python - 将一个匀称的多边形切割成 N 个大小相等的多边形

Question

我有一个匀称的多边形。我想将这些多边形切割成n 个多边形，它们都具有或多或少相同大小的区域。大小相同是最好的，但近似值也可以。

我尝试使用这里描述的两种方法，这两种方法都是朝着正确方向迈出的一步，而不是我需要的。两者都不允许目标n

我调查了voronoi，对此我很陌生。此分析给出的最终形状将是理想的，但它需要点，而不是形状作为输入。

标签： pythonshapes

Answer 1

这是我能做到的最好的。它不会导致每个多边形的表面积相等，但事实证明它可以满足我的需要。这会使用特定数量的点填充形状（如果参数保持不变，点数也会保持不变）。然后将这些点转换为 voronoi，然后将其转换为三角形。

from shapely import affinity
from shapely.geometry.multipolygon import MultiPolygon
from scipy.spatial import Voronoi

# Voronoi doesn't work properly with points below (0,0) so set lowest point to (0,0)
shape = affinity.translate(shape, -shape_a.bounds[0], -shape_a.bounds[1])

points = shape_to_points(shape)

vor = points_to_voronoi(points)

triangles = MultiPolygon(triangulate(MultiLineString(vor)))



def shape_to_points(shape, num = 10, smaller_versions = 10):
    points = []

    # Take the shape, shrink it by a factor (first iteration factor=1), and then 
    # take points around the contours
    for shrink_factor in range(0,smaller_versions,1):
        # calculate the shrinking factor
        shrink_factor = smaller_versions - shrink_factor
        shrink_factor = shrink_factor / float(smaller_versions)
        # actually shrink - first iteration it remains at 1:1
        smaller_shape = affinity.scale(shape, shrink_factor, shrink_factor)
        # Interpolate numbers around the boundary of the shape
        for i in range(0,int(num*shrink_factor),1):
            i = i / int(num*shrink_factor)
            x,y =  smaller_shape.interpolate(i, normalized=True).xy
            points.append( (x[0],y[0]))
    
    # add the origin
    x,y = smaller_shape.centroid.xy
    points.append( (x[0], y[0]) ) # near, but usually not add (0,0)
    
    points = np.array(points)
    return points


def points_to_voronoi(points):
    vor = Voronoi(points)
    vertices = [ x for x in vor.ridge_vertices if -1 not in x]
    # For some reason, some vertices were seen as super, super long. Probably also infinite lines, so take them out
    lines = [ LineString(vor.vertices[x]) for x in vertices if not vor.vertices[x].max() > 50000]
    return MultiLineString(lines)

这是输入形状：

这是之后shape_to_points：

这是之后points_to_voronoi

然后我们可以对 voronoi 进行三角测量：