Question

我正在实施Voronoi tesselation，然后进行平滑处理。为了平滑，我打算做劳埃德放松，但我遇到了一个问题。

我正在使用以下模块来计算Voronoi边：

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

对于平滑，我需要知道每个多边形的边缘，这样我才能计算出中心，遗憾的是这段代码没有提供。

我可以访问的信息包括：

所有节点的列表
所有边的列表（但只是在哪里它们不是与它们相关联的节点。）

有人能看到一种相对简单的计算方法吗？

Answer 1

要查找质心，可以使用the formula described on wikipedia：

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gift-wrapping algorithm来查找外部点（a.k.a。convex hull）。有很多方法可以做到这一点，但礼品包装很好，因为它的概念和实用的简单性。这是一个动画gif解释这个特定的实现：

step-by-step animated gif for counter-clockwise gift-wrapping, starting at the bottommost node

这里有一些天真的代码，用于根据voronoi图的节点和边缘集合查找各个voronoi单元的质心。它引入了一种方法来查找属于一个节点的边，并依赖于先前的质心和凸壳代码：

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'])

下面是脚本结果的图像。蓝线是边缘。黑色方块是节点。红色正方形是蓝色线派生的顶点。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 tesselation，但是用于获得质心的方法应该适用于由凸细胞组成的tessalations：

triangulated point cloud with calculated centroids and arbitrarily-chosen approximate centers

这里是用于渲染图像的html：

<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>

这可能会完成工作。用于找到属于单元格的边缘的更稳健的算法将是使用反向礼品包装方法，其中边缘端对端地链接并且分割处的路径选择将由角度确定。该方法不会对凹面多边形产生敏感性，并且它具有不依赖节点的额外好处。

Answer 2

这是@ mgamba的答案，用更多的python风格重写。特别是，它在点上使用itertools.cycle，因此可以以更自然的方式将“一加一最后一点”视为第一点。

import itertools as IT

def area_of_polygon(x, y):
    """Calculates the signed area of an arbitrary polygon given its verticies
    http://stackoverflow.com/a/4682656/190597 (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):
    """
    http://stackoverflow.com/a/14115494/190597 (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()

Answer 3

也许这可以帮到你： https://github.com/Bennyelg/geo_polygon_finder 此存储库接收城市列表并将其转换为多边形。

使用有限的数据查找多边形的中心

3 个答案: