查找点是否位于线段上

Question

我有由两个点A（x1，y1，z1）和B（x2，y2，z2）和点p（x，y，z）定义的线段。如何检查点是否位于线段上？

Answer 1

从线端点A，B中找出点P的距离。如果AB = AP + PB，则P位于线段AB上。

AB = sqrt((x2-x1)*(x2-x1)+(y2-y1)*(y2-y1)+(z2-z1)*(z2-z1));
AP = sqrt((x-x1)*(x-x1)+(y-y1)*(y-y1)+(z-z1)*(z-z1));
PB = sqrt((x2-x)*(x2-x)+(y2-y)*(y2-y)+(z2-z)*(z2-z));
if(AB == AP + PB)
    return true;

Answer 2

如果该行 点，则：

(x - x1) / (x2 - x1) = (y - y1) / (y2 - y1) = (z - z1) / (z2 - z1)

计算所有三个值，如果它们相同（在某种程度的容差范围内），那么你的点就行了。

要测试该点是否在线段中，而不仅仅是在线上，您可以检查

x1 < x < x2, assuming x1 < x2, or
y1 < y < y2, assuming y1 < y2, or
z1 < z < z2, assuming z1 < z2

Answer 3

首先take the cross product of AB and AP。如果它们是共线的，则它将为0。

此时，它仍然可以在延伸过B或A之前的更大线上，所以我认为你应该能够检查pz是否介于az和bz之间。

实际上，这是appears to be a duplicate，并且作为答案之一，它位于Beautiful Code。

Answer 4

您的线段最好由参数方程

定义

对于您的细分中的所有点，以下等式成立： x = x1 +（x2-x1）* p y = y1 +（y2-y1）* p z = z1 +（z2-z1）* p

其中p是[0; 1]

中的数字

所以，如果有一个p使你的点坐标满足那些 3个等式，你的观点是在这一行。并且它在0和1之间 - 它也在线段

Answer 5

以防有人查找内联版本：

public static bool PointOnLine2D (this Vector2 p, Vector2 a, Vector2 b, float t = 1E-03f)
{
    // ensure points are collinear
    var zero = (b.x - a.x) * (p.y - a.y) - (p.x - a.x) * (b.y - a.y);
    if (zero > t || zero < -t) return false;

    // check if x-coordinates are not equal
    if (a.x - b.x > t || b.x - a.x > t)
        // ensure x is between a.x & b.x (use tolerance)
        return a.x > b.x
            ? p.x + t > b.x && p.x - t < a.x
            : p.x + t > a.x && p.x - t < b.x;

    // ensure y is between a.y & b.y (use tolerance)
    return a.y > b.y
        ? p.y + t > b.y && p.y - t < a.y
        : p.y + t > a.y && p.y - t < b.y;
}

Answer 6

这里是2D案例的一些C＃代码：

public static bool PointOnLineSegment(PointD pt1, PointD pt2, PointD pt, double epsilon = 0.001)
{
  if (pt.X - Math.Max(pt1.X, pt2.X) > epsilon || 
      Math.Min(pt1.X, pt2.X) - pt.X > epsilon || 
      pt.Y - Math.Max(pt1.Y, pt2.Y) > epsilon || 
      Math.Min(pt1.Y, pt2.Y) - pt.Y > epsilon)
    return false;

  if (Math.Abs(pt2.X - pt1.X) < epsilon)
    return Math.Abs(pt1.X - pt.X) < epsilon || Math.Abs(pt2.X - pt.X) < epsilon;
  if (Math.Abs(pt2.Y - pt1.Y) < epsilon)
    return Math.Abs(pt1.Y - pt.Y) < epsilon || Math.Abs(pt2.Y - pt.Y) < epsilon;

  double x = pt1.X + (pt.Y - pt1.Y) * (pt2.X - pt1.X) / (pt2.Y - pt1.Y);
  double y = pt1.Y + (pt.X - pt1.X) * (pt2.Y - pt1.Y) / (pt2.X - pt1.X);

  return Math.Abs(pt.X - x) < epsilon || Math.Abs(pt.Y - y) < epsilon;
}

Answer 7

叉积（B - A）×（p - A）应比B - A短得多。理想情况下，叉积为零，但在有限精度浮点硬件上这不太可能。

Answer 8

根据Konstantin上面的答案，这里有一些C代码可以找到一个点是否实际在FINITE线段上。这考虑了水平/垂直线段。这也考虑到浮点数在彼此比较时从不真正“精确”。在大多数情况下，0.001f的默认epsilon就足够了。这适用于2D线......添加“Z”将是微不足道的。 PointF类来自GDI +，基本上只是：struct PointF{float X,Y};

