多边形的重心

Question

我正在尝试编写一个PHP函数来计算多边形的重心。

我看过其他类似的问题，但我似乎无法找到解决方案。

我的问题是我需要能够计算规则和不规则多边形甚至自相交多边形的重心。

这可能吗？

我也读过：http://paulbourke.net/geometry/polyarea/ 但这仅限于非自相交多边形。

我该怎么做？你能指出我正确的方向吗？

Answer 1

重心（也称为“质心”或“质心”）可以使用以下公式计算：

X = SUM[(Xi + Xi+1) * (Xi * Yi+1 - Xi+1 * Yi)] / 6 / A
Y = SUM[(Yi + Yi+1) * (Xi * Yi+1 - Xi+1 * Yi)] / 6 / A

摘自Wikipedia： 由n个顶点（x0，y0），（x1，y1），...，（xn-1，yn-1）定义的非自相交闭合多边形的质心是点（Cx，Cy），其中
    
    
其中A是多边形的有符号区域，
    

使用VBasic的示例：

' Find the polygon's centroid.
Public Sub FindCentroid(ByRef X As Single, ByRef Y As _
    Single)
Dim pt As Integer
Dim second_factor As Single
Dim polygon_area As Single

    ' Add the first point at the end of the array.
    ReDim Preserve m_Points(1 To m_NumPoints + 1)
    m_Points(m_NumPoints + 1) = m_Points(1)

    ' Find the centroid.
    X = 0
    Y = 0
    For pt = 1 To m_NumPoints
        second_factor = _
            m_Points(pt).X * m_Points(pt + 1).Y - _
            m_Points(pt + 1).X * m_Points(pt).Y
        X = X + (m_Points(pt).X + m_Points(pt + 1).X) * _
            second_factor
        Y = Y + (m_Points(pt).Y + m_Points(pt + 1).Y) * _
            second_factor
    Next pt

    ' Divide by 6 times the polygon's area.
    polygon_area = PolygonArea
    X = X / 6 / polygon_area
    Y = Y / 6 / polygon_area

    ' If the values are negative, the polygon is
    ' oriented counterclockwise. Reverse the signs.
    If X < 0 Then
        X = -X
        Y = -Y
    End If
End Sub

有关详情，请查看此website或Wikipedia。

希望它有所帮助。

问候！

Answer 2

在冷c ++中，并假设你有一个带有x和y属性的Vec2结构：

const Vec2 findCentroid(Vec2* pts, size_t nPts){
    Vec2 off = pts[0];
    float twicearea = 0;
    float x = 0;
    float y = 0;
    Vec2 p1, p2;
    float f;
    for (int i = 0, j = nPts - 1; i < nPts; j = i++) {
        p1 = pts[i];
        p2 = pts[j];
        f = (p1.x - off.x) * (p2.y - off.y) - (p2.x - off.x) * (p1.y - off.y);
        twicearea += f;
        x += (p1.x + p2.x - 2 * off.x) * f;
        y += (p1.y + p2.y - 2 * off.y) * f;
    }

    f = twicearea * 3;

    return Vec2(x / f + off.x, y / f + off.y);
}

并在javascript中：

function findCentroid(pts, nPts) {
    var off = pts[0];
    var twicearea = 0;
    var x = 0;
    var y = 0;
    var p1,p2;
    var f;
    for (var i = 0, j = nPts - 1; i < nPts; j = i++) {
        p1 = pts[i];
        p2 = pts[j];
        f = (p1.lat - off.lat) * (p2.lng - off.lng) - (p2.lat - off.lat) * (p1.lng - off.lng);
        twicearea += f;
        x += (p1.lat + p2.lat - 2 * off.lat) * f;
        y += (p1.lng + p2.lng - 2 * off.lng) * f;
    }
    f = twicearea * 3;
    return {
    X: x / f + off.lat,
    Y: y / f + off.lng
    };
}

