Calculate bounding polygon of alpha shape from the

问题:

Given a set of points in the plane, a notion of alpha-shape, for a given positive number alpha, is defined by finding the Delaunay triangulation and deleting any triangles for which at least one edge exceeds alpha in length. Here's an example using d3:

http://bl.ocks.org/gka/1552725

The problem is that, when there are thousands of points, simply drawing all the interior triangles is too slow for an interactive visualization, so I'd like to just find the bounding polygons. This isn't so simple, because as you can see from that example sometimes there might be two such polygons.

As a simplification, suppose some clustering has been performed so that there's guaranteed to be a unique bounding polygon for each triangulation. What's the best way to find this bounding polygon? In particular, the edges have to be ordered consistently and it must support the possibility of "holes" (think a torus or donut shape--this is expressible in GeoJSON).

回答1:

1. Create a graph in which the nodes correspond to Delaunay triangles and in which there is a graph edge between two triangles if and only if they share two vertices.

2. Compute the connected components of the graph.

3. For each connected component, find all of the nodes that have less than three adjacent nodes (that is, those that have degree 0, 1, or 2). These correspond to the boundary triangles. We define the edges of a boundary triangle that are not shared with another triangle to be the boundary edges of that boundary triangle.

As an example, I have highlighted these boundary triangles in your example "question mark" Delaunay triangulation:

By definition, every boundary triangle is adjacent to at most two other boundary triangles. The boundary edges of boundary triangles form cycles. You can simply traverse those cycles to determine polygon shapes for the boundaries. This will also work for polygons with holes if you keep them in mind in your implementation.

回答2:

Here is some Python code that does what you need. I modified the alpha-shape (concave hull) computation from here so that it doesn't insert inner edges (the only_outer parameter). I also made it self-contained so it doesn't depend on an outside library.

from scipy.spatial import Delaunay
import numpy as np


def alpha_shape(points, alpha, only_outer=True):
    """
    Compute the alpha shape (concave hull) of a set of points.
    :param points: np.array of shape (n,2) points.
    :param alpha: alpha value.
    :param only_outer: boolean value to specify if we keep only the outer border
    or also inner edges.
    :return: set of (i,j) pairs representing edges of the alpha-shape. (i,j) are
    the indices in the points array.
    """
    assert points.shape[0] > 3, "Need at least four points"

    def add_edge(edges, i, j):
        """
        Add a line between the i-th and j-th points,
        if not in the list already
        """
        if (i, j) in edges or (j, i) in edges:
            # already added
            assert (j, i) in edges, "Can't go twice over same directed edge right?"
            if only_outer:
                # if both neighboring triangles are in shape, it's not a boundary edge
                edges.remove((j, i))
            return
        edges.add((i, j))

    tri = Delaunay(points)
    edges = set()
    # Loop over triangles:
    # ia, ib, ic = indices of corner points of the triangle
    for ia, ib, ic in tri.vertices:
        pa = points[ia]
        pb = points[ib]
        pc = points[ic]
        # Computing radius of triangle circumcircle
        # www.mathalino.com/reviewer/derivation-of-formulas/derivation-of-formula-for-radius-of-circumcircle
        a = np.sqrt((pa[0] - pb[0]) ** 2 + (pa[1] - pb[1]) ** 2)
        b = np.sqrt((pb[0] - pc[0]) ** 2 + (pb[1] - pc[1]) ** 2)
        c = np.sqrt((pc[0] - pa[0]) ** 2 + (pc[1] - pa[1]) ** 2)
        s = (a + b + c) / 2.0
        area = np.sqrt(s * (s - a) * (s - b) * (s - c))
        circum_r = a * b * c / (4.0 * area)
        if circum_r < alpha:
            add_edge(edges, ia, ib)
            add_edge(edges, ib, ic)
            add_edge(edges, ic, ia)
    return edges

If you run it with the following test code you will get this figure:

from matplotlib.pyplot import *

# Constructing the input point data
np.random.seed(0)
x = 3.0 * np.random.rand(2000)
y = 2.0 * np.random.rand(2000) - 1.0
inside = ((x ** 2 + y ** 2 > 1.0) & ((x - 3) ** 2 + y ** 2 > 1.0) & ((x - 1.5) ** 2 + y ** 2 > 0.09))
points = np.vstack([x[inside], y[inside]]).T

# Computing the alpha shape
edges = alpha_shape(points, alpha=0.25, only_outer=True)

# Plotting the output
figure()
axis('equal')
plot(points[:, 0], points[:, 1], '.')
for i, j in edges:
    plot(points[[i, j], 0], points[[i, j], 1])
show()

回答3:

It turns out TopoJSON has a merge algorithm which performs just this task: https://github.com/mbostock/topojson/wiki/API-Reference#merge

There's even an example showing it in action: http://bl.ocks.org/mbostock/9927735

In my case, it was easy for me to generate TopoJSON data, and this library function accomplished the task perfectly for me.

