我试图创建一个快速的2D点内多边形算法,用于命中测试(例如多边形.contains(p:点))。对有效技术的建议将不胜感激。


当前回答

VBA版本:

注意:请记住,如果你的多边形是地图中的一个区域,纬度/经度是Y/X值,而不是X/Y(纬度= Y,经度= X),因为从我的理解来看,这是历史含义,因为经度不是一个测量值。

类模块:CPoint

Private pXValue As Double
Private pYValue As Double

'''''X Value Property'''''

Public Property Get X() As Double
    X = pXValue
End Property

Public Property Let X(Value As Double)
    pXValue = Value
End Property

'''''Y Value Property'''''

Public Property Get Y() As Double
    Y = pYValue
End Property

Public Property Let Y(Value As Double)
    pYValue = Value
End Property

模块:

Public Function isPointInPolygon(p As CPoint, polygon() As CPoint) As Boolean

    Dim i As Integer
    Dim j As Integer
    Dim q As Object
    Dim minX As Double
    Dim maxX As Double
    Dim minY As Double
    Dim maxY As Double
    minX = polygon(0).X
    maxX = polygon(0).X
    minY = polygon(0).Y
    maxY = polygon(0).Y

    For i = 1 To UBound(polygon)
        Set q = polygon(i)
        minX = vbMin(q.X, minX)
        maxX = vbMax(q.X, maxX)
        minY = vbMin(q.Y, minY)
        maxY = vbMax(q.Y, maxY)
    Next i

    If p.X < minX Or p.X > maxX Or p.Y < minY Or p.Y > maxY Then
        isPointInPolygon = False
        Exit Function
    End If


    ' SOURCE: http://www.ecse.rpi.edu/Homepages/wrf/Research/Short_Notes/pnpoly.html

    isPointInPolygon = False
    i = 0
    j = UBound(polygon)

    Do While i < UBound(polygon) + 1
        If (polygon(i).Y > p.Y) Then
            If (polygon(j).Y < p.Y) Then
                If p.X < (polygon(j).X - polygon(i).X) * (p.Y - polygon(i).Y) / (polygon(j).Y - polygon(i).Y) + polygon(i).X Then
                    isPointInPolygon = True
                    Exit Function
                End If
            End If
        ElseIf (polygon(i).Y < p.Y) Then
            If (polygon(j).Y > p.Y) Then
                If p.X < (polygon(j).X - polygon(i).X) * (p.Y - polygon(i).Y) / (polygon(j).Y - polygon(i).Y) + polygon(i).X Then
                    isPointInPolygon = True
                    Exit Function
                End If
            End If
        End If
        j = i
        i = i + 1
    Loop   
End Function

Function vbMax(n1, n2) As Double
    vbMax = IIf(n1 > n2, n1, n2)
End Function

Function vbMin(n1, n2) As Double
    vbMin = IIf(n1 > n2, n2, n1)
End Function


Sub TestPointInPolygon()

    Dim i As Integer
    Dim InPolygon As Boolean

'   MARKER Object
    Dim p As CPoint
    Set p = New CPoint
    p.X = <ENTER X VALUE HERE>
    p.Y = <ENTER Y VALUE HERE>

'   POLYGON OBJECT
    Dim polygon() As CPoint
    ReDim polygon(<ENTER VALUE HERE>) 'Amount of vertices in polygon - 1
    For i = 0 To <ENTER VALUE HERE> 'Same value as above
       Set polygon(i) = New CPoint
       polygon(i).X = <ASSIGN X VALUE HERE> 'Source a list of values that can be looped through
       polgyon(i).Y = <ASSIGN Y VALUE HERE> 'Source a list of values that can be looped through
    Next i

    InPolygon = isPointInPolygon(p, polygon)
    MsgBox InPolygon

End Sub

其他回答

这只适用于凸形状,但是Minkowski Portal Refinement和GJK也是测试一个点是否在多边形中的很好的选择。您使用闵可夫斯基减法从多边形中减去点,然后运行这些算法来查看多边形是否包含原点。

另外,有趣的是,你可以用支持函数更隐式地描述你的形状,它以一个方向向量作为输入,并输出沿该向量的最远点。这可以让你描述任何凸形状..弯曲的,由多边形制成的,或混合的您还可以执行一些操作,将简单支持函数的结果组合起来,以生成更复杂的形状。