或在旧的c中，并假设您有一个带有x和y属性的Point结构：

const Point centroidForPoly(const int numVerts, const Point* verts)
{
    float sum = 0.0f;
    Point vsum = 0;

    for (int i = 0; i<numVerts; i++){
        Point v1 = verts[i];
        Point v2 = verts[(i + 1) % numVerts];
        float cross = v1.x*v2.y - v1.y*v2.x;
        sum += cross;
        vsum = Point(((v1.x + v2.x) * cross) + vsum.x, ((v1.y + v2.y) * cross) + vsum.y);
    }

    float z = 1.0f / (3.0f * sum);
    return Point(vsum.x * z, vsum.y * z);
}

Answer 3

迅速4，基于上面给出的c答案

/// Given an array of points, find the "center of gravity" of the points
/// - Parameters:
///     - points: Array of points
/// - Returns:
///     - Point or nil if input points count < 3
static func centerOfPoints(points: [CGPoint]) -> CGPoint? {
    if points.count < 3 {
        return nil
    }

    var sum: CGFloat = 0
    var pSum: CGPoint = .zero

    for i in 0..<points.count {
        let p1 = points[i]
        let p2 = points[(i+1) % points.count]
        let cross = p1.x * p2.y - p1.y * p2.x
        sum += cross
        pSum = CGPoint(x:((p1.x + p2.x) * cross) + pSum.x,
                       y:((p1.y + p2.y) * cross) + pSum.y)
    }

    let z = 1 / (3 * sum)
    return CGPoint(x:pSum.x * z,
                   y:pSum.y * z)
}

Answer 4

在php中：

// Find the polygon's centroid.
function getCenter($polygon)
{ 
    $NumPoints = count($polygon);

    if($polygon[$NumPoints-1] == $polygon[0]){
        $NumPoints--;
    }else{
        //Add the first point at the end of the array.
        $polygon[$NumPoints] = $polygon[0];
    }

    // Find the centroid.
    $X = 0;
    $Y = 0;
    For ($pt = 0 ;$pt<= $NumPoints-1;$pt++){
        $factor = $polygon[$pt][0] * $polygon[$pt + 1][1] - $polygon[$pt + 1][0] * $polygon[$pt][1];
        $X += ($polygon[$pt][0] + $polygon[$pt + 1][0]) * $factor;
        $Y += ($polygon[$pt][1] + $polygon[$pt + 1][1]) * $factor;
    }

    // Divide by 6 times the polygon's area.
    $polygon_area = ComputeArea($polygon);
    $X = $X / 6 / $polygon_area;
    $Y = $Y / 6 / $polygon_area;

    return array($X, $Y);
}


function ComputeArea($polygon)
{ 
    $NumPoints = count($polygon);

    if($polygon[$NumPoints-1] == $polygon[0]){
        $NumPoints--;
    }else{
        //Add the first point at the end of the array.
        $polygon[$NumPoints] = $polygon[0];
    }

    $area = 0;

    for ($i = 0; $i < $NumPoints; $i++) {
      $i1 = ($i + 1) % $NumPoints;
      $area += ($polygon[$i][1] + $polygon[$i1][1]) * ($polygon[$i1][0] - $polygon[$i][0]);
    }

    $area /= 2;
    return $area;
}

阅读更多内容：

PHP: How to determine the center of a Polygon

Answer 5

这是我用Java实现的公认解决方案，我添加了一个额外的条件检查，因为我的一些多边形是扁平的，没有区域，而不是给我中点，它返回（0,0）。因此，在这种情况下，我引用了一种简单平均顶点的不同方法。最后的舍入是因为我想将输出对象保持为整数，即使它不精确，但我欢迎你删除那一点。此外，因为我的所有积分都是正整数，所以检查对我来说很有意义，但对你来说，添加区域检查== 0也是有道理的。

private Vertex getCentroid() {

        double xsum = 0, ysum = 0, A = 0;
        for (int i = 0; i < corners.size() ; i++) {

