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''' | |
Copyright (C) 2018 Andre Lester Kruger | |
ConcaveHull.py is free software: you can redistribute it and/or modify | |
it under the terms of the GNU General Public License as published by | |
the Free Software Foundation, either version 2 of the License, or | |
(at your option) any later version. | |
ConcaveHull.py is distributed in the hope that it will be useful, | |
but WITHOUT ANY WARRANTY; without even the implied warranty of | |
MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the | |
GNU General Public License for more details. | |
You should have received a copy of the GNU General Public License | |
along with ConcaveHull.py. If not, see <http://www.gnu.org/licenses/>. | |
''' | |
import bisect | |
from collections import OrderedDict | |
import math | |
#import numpy as np | |
import matplotlib.tri as tri | |
from shapely.geometry import LineString | |
from shapely.geometry import Polygon | |
from shapely.ops import linemerge | |
class ConcaveHull: | |
def __init__(self): | |
self.triangles = {} | |
self.crs = {} | |
def loadpoints(self, points): | |
#self.points = np.array(points) | |
self.points = points | |
def edge(self, key, triangle): | |
'''Calculate the length of the triangle's outside edge | |
and returns the [length, key]''' | |
pos = triangle[1].index(-1) | |
if pos==0: | |
x1, y1 = self.points[triangle[0][0]] | |
x2, y2 = self.points[triangle[0][1]] | |
elif pos==1: | |
x1, y1 = self.points[triangle[0][1]] | |
x2, y2 = self.points[triangle[0][2]] | |
elif pos==2: | |
x1, y1 = self.points[triangle[0][0]] | |
x2, y2 = self.points[triangle[0][2]] | |
length = ((x1-x2)**2+(y1-y2)**2)**0.5 | |
rec = [length, key] | |
return rec | |
def triangulate(self): | |
if len(self.points) < 2: | |
raise Exception('CountError: You need at least 3 points to Triangulate') | |
temp = list(zip(*self.points)) | |
x, y = list(temp[0]), list(temp[1]) | |
del(temp) | |
triang = tri.Triangulation(x, y) | |
self.triangles = {} | |
for i, triangle in enumerate(triang.triangles): | |
self.triangles[i] = [list(triangle), list(triang.neighbors[i])] | |
def calculatehull(self, tol=50): | |
self.tol = tol | |
if len(self.triangles) == 0: | |
self.triangulate() | |
# All triangles with one boundary longer than the tolerance (self.tol) | |
# is added to a sorted deletion list. | |
# The list is kept sorted from according to the boundary edge's length | |
# using bisect | |
deletion = [] | |
self.boundary_vertices = set() | |
for i, triangle in self.triangles.items(): | |
if -1 in triangle[1]: | |
for pos, neigh in enumerate(triangle[1]): | |
if neigh == -1: | |
if pos == 0: | |
self.boundary_vertices.add(triangle[0][0]) | |
self.boundary_vertices.add(triangle[0][1]) | |
elif pos == 1: | |
self.boundary_vertices.add(triangle[0][1]) | |
self.boundary_vertices.add(triangle[0][2]) | |
elif pos == 2: | |
self.boundary_vertices.add(triangle[0][0]) | |
self.boundary_vertices.add(triangle[0][2]) | |
if -1 in triangle[1] and triangle[1].count(-1) == 1: | |
rec = self.edge(i, triangle) | |
if rec[0] > self.tol and triangle[1].count(-1) == 1: | |
bisect.insort(deletion, rec) | |
while len(deletion) != 0: | |
# The triangles with the longest boundary edges will be | |
# deleted first | |
item = deletion.pop() | |
ref = item[1] | |
flag = 0 | |
# Triangle will not be deleted if it already has two boundary edges | |
if self.triangles[ref][1].count(-1) > 1: | |
continue | |
# Triangle will not be deleted if the inside node which is not | |
# on this triangle's boundary is already on the boundary of | |
# another triangle | |
adjust = {0: 2, 1: 0, 2: 1} | |
for i, neigh in enumerate(self.triangles[ref][1]): | |
j = adjust[i] | |
if neigh == -1 and self.triangles[ref][0][j] in self.boundary_vertices: | |
flag = 1 | |
break | |
if flag == 1: | |
continue | |
for i, neigh in enumerate(self.triangles[ref][1]): | |
if neigh == -1: | |
continue | |
pos = self.triangles[neigh][1].index(ref) | |
self.triangles[neigh][1][pos] = -1 | |
rec = self.edge(neigh, self.triangles[neigh]) | |
if rec[0] > self.tol and self.triangles[rec[1]][1].count(-1) == 1: | |
bisect.insort(deletion, rec) | |
for pt in self.triangles[ref][0]: | |
self.boundary_vertices.add(pt) | |
del self.triangles[ref] | |
self.polygon() | |
def polygon(self): | |
edgelines = [] | |
for i, triangle in self.triangles.items(): | |
if -1 in triangle[1]: | |
for pos, value in enumerate(triangle[1]): | |
if value == -1: | |
if pos==0: | |
x1, y1 = self.points[triangle[0][0]] | |
x2, y2 = self.points[triangle[0][1]] | |
elif pos==1: | |
x1, y1 = self.points[triangle[0][1]] | |
x2, y2 = self.points[triangle[0][2]] | |
elif pos==2: | |
x1, y1 = self.points[triangle[0][0]] | |
x2, y2 = self.points[triangle[0][2]] | |
line = LineString([(x1, y1), (x2, y2)]) | |
edgelines.append(line) | |
bound = linemerge(edgelines) | |
self.boundary = Polygon(bound.coords) | |
#if __name__ == '__main__': |
Thanks for the clarification. Matter of fact I was also using degrees in the test case where I'm trying your code.
It actually makes a lot of sense now. You could have it in degrees, it is just a question of changing the tol
value to a meaningful value of distance, if you would convert it to degrees (e.g. divide tol
in km it by 111.045 to have a rough measure in degrees).
As this will depend on some triangulation of a set of points, eventually tol
could be made as a fraction of the total width (or height) span of the collection of points.
@AndreLester, Thanks a lot for the code! I am trying to implement a offset value into this function, do you have any idea how this would work?
Like in the image above the green is original shape and the blue is the offset shape.
@Joshua-Mursic I would use the buffer function from shapely on the resultant polygon. Play around with the options to get the desired result.
I have updated, modified and documented this script. It can be found at https://github.com/civildot/cdBoundary and installed with:
pip install cdBoundary
@cvr The "tol" variable presents the length of the longest edge you wish to have in the respective units. Usually in feet or metres. It is not a factor between zero and one as other concave hull algorithms. I prefer it like this.
The problems above where the people got convex hulls was probably due to the fact that their units was in degrees.
Note: Apologies. I was unaware that my "gist" blew up like this.