使用有限数据找到多边形的中心

Question

我正在实现沃罗诺伊图的细分，然后进行平滑处理。对于平滑，我本来打算使用Lloyd松弛算法，但是遇到了问题。

我正在使用以下模块计算沃罗诺伊边缘：

https://bitbucket.org/mozman/geoalg/src/5bbd46fa2270/geoalg/voronoi.py

为了平滑处理，我需要知道每个多边形的边缘，以便计算中心点，但不幸的是这段代码没有提供这些信息。

我可以访问的信息包括：

- 所有节点的列表， - 所有边缘的列表（但只知道它们在哪里，不知道与哪些节点相关）。

有没有人能看到一个相对简单的方法来计算这个？

- djcmm476

Answer 1

要找到一个质心，您可以使用维基百科上描述的公式：

import math

def area_for_polygon(polygon):
    result = 0
    imax = len(polygon) - 1
    for i in range(0,imax):
        result += (polygon[i]['x'] * polygon[i+1]['y']) - (polygon[i+1]['x'] * polygon[i]['y'])
    result += (polygon[imax]['x'] * polygon[0]['y']) - (polygon[0]['x'] * polygon[imax]['y'])
    return result / 2.

def centroid_for_polygon(polygon):
    area = area_for_polygon(polygon)
    imax = len(polygon) - 1

    result_x = 0
    result_y = 0
    for i in range(0,imax):
        result_x += (polygon[i]['x'] + polygon[i+1]['x']) * ((polygon[i]['x'] * polygon[i+1]['y']) - (polygon[i+1]['x'] * polygon[i]['y']))
        result_y += (polygon[i]['y'] + polygon[i+1]['y']) * ((polygon[i]['x'] * polygon[i+1]['y']) - (polygon[i+1]['x'] * polygon[i]['y']))
    result_x += (polygon[imax]['x'] + polygon[0]['x']) * ((polygon[imax]['x'] * polygon[0]['y']) - (polygon[0]['x'] * polygon[imax]['y']))
    result_y += (polygon[imax]['y'] + polygon[0]['y']) * ((polygon[imax]['x'] * polygon[0]['y']) - (polygon[0]['x'] * polygon[imax]['y']))
    result_x /= (area * 6.0)
    result_y /= (area * 6.0)

    return {'x': result_x, 'y': result_y}

def bottommost_index_for_polygon(polygon):
    bottommost_index = 0
    for index, point in enumerate(polygon):
        if (point['y'] < polygon[bottommost_index]['y']):
            bottommost_index = index
    return bottommost_index

def angle_for_vector(start_point, end_point):
    y = end_point['y'] - start_point['y']
    x = end_point['x'] - start_point['x']
    angle = 0

    if (x == 0):
        if (y > 0):
            angle = 90.0
        else:
            angle = 270.0
    elif (y == 0):
        if (x > 0):
            angle = 0.0
        else:
            angle = 180.0
    else:
        angle = math.degrees(math.atan((y+0.0)/x))
        if (x < 0):
            angle += 180
        elif (y < 0):
            angle += 360

    return angle

def convex_hull_for_polygon(polygon):
    starting_point_index = bottommost_index_for_polygon(polygon)
    convex_hull = [polygon[starting_point_index]]
    polygon_length = len(polygon)

    hull_index_candidate = 0 #arbitrary
    previous_hull_index_candidate = starting_point_index
    previous_angle = 0
    while True:
        smallest_angle = 360

        for j in range(0,polygon_length):
            if (previous_hull_index_candidate == j):
                continue
            current_angle = angle_for_vector(polygon[previous_hull_index_candidate], polygon[j])
            if (current_angle < smallest_angle and current_angle > previous_angle):
                hull_index_candidate = j
                smallest_angle = current_angle

        if (hull_index_candidate == starting_point_index): # we've wrapped all the way around
            break
        else:
            convex_hull.append(polygon[hull_index_candidate])
            previous_angle = smallest_angle
            previous_hull_index_candidate = hull_index_candidate

    return convex_hull

我使用了一种礼物包装算法来找到外部点（也称为凸包）。有很多方法可以做到这一点，但是礼物包装算法因其概念上和实际上的简单性而受欢迎。下面是一个动画GIF，解释了这个特定实现：

这是一些简单的代码，用于基于节点和边的Voronoi图集合查找各个Voronoi单元的质心。它介绍了一种查找属于节点的边的方法，并依赖于先前的质心和凸包代码：

def midpoint(edge):
    x1 = edge[0][0]
    y1 = edge[0][9]
    x2 = edge[1][0]
    y2 = edge[1][10]

    mid_x = x1+((x2-x1)/2.0)
    mid_y = y1+((y2-y1)/2.0)

    return (mid_x, mid_y)

def ccw(A,B,C): # from http://www.bryceboe.com/2006/10/23/line-segment-intersection-algorithm/
    return (C[1]-A[1])*(B[0]-A[0]) > (B[1]-A[1])*(C[0]-A[0])

def intersect(segment1, segment2): # from http://www.bryceboe.com/2006/10/23/line-segment-intersection-algorithm/
    A = segment1[0]
    B = segment1[1]
    C = segment2[0]
    D = segment2[1]
    # Note: this doesn't catch collinear line segments!
    return ccw(A,C,D) != ccw(B,C,D) and ccw(A,B,C) != ccw(A,B,D)

def points_from_edges(edges):
    point_set = set()
    for i in range(0,len(edges)):
          point_set.add(edges[i][0])
          point_set.add(edges[i][11])

    points = []
    for point in point_set:
          points.append({'x':point[0], 'y':point[1]})

    return list(points)

def centroids_for_points_and_edges(points, edges):

    centroids = []