希望这有帮助！

#define DEFFLEQEPSILON 0.001
#define FLOAT_EQE(x,v,e)((((v)-(e))<(x))&&((x)<((v)+(e))))

static bool Within(float fl, float flLow, float flHi, float flEp=DEFFLEQEPSILON){
    if((fl>flLow) && (fl<flHi)){ return true; }
    if(FLOAT_EQE(fl,flLow,flEp) || FLOAT_EQE(fl,flHi,flEp)){ return true; }
    return false;
}

static bool PointOnLine(const PointF& ptL1, const PointF& ptL2, const PointF& ptTest, float flEp=DEFFLEQEPSILON){
    bool bTestX = true;
    const float flX = ptL2.X-ptL1.X;
    if(FLOAT_EQE(flX,0.0f,flEp)){
        // vertical line -- ptTest.X must equal ptL1.X to continue
        if(!FLOAT_EQE(ptTest.X,ptL1.X,flEp)){ return false; }
        bTestX = false;
    }
    bool bTestY = true;
    const float flY = ptL2.Y-ptL1.Y;
    if(FLOAT_EQE(flY,0.0f,flEp)){
        // horizontal line -- ptTest.Y must equal ptL1.Y to continue
        if(!FLOAT_EQE(ptTest.Y,ptL1.Y,flEp)){ return false; }
        bTestY = false;
    }
    // found here: http://stackoverflow.com/a/7050309
    // x = x1 + (x2 - x1) * p
    // y = y1 + (y2 - y1) * p
    // solve for p:
    const float pX = bTestX?((ptTest.X-ptL1.X)/flX):0.5f;
    const float pY = bTestY?((ptTest.Y-ptL1.Y)/flY):0.5f;
    return Within(pX,0.0f,1.0f,flEp) && Within(pY,0.0f,1.0f,flEp);
}

Answer 9

我用它来计算点a和b之间的距离AB。

static void Main(string[] args)
{
        double AB = segment(0, 1, 0, 4);
        Console.WriteLine("Length of segment AB: {0}",AB);
}

static double segment (int ax,int ay, int bx, int by)
{
    Vector a = new Vector(ax,ay);
    Vector b = new Vector(bx,by);
    Vector c = (a & b);
    return Math.Sqrt(c.X + c.Y);
}

struct Vector
{
    public readonly float X;
    public readonly float Y;

    public Vector(float x, float y)
    {
        this.X = x;
        this.Y = y;
    }
    public static Vector operator &(Vector a, Vector b)  
    {
        return new Vector((b.X - a.X) * (b.X - a.X), (b.Y - a.Y) * (b.Y - a.Y));
    }
}

基于Calculate a point along the line A-B at a given distance from A

Answer 10

设V1为向量（B-A），V2 =（p-A），对V1和V2进行归一化。

如果V1 ==（ - V2）则点p在线上，但在A之前，＆amp;因此不在细分市场。 如果V1 == V2，则点p在线上。得到（p-A）的长度并检查它是否小于或等于（B-A）的长度，如果是，则该点在该段上，否则它超过B。

Answer 11

或者如果使用Visual Studio使用GraphicsPath，则让dotnet为您完成繁重的工作

这也允许您添加公差，只要在行外单击即可。

using (Drawing2D.GraphicsPath gp = new Drawing2D.GraphicsPath())
{
    gp.AddLine(new Point(x1, y1), new Point(x2, y2));

    // Make the line as wide as needed (make this larger to allow clicking slightly outside the line) 
    using (Pen objPen = new Pen(Color.Black, 6))
    {
        gp.Widen(objPen);
    }

    if (gp.IsVisible(Mouse.x, Mouse.y))
    {
        // The line was clicked
    }
}

Answer 12

这是我可以在WPF中运行的代码

    public static class Math2DExtensions
    {
        public static bool CheckIsPointOnLineSegment(Point point, Line line, double epsilon = 0.1)
        {
            // Thank you @Rob Agar           
            // (x - x1) / (x2 - x1) = (y - y1) / (y2 - y1)
            // x1 < x < x2, assuming x1 < x2
            // y1 < y < y2, assuming y1 < y2          

            var minX = Math.Min(line.APoint.X, line.BPoint.X);
            var maxX = Math.Max(line.APoint.X, line.BPoint.X);

            var minY = Math.Min(line.APoint.Y, line.BPoint.Y);
            var maxY = Math.Max(line.APoint.Y, line.BPoint.Y);

            if (!(minX <= point.X) || !(point.X <= maxX) || !(minY <= point.Y) || !(point.Y <= maxY))
            {
                return false;
            }
            
            if (Math.Abs(line.APoint.X - line.BPoint.X) < epsilon)
            {
                return Math.Abs(line.APoint.X - point.X) < epsilon || Math.Abs(line.BPoint.X - point.X) < epsilon;
            }

            if (Math.Abs(line.APoint.Y - line.BPoint.Y) < epsilon)
            {
                return Math.Abs(line.APoint.Y - point.Y) < epsilon || Math.Abs(line.BPoint.Y - point.Y) < epsilon;
            }

            if (Math.Abs((point.X - line.APoint.X) / (line.BPoint.X - line.APoint.X) - (point.Y - line.APoint.Y) / (line.BPoint.Y - line.APoint.Y)) < epsilon)
            {
                return true;
            }
            else
            {
                return false;
            }
        }
    }

    public record Line
    {
        public Point APoint { get; init; }

        public Point BPoint { get; init; }
    }

我的代码在 github

谢谢@Rob Agar和@MetaMapper

Answer 13

您可以检查点是否位于由point1和point2定义的两个平面与线方向之间：

///  Returns the closest point from @a point to this line on this line.
vector3 <Type>
line3d <Type>::closest_point (const vector3 <Type> & point) const
{
    return this -> point () + direction () * dot (point - this -> point (), direction ());
}

///  Returns true if @a point lies between point1 and point2.
template <class Type>
bool
line_segment3 <Type>::is_between (const vector3 <Type> & point) const
{
    const auto closest = line () .closest_point (point);
    return abs ((closest - point0 ()) + (closest - point1 ())) <= abs (point0 () - point1 ());
}