更多信息: http://xenocollide.snethen.com/mpr2d.html

此外,game programming gems 7讨论了如何在3d中做到这一点(:

bobobobo引用的Eric Haines的文章真的很棒。特别有趣的是比较算法性能的表格;角度求和法和其他方法比起来真的很差。同样有趣的是,使用查找网格将多边形进一步细分为“in”和“out”扇区的优化可以使测试非常快,即使是在> 1000条边的多边形上。

不管怎样,现在还为时过早,但我的投票倾向于“交叉”方法,我认为这几乎就是Mecki所描述的。然而,我发现大卫·伯克(David Bourke)对它进行了最简洁的描述和编纂。我喜欢它不需要真正的三角函数,它适用于凸和凹,而且随着边数的增加,它的表现也相当不错。

顺便说一下,这是Eric Haines文章中的一个性能表,在随机多边形上进行测试。

                       number of edges per polygon
                         3       4      10      100    1000
MacMartin               2.9     3.2     5.9     50.6    485
Crossings               3.1     3.4     6.8     60.0    624
Triangle Fan+edge sort  1.1     1.8     6.5     77.6    787
Triangle Fan            1.2     2.1     7.3     85.4    865
Barycentric             2.1     3.8    13.8    160.7   1665
Angle Summation        56.2    70.4   153.6   1403.8  14693

Grid (100x100)          1.5     1.5     1.6      2.1      9.8
Grid (20x20)            1.7     1.7     1.9      5.7     42.2
Bins (100)              1.8     1.9     2.7     15.1    117
Bins (20)               2.1     2.2     3.7     26.3    278

在C语言的多边形测试中,有一个点没有使用光线投射。它可以用于重叠区域(自我交叉),请参阅use_holes参数。

/* math lib (defined below) */
static float dot_v2v2(const float a[2], const float b[2]);
static float angle_signed_v2v2(const float v1[2], const float v2[2]);
static void copy_v2_v2(float r[2], const float a[2]);

/* intersection function */
bool isect_point_poly_v2(const float pt[2], const float verts[][2], const unsigned int nr,
                         const bool use_holes)
{
    /* we do the angle rule, define that all added angles should be about zero or (2 * PI) */
    float angletot = 0.0;
    float fp1[2], fp2[2];
    unsigned int i;
    const float *p1, *p2;

    p1 = verts[nr - 1];

    /* first vector */
    fp1[0] = p1[0] - pt[0];
    fp1[1] = p1[1] - pt[1];

    for (i = 0; i < nr; i++) {
        p2 = verts[i];

        /* second vector */
        fp2[0] = p2[0] - pt[0];
        fp2[1] = p2[1] - pt[1];

        /* dot and angle and cross */
        angletot += angle_signed_v2v2(fp1, fp2);

        /* circulate */
        copy_v2_v2(fp1, fp2);
        p1 = p2;
    }

    angletot = fabsf(angletot);
    if (use_holes) {
        const float nested = floorf((angletot / (float)(M_PI * 2.0)) + 0.00001f);
        angletot -= nested * (float)(M_PI * 2.0);
        return (angletot > 4.0f) != ((int)nested % 2);
    }
    else {
        return (angletot > 4.0f);
    }
}

/* math lib */

static float dot_v2v2(const float a[2], const float b[2])
{
    return a[0] * b[0] + a[1] * b[1];
}

static float angle_signed_v2v2(const float v1[2], const float v2[2])
{
    const float perp_dot = (v1[1] * v2[0]) - (v1[0] * v2[1]);
    return atan2f(perp_dot, dot_v2v2(v1, v2));
}

static void copy_v2_v2(float r[2], const float a[2])
{
    r[0] = a[0];
    r[1] = a[1];
}