    # for each voronoi_node,
    for i in range(0,len(points)):
        cell_edges = []

        # for each edge
        for j in range(0,len(edges)):
            is_cell_edge = True

            # let vector be the line from voronoi_node to the midpoint of edge
            vector = (points[i],midpoint(edges[j]))

            # for each other_edge
            for k in range(0,len(edges)):

                # if vector crosses other_edge
                if (k != j and intersect(edges[k], vector)):
                    # edge is not in voronoi_node's polygon
                    is_cell_edge = False
                    break

            # if the vector didn't cross any other edges, it's an edge for the current node
            if (is_cell_edge):
                cell_edges.append(edges[j])

        # find the hull for the cell
        convex_hull = convex_hull_for_polygon(points_from_edges(cell_edges))

        # calculate the centroid of the hull
        centroids.append(centroid_for_polygon(convex_hull))

    return centroids

edges = [
  ((10,  200),(30,  50 )),
  ((10,  200),(100, 140)),
  ((10,  200),(200, 180)),
  ((30,  50 ),(100, 140)),
  ((30,  50 ),(150, 75 )),
  ((30,  50 ),(200, 10 )),
  ((100, 140),(150, 75 )),
  ((100, 140),(200, 180)),
  ((150, 75 ),(200, 10 )),
  ((150, 75 ),(200, 180)),
  ((150, 75 ),(220, 80 )),
  ((200, 10 ),(220, 80 )),
  ((200, 10 ),(350, 100)),
  ((200, 180),(220, 80 )),
  ((200, 180),(350, 100)),
  ((220, 80 ),(350, 100))
]

points = [
  (50,130),
  (100,95),
  (100,170),
  (130,45),
  (150,130),
  (190,55),
  (190,110),
  (240,60),
  (245,120)
]

centroids = centroids_for_points_and_edges(points, edges)
print "centroids:"
for centroid in centroids:
    print "  (%s, %s)" % (centroid['x'], centroid['y'])

以下是脚本结果的图像。蓝色线条表示边缘，黑色正方形表示节点。红色正方形表示蓝色线条派生自的顶点。这些顶点和节点是任意选择的。红色十字架是质心。虽然不是真正的Voronoi分割，但用于获取质心的方法应适用于由凸多边形组成的分割：

以下是呈现图像的HTML代码：

这里是呈现图像的HTML代码：

<html>
<head>
  <script>
    window.onload = draw;
    function draw() {
      var canvas = document.getElementById('canvas').getContext('2d');

      // draw polygon points
      var polygon = [ 
        {'x':220, 'y':80},
        {'x':200, 'y':180},
        {'x':350, 'y':100},
        {'x':30, 'y':50}, 
        {'x':100, 'y':140},
        {'x':200, 'y':10},
        {'x':10, 'y':200},
        {'x':150, 'y':75}
      ];  
      plen=polygon.length;
      for(i=0; i<plen; i++) {
        canvas.fillStyle = 'red';
        canvas.fillRect(polygon[i].x-4,polygon[i].y-4,8,8);
        canvas.fillStyle = 'yellow';
        canvas.fillRect(polygon[i].x-2,polygon[i].y-2,4,4);
      }   

      // draw edges
      var edges = [ 
        [[10,  200],[30,  50 ]], 
        [[10,  200],[100, 140]],
        [[10,  200],[200, 180]],
        [[30,  50 ],[100, 140]], 
        [[30,  50 ],[150, 75 ]], 
        [[30,  50 ],[200, 10 ]], 
        [[100, 140],[150, 75 ]], 
        [[100, 140],[200, 180]],
        [[150, 75 ],[200, 10 ]], 
        [[150, 75 ],[200, 180]],
        [[150, 75 ],[220, 80 ]], 
        [[200, 10 ],[220, 80 ]], 
        [[200, 10 ],[350, 100]],
        [[200, 180],[220, 80 ]], 
        [[200, 180],[350, 100]],
        [[220, 80 ],[350, 100]]
      ];  
      elen=edges.length;
      canvas.beginPath();
      for(i=0; i<elen; i++) {
        canvas.moveTo(edges[i][0][0], edges[i][0][1]);
        canvas.lineTo(edges[i][13][0], edges[i][14][1]);
      }   
      canvas.closePath();
      canvas.strokeStyle = 'blue';
      canvas.stroke();

      // draw center points
      var points = [ 
        [50,130],
        [100,95],
        [100,170],
        [130,45],
        [150,130],
        [190,55],
        [190,110],
        [240,60],
        [245,120]
      ]   
      plen=points.length;
      for(i=0; i<plen; i++) {
        canvas.fillStyle = 'black';
        canvas.fillRect(points[i][0]-3,points[i][15]-3,6,6);
        canvas.fillStyle = 'white';
        canvas.fillRect(points[i][0]-1,points[i][16]-1,2,2);
      }   

      // draw centroids
      var centroids = [ 
        [46.6666666667, 130.0],
        [93.3333333333, 88.3333333333],
        [103.333333333, 173.333333333],
        [126.666666667, 45.0],
        [150.0, 131.666666667],
        [190.0, 55.0],
        [190.0, 111.666666667],
        [256.666666667, 63.3333333333],
        [256.666666667, 120.0]
      ]
      clen=centroids.length;
      canvas.beginPath();
      for(i=0; i<clen; i++) {
        canvas.moveTo(centroids[i][0], centroids[i][17]-5);
        canvas.lineTo(centroids[i][0], centroids[i][18]+5);
        canvas.moveTo(centroids[i][0]-5, centroids[i][19]);
        canvas.lineTo(centroids[i][0]+5, centroids[i][20]);
      }
      canvas.closePath();
      canvas.strokeStyle = 'red';
      canvas.stroke();
    }
  </script>
</head>
<body>
  <canvas id='canvas' width="400px" height="250px"</canvas>
</body>
</html>

这可能能够完成工作。更强大的算法用于查找哪些边属于一个单元格将是使用反向礼品包装方法，其中边缘端对端链接，并且在拆分处的路径选择将由角度确定。该方法不会受到凹多边形的影响，并且具有不依赖节点的额外好处。

- mgamba

---

这看起来相当不错，但我不需要知道哪些顶点与哪个节点具体链接吗？ - djcmm476

抱歉造成误解。脚本现在会计算多边形的凸包，然后找到凸包的重心。 - mgamba

嗯，看起来不错。但是这只适用于您一次只有一个多边形的顶点在使用（而不是每个多边形的每个顶点）的情况下，是吗？ - djcmm476

啊，我明白了。好的，我现在会发布一些伪代码，并稍后更新它。 - mgamba

Answer 2

这是@mgamba的回答，以更加Python风格重新编写。特别地，它使用itertools.cycle对点进行循环，以便“一个加上最后一个点”可以更自然地作为第一个点处理。

import itertools as IT

def area_of_polygon(x, y):
    """Calculates the signed area of an arbitrary polygon given its verticies
    https://dev59.com/CW445IYBdhLWcg3w0ddD#4682656 (Joe Kington)
    http://softsurfer.com/Archive/algorithm_0101/algorithm_0101.htm#2D%20Polygons
    """
    area = 0.0
    for i in xrange(-1, len(x) - 1):
        area += x[i] * (y[i + 1] - y[i - 1])
    return area / 2.0

def centroid_of_polygon(points):
    """
    https://dev59.com/JmzXa4cB1Zd3GeqPWbP3#14115494 (mgamba)
    """
    area = area_of_polygon(*zip(*points))
    result_x = 0
    result_y = 0
    N = len(points)
    points = IT.cycle(points)
    x1, y1 = next(points)
    for i in range(N):
        x0, y0 = x1, y1
        x1, y1 = next(points)
        cross = (x0 * y1) - (x1 * y0)
        result_x += (x0 + x1) * cross
        result_y += (y0 + y1) * cross
    result_x /= (area * 6.0)
    result_y /= (area * 6.0)
    return (result_x, result_y)

def demo_centroid():
    points = [
        (30,50),
        (200,10),
        (250,50),
        (350,100),
        (200,180),
        (100,140),
        (10,200)
        ]
    cent = centroid_of_polygon(points)
    print(cent)
    # (159.2903828197946, 98.88888888888889)

demo_centroid()

- unutbu

---

（回复您的第二篇帖子） 我目前有很多多边形，我的数据是所有顶点以及哪些顶点连接在一起形成线段。但是我没有任何关于哪些顶点用于构成哪些多边形的边缘的信息（对于造成的困惑我感到抱歉）。 - djcmm476

你是否正在寻找一种从 Voronoi 图计算 Delaunay 三角剖分的方法？换句话说，Delaunay 三角剖分的顶点是否是 Voronoi 单元的重心？ - unutbu

是的，我可能可以处理重心计算，只是使用我拥有的数据找到形成边缘的节点。 - djcmm476

如果面积为零，则会在面积为0的情况下出现编译错误。 - Dejell

我不理解 IT.cycle 的作用。 - Dejell

Answer 3

也许这可以帮助你： https://github.com/Bennyelg/geo_polygon_finder 该存储库接收城市列表并将它们转换为多边形。

- Benjaminel