问题
I have been searching for an answer to this question but cannot find anything useful.
I am working with the python scientific computing stack (scipy,numpy,matplotlib) and I have a set of 2 dimensional points, for which I compute the Delaunay traingulation (wiki) using scipy.spatial.Delaunay.
I need to write a function that, given any point a, will return all other points which are vertices of any simplex (i.e. triangle) that a is also a vertex of (the neighbors of a in the triangulation). However, the documentation for scipy.spatial.Delaunay (here) is pretty bad, and I can't for the life of me understand how the simplices are being specified or I would go about doing this. Even just an explanation of how the neighbors, vertices and vertex_to_simplex arrays in the Delaunay output are organized would be enough to get me going.
Much thanks for any help.
回答1:
I figured it out on my own, so here's an explanation for anyone future person who is confused by this.
As an example, let's use the simple lattice of points that I was working with in my code, which I generate as follows
import numpy as np
import itertools as it
from matplotlib import pyplot as plt
import scipy as sp
inputs = list(it.product([0,1,2],[0,1,2]))
i = 0
lattice = range(0,len(inputs))
for pair in inputs:
lattice[i] = mksite(pair[0], pair[1])
i = i +1
Details here not really important, suffice to say it generates a regular triangular lattice in which the distance between a point and any of its six nearest neighbors is 1.
To plot it
plt.plot(*np.transpose(lattice), marker = 'o', ls = '')
axes().set_aspect('equal')
Now compute the triangulation:
dela = sp.spatial.Delaunay
triang = dela(lattice)
Let's look at what this gives us.
triang.points
output:
array([[ 0. , 0. ],
[ 0.5 , 0.8660254 ],
[ 1. , 1.73205081],
[ 1. , 0. ],
[ 1.5 , 0.8660254 ],
[ 2. , 1.73205081],
[ 2. , 0. ],
[ 2.5 , 0.8660254 ],
[ 3. , 1.73205081]])
simple, just an array of all nine points in the lattice illustrated above. How let's look at:
triang.vertices
output:
array([[4, 3, 6],
[5, 4, 2],
[1, 3, 0],
[1, 4, 2],
[1, 4, 3],
[7, 4, 6],
[7, 5, 8],
[7, 5, 4]], dtype=int32)
In this array, each row represents one simplex (triangle) in the triangulation. The three entries in each row are the indices of the vertices of that simplex in the points array we just saw. So for example the first simplex in this array, [4, 3, 6] is composed of the points:
[ 1.5 , 0.8660254 ]
[ 1. , 0. ]
[ 2. , 0. ]
Its easy to see this by drawing the lattice on a piece of paper, labeling each point according to its index, and then tracing through each row in triang.vertices.
This is all the information we need to write the function I specified in my question. It looks like:
def find_neighbors(pindex, triang):
neighbors = list()
for simplex in triang.vertices:
if pindex in simplex:
neighbors.extend([simplex[i] for i in range(len(simplex)) if simplex[i] != pindex])
'''
this is a one liner for if a simplex contains the point we`re interested in,
extend the neighbors list by appending all the *other* point indices in the simplex
'''
#now we just have to strip out all the dulicate indices and return the neighbors list:
return list(set(neighbors))
And that's it! I'm sure the function above could do with some optimization, its just what I came up with in a few minutes. If anyone has any suggestions, feel free to post them. Hopefully this helps somebody in the future who is as confused about this as I was.
回答2:
The methods described above cycle through all the simplices, which could take very long, in case there's a large number of points. A better way might be to use Delaunay.vertex_neighbor_vertices, which already contains all the information about the neighbors. Unfortunately, extracting the information
def find_neighbors(pindex, triang):
return triang.vertex_neighbor_vertices[1][triang.vertex_neighbor_vertices[0][pindex]:triang.vertex_neighbor_vertices[0][pindex+1]]
The following code demonstrates how to get the indices of some vertex (number 17, in this example):
import scipy.spatial
import numpy
import pylab
x_list = numpy.random.random(200)
y_list = numpy.random.random(200)
tri = scipy.spatial.Delaunay(numpy.array([[x,y] for x,y in zip(x_list, y_list)]))
pindex = 17
neighbor_indices = find_neighbors(pindex,tri)
pylab.plot(x_list, y_list, 'b.')
pylab.plot(x_list[pindex], y_list[pindex], 'dg')
pylab.plot([x_list[i] for i in neighbor_indices],
[y_list[i] for i in neighbor_indices], 'ro')
pylab.show()
回答3:
Here is also a simple one line version of James Porter's own answer using list comprehension:
find_neighbors = lambda x,triang: list(set(indx for simplex in triang.simplices if x in simplex for indx in simplex if indx !=x))
回答4:
Here is an ellaboration on @astrofrog answer. This works also in more than 2D.
It took about 300 ms on set of 2430 points in 3D (about 16000 simplices).
from collections import defaultdict
def find_neighbors(tess):
neighbors = defaultdict(set)
for simplex in tess.simplices:
for idx in simplex:
other = set(simplex)
other.remove(idx)
neighbors[idx] = neighbors[idx].union(other)
return neighbors
回答5:
I needed this too and came across the following answer. It turns out that if you need the neighbors for all initial points, it's much more efficient to produce a dictionary of neighbors in one go (the following example is for 2D):
def find_neighbors(tess, points):
neighbors = {}
for point in range(points.shape[0]):
neighbors[point] = []
for simplex in tess.simplices:
neighbors[simplex[0]] += [simplex[1],simplex[2]]
neighbors[simplex[1]] += [simplex[2],simplex[0]]
neighbors[simplex[2]] += [simplex[0],simplex[1]]
return neighbors
The neighbors of point v are then neighbors[v]. For 10,000 points in this takes 370ms to run on my laptop. Maybe others have ideas on optimizing this further?
回答6:
I know it's been a while since this question was posed. However, I just had the same problem and figured out how to solve it. Just use the (somewhat poorly documented) method 'vertex_neighbor_vertices' of your Delaunay triangulation object (let us call it 'tri'). It will return two arrays:
def get_neighbor_vertex_ids_from_vertex_id(vertex_id,tri):
#use a less awful function name
helper = tri.vertex_neighbor_vertices
index_pointers = helper[0]
indices = helper[1]
result_ids = indices[index_pointers[vertex_id]:index_pointers[vertex_id+1]]
return result_ids
The neighbor vertices to the point with the index vertex_id are stored somewhere in the second array that I named 'indices'. But where? This is where the first array (which I called 'index_pointers') comes in. The starting position (for the second array 'indices') is index_pointers[vertex_id], the first position past the relevant sub-array is index_pointers[vertex_id+1]. So the solution is indices[index_pointers[vertex_id]:index_pointers[vertex_id+1]]
回答7:
All the answers here are focused on getting the neighbors for one point (except astrofrog, but that is in 2D and this is 6x faster), however, it's equally expensive to get a mapping for all of the points → all neighbors.
You can do this with
from collections import defaultdict
from itertools import permutations
tri = Delaunay(...)
_neighbors = defaultdict(set)
for simplex in tri.vertices:
for i, j in permutations(simplex, 2):
_neighbors[i].add(j)
points = [tuple(p) for p in tri.points]
neighbors = {}
for k, v in _neighbors.items():
neighbors[points[k]] = [points[i] for i in v]
This works in any dimension and this solution, finding all neighbors of all points, is faster than finding only the neighbors of one point (the excepted answer of James Porter).
来源：https://stackoverflow.com/questions/12374781/how-to-find-all-neighbors-of-a-given-point-in-a-delaunay-triangulation-using-sci