# Triangulation of a set of points:
points = np.array([[0, 0], [0, 1.1], [1, 0], [1, 1]])
from scipy.spatial import Delaunay
tri = Delaunay(points)
# We can plot it:
import matplotlib.pyplot as plt
plt.triplot(points[:,0], points[:,1], tri.simplices.copy())
plt.plot(points[:,0], points[:,1], 'o')
plt.show()
# Point indices and coordinates for the two triangles forming the
# triangulation:
tri.simplices
# array([[2, 3, 0], # may vary
# [3, 1, 0]], dtype=int32)
# Note that depending on how rounding errors go, the simplices may
# be in a different order than above.
points[tri.simplices]
# array([[[ 1. , 0. ], # may vary
# [ 1. , 1. ],
# [ 0. , 0. ]],
# [[ 1. , 1. ],
# [ 0. , 1.1],
# [ 0. , 0. