X, pt2.X) && IsBetween(y, pt1.Y, pt2.Y)) {
intersection = new Point((int)x, (int)y);
R = Distance(intersection, pt); }
else {
double d1 = Distance(pt, pt1), d2 = Distance(pt, pt2);
if (d1 < d2) { intersection = pt1; R = d1; }
else { intersection = pt2; R = d2; }}
return R;
}
省道和颤振的解决方法:
import 'dart:math' as math;
class Utils {
static double shortestDistance(Point p1, Point p2, Point p3){
double px = p2.x - p1.x;
double py = p2.y - p1.y;
double temp = (px*px) + (py*py);
double u = ((p3.x - p1.x)*px + (p3.y - p1.y)* py) /temp;
if(u>1){
u=1;
}
else if(u<0){
u=0;
}
double x = p1.x + u*px;
double y = p1.y + u*py;
double dx = x - p3.x;
double dy = y - p3.y;
double dist = math.sqrt(dx*dx+dy*dy);
return dist;
}
}
class Point {
double x;
double y;
Point(this.x, this.y);
}
我需要一个Godot (GDscript)的实现,所以我写了一个基于grumdrig接受的答案:
func minimum_distance(v: Vector2, w: Vector2, p: Vector2):
# Return minimum distance between line segment vw and point p
var l2: float = (v - w).length_squared() # i.e. |w-v|^2 - avoid a sqrt
if l2 == 0.0:
return p.distance_to(v) # v == w case
# 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
# We clamp t from [0,1] to handle points outside the segment vw.
var t: float = max(0, min(1, (p - v).dot(w - v) / l2))
var projection: Vector2 = v + t * (w - v) # Projection falls on the segment
return p.distance_to(projection)