注意:这是一个不太理想的方法,因为它包含很多对atan2f的调用,但它可能会引起阅读这个线程的开发人员的兴趣(在我的测试中,它比使用线交方法慢23倍)。

令人惊讶的是之前没有人提出这个问题,但是对于需要数据库的实用主义者来说:MongoDB对Geo查询提供了出色的支持,包括这个查询。

你需要的是:

db.neighborhoods。findOne({geometry: {$geoIntersects: {$geometry: { type: "Point",坐标:["经度","纬度"]}}} })

communities是存储一个或多个标准GeoJson格式多边形的集合。如果查询返回null,则表示不相交,否则为。

这里有详细的记录: https://docs.mongodb.com/manual/tutorial/geospatial-tutorial/

在330个不规则多边形网格中,超过6000个点分类的性能不到一分钟,没有任何优化,包括用各自的多边形更新文档的时间。

VBA版本:

注意:请记住,如果你的多边形是地图中的一个区域,纬度/经度是Y/X值,而不是X/Y(纬度= Y,经度= X),因为从我的理解来看,这是历史含义,因为经度不是一个测量值。

类模块:CPoint

Private pXValue As Double
Private pYValue As Double

'''''X Value Property'''''

Public Property Get X() As Double
    X = pXValue
End Property

Public Property Let X(Value As Double)
    pXValue = Value
End Property

'''''Y Value Property'''''

Public Property Get Y() As Double
    Y = pYValue
End Property

Public Property Let Y(Value As Double)
    pYValue = Value
End Property

模块:

Public Function isPointInPolygon(p As CPoint, polygon() As CPoint) As Boolean

    Dim i As Integer
    Dim j As Integer
    Dim q As Object
    Dim minX As Double
    Dim maxX As Double
    Dim minY As Double
    Dim maxY As Double
    minX = polygon(0).X
    maxX = polygon(0).X
    minY = polygon(0).Y
    maxY = polygon(0).Y

    For i = 1 To UBound(polygon)
        Set q = polygon(i)
        minX = vbMin(q.X, minX)
        maxX = vbMax(q.X, maxX)
        minY = vbMin(q.Y, minY)
        maxY = vbMax(q.Y, maxY)
    Next i

    If p.X < minX Or p.X > maxX Or p.Y < minY Or p.Y > maxY Then
        isPointInPolygon = False
        Exit Function
    End If


    ' SOURCE: http://www.ecse.rpi.edu/Homepages/wrf/Research/Short_Notes/pnpoly.html

    isPointInPolygon = False
    i = 0
    j = UBound(polygon)

    Do While i < UBound(polygon) + 1
        If (polygon(i).Y > p.Y) Then
            If (polygon(j).Y < p.Y) Then
                If p.X < (polygon(j).X - polygon(i).X) * (p.Y - polygon(i).Y) / (polygon(j).Y - polygon(i).Y) + polygon(i).X Then
                    isPointInPolygon = True
                    Exit Function
                End If
            End If
        ElseIf (polygon(i).Y < p.Y) Then
            If (polygon(j).Y > p.Y) Then
                If p.X < (polygon(j).X - polygon(i).X) * (p.Y - polygon(i).Y) / (polygon(j).Y - polygon(i).Y) + polygon(i).X Then
                    isPointInPolygon = True
                    Exit Function
                End If
            End If
        End If
        j = i
        i = i + 1
    Loop   
End Function

Function vbMax(n1, n2) As Double
    vbMax = IIf(n1 > n2, n1, n2)
End Function

Function vbMin(n1, n2) As Double
    vbMin = IIf(n1 > n2, n2, n1)
End Function


Sub TestPointInPolygon()

    Dim i As Integer
    Dim InPolygon As Boolean

'   MARKER Object
    Dim p As CPoint
    Set p = New CPoint
    p.X = <ENTER X VALUE HERE>
    p.Y = <ENTER Y VALUE HERE>

'   POLYGON OBJECT
    Dim polygon() As CPoint
    ReDim polygon(<ENTER VALUE HERE>) 'Amount of vertices in polygon - 1
    For i = 0 To <ENTER VALUE HERE> 'Same value as above
       Set polygon(i) = New CPoint
       polygon(i).X = <ASSIGN X VALUE HERE> 'Source a list of values that can be looped through
       polgyon(i).Y = <ASSIGN Y VALUE HERE> 'Source a list of values that can be looped through
    Next i

    InPolygon = isPointInPolygon(p, polygon)
    MsgBox InPolygon

End Sub