回答4:

Building up on @Timothy's answer, I used the following algorithm to calculate the boundary rings of a Delaunay triangulation.

from matplotlib.tri import Triangulation
import numpy as np

def get_boundary_rings(x, y, elements):
    mpl_tri = Triangulation(x, y, elements)
    idxs = np.vstack(list(np.where(mpl_tri.neighbors == -1))).T
    unique_edges = list()
    for i, j in idxs:
        unique_edges.append((mpl_tri.triangles[i, j],
                             mpl_tri.triangles[i, (j+1) % 3]))
    unique_edges = np.asarray(unique_edges)
    ring_collection = list()
    initial_idx = 0
    for i in range(1, len(unique_edges)-1):
        if unique_edges[i-1, 1] != unique_edges[i, 0]:
            try:
                idx = np.where(
                    unique_edges[i-1, 1] == unique_edges[i:, 0])[0][0]
                unique_edges[[i, idx+i]] = unique_edges[[idx+i, i]]
            except IndexError:
                ring_collection.append(unique_edges[initial_idx:i, :])
                initial_idx = i
                continue
    # if there is just one ring, the exception is never reached,
    # so populate ring_collection before returning.
    if len(ring_collection) == 0:
        ring_collection.append(np.asarray(unique_edges))
    return ring_collection

回答5:

Slightly revise Hanniel's answer for 3d point case (tetrahedron).

def alpha_shape(points, alpha, only_outer=True):
    """
    Compute the alpha shape (concave hull) of a set of points.
    :param points: np.array of shape (n, 3) points.
    :param alpha: alpha value.
    :param only_outer: boolean value to specify if we keep only the outer border
    or also inner edges.
    :return: set of (i,j) pairs representing edges of the alpha-shape. (i,j) are
    the indices in the points array.
    """
    assert points.shape[0] > 3, "Need at least four points"

    def add_edge(edges, i, j):
        """
        Add a line between the i-th and j-th points,
        if not in the set already
        """
        if (i, j) in edges or (j, i) in edges:
            # already added
            if only_outer:
                # if both neighboring triangles are in shape, it's not a boundary edge
                if (j, i) in edges:
                    edges.remove((j, i))
            return
        edges.add((i, j))

    tri = Delaunay(points)
    edges = set()
    # Loop over triangles:
    # ia, ib, ic, id = indices of corner points of the tetrahedron
    print(tri.vertices.shape)
    for ia, ib, ic, id in tri.vertices:
        pa = points[ia]
        pb = points[ib]
        pc = points[ic]
        pd = points[id]

        # Computing radius of tetrahedron Circumsphere
        # http://mathworld.wolfram.com/Circumsphere.html

        pa2 = np.dot(pa, pa)
        pb2 = np.dot(pb, pb)
        pc2 = np.dot(pc, pc)
        pd2 = np.dot(pd, pd)

        a = np.linalg.det(np.array([np.append(pa, 1), np.append(pb, 1), np.append(pc, 1), np.append(pd, 1)]))

        Dx = np.linalg.det(np.array([np.array([pa2, pa[1], pa[2], 1]),
                                     np.array([pb2, pb[1], pb[2], 1]),
                                     np.array([pc2, pc[1], pc[2], 1]),
                                     np.array([pd2, pd[1], pd[2], 1])]))

        Dy = - np.linalg.det(np.array([np.array([pa2, pa[0], pa[2], 1]),
                                       np.array([pb2, pb[0], pb[2], 1]),
                                       np.array([pc2, pc[0], pc[2], 1]),
                                       np.array([pd2, pd[0], pd[2], 1])]))

        Dz = np.linalg.det(np.array([np.array([pa2, pa[0], pa[1], 1]),
                                     np.array([pb2, pb[0], pb[1], 1]),
                                     np.array([pc2, pc[0], pc[1], 1]),
                                     np.array([pd2, pd[0], pd[1], 1])]))

        c = np.linalg.det(np.array([np.array([pa2, pa[0], pa[1], pa[2]]),
                                    np.array([pb2, pb[0], pb[1], pb[2]]),
                                    np.array([pc2, pc[0], pc[1], pc[2]]),
                                    np.array([pd2, pd[0], pd[1], pd[2]])]))

        circum_r = math.sqrt(math.pow(Dx, 2) + math.pow(Dy, 2) + math.pow(Dz, 2) - 4 * a * c) / (2 * abs(a))
        if circum_r < alpha:
            add_edge(edges, ia, ib)
            add_edge(edges, ib, ic)
            add_edge(edges, ic, id)
            add_edge(edges, id, ia)
            add_edge(edges, ia, ic)
            add_edge(edges, ib, id)
    return edges

回答6:

There now exists a python package alphashape which is extremely easy to use, and can be installed by pip or conda.

The main function has similar inputs to the answer given by @Iddo Hanniel, adjusting the second positional argument would give you the desired outline. Alternatively, you could leave the seconda positional argument blank and the function would optimize that parameter for you to give you the best concave hull. Beware, the computational time is increased greatly if you let the function optimize the value.

回答7:

Alpha shapes is defined as a delaunay triangulation without edges exceeding alpha. First of remove all interior triangles and then all edges exceeding alpha.