            int iPlusOne = (i==corners.size()-1)?0:i+1;

            xsum += (corners.get(i).getX() + corners.get(iPlusOne).getX()) * (corners.get(i).getX() * corners.get(iPlusOne).getY() - corners.get(iPlusOne).getX() * corners.get(i).getY());
            ysum += (corners.get(i).getY() + corners.get(iPlusOne).getY()) * (corners.get(i).getX() * corners.get(iPlusOne).getY() - corners.get(iPlusOne).getX() * corners.get(i).getY());
            A += (corners.get(i).getX() * corners.get(iPlusOne).getY() - corners.get(iPlusOne).getX() * corners.get(i).getY());
        }
        A = A / 2;
        if(xsum==0 &&ysum==0)
        {
            area = averageHeight/2;
            return getMidpointCenter();
        }
        double x = xsum / (6 * A);
        double y = ysum / (6 * A);
        area = A;


        return new Vertex((int) Math.round(x), (int) Math.round(y));
    }

Answer 6

由于我们在使用不同语言实现这个算法时非常有趣，所以这是我为Python编写的版本：

def polygon_centre_area(vertices: Sequence[Sequence[float]]) -> Tuple[Sequence[float], float]:
    x_cent = y_cent = area = 0
    v_local = vertices + [vertices[0]]

    for i in range(len(v_local) - 1):
        factor = v_local[i][0] * v_local[i+1][1] - v_local[i+1][0] * v_local[i][1]
        area += factor
        x_cent += (v_local[i][0] + v_local[i+1][0]) * factor
        y_cent += (v_local[i][1] + v_local[i+1][1]) * factor

    area /= 2.0
    x_cent /= (6 * area)
    y_cent /= (6 * area)

    area = math.fabs(area)

    return ([x_cent, y_cent], area)

Answer 7

这是我在Python中的实现，它基于Joseph的C ++实现。我认为这比其他python答案更清楚。

def find_centroid(polygon):
    """ Computes the centroid (a.k.a. center of gravity) for a non-self-intersecting polygon.

    Parameters
    ----------
    polygon : list of two-dimensional points (points are array-like with two elements)
        Non-self-intersecting polygon (orientation does not matter).

    Returns
    -------
    center_of_gravity : list with 2 elements
        Coordinates (or vector) to the centroid of the polygon.
    """
    offset = polygon[0]
    center_of_gravity = [0.0, 0.0]
    double_area = 0.0
    for ii in range(len(polygon)):
        p1 = polygon[ii]
        p2 = polygon[ii-1]
        f = (p1[0]-offset[0])*(p2[1]-offset[1]) - (p2[0]-offset[0])*(p1[1]-offset[1])
        double_area += f
        center_of_gravity[0] += (p1[0] + p2[0] - 2*offset[0]) * f
        center_of_gravity[1] += (p1[1] + p2[1] - 2*offset[1]) * f
    center_of_gravity[0] = center_of_gravity[0] / (3*double_area) + offset[0]
    center_of_gravity[1] = center_of_gravity[1] / (3*double_area) + offset[1]
    return center_of_gravity
    # If you want to return both the CoG and the area, comment the return above
    return center_of_gravity, abs(double_area/2)

Answer 8

根据this answer

在 C# 中：

public static Point findCentroid(List<Point> pts)
    {
        Point off = pts[0];
        double twicearea = 0;
        double x = 0;
        double y = 0;
        Point p1, p2;
        double f;
        for (int i = 0, j = pts.Count - 1; i < pts.Count; j = i++)
        {
            p1 = pts[i];
            p2 = pts[j];
            f = (p1.x - off.x) * (p2.y - off.y) - (p2.x - off.x) * (p1.y - off.y);
            twicearea += f;
            x += (p1.x + p2.x - 2 * off.x) * f;
            y += (p1.y + p2.y - 2 * off.y) * f;
        }

        f = twicearea * 3;

        return new Point(x / f + off.x, y / f + off.y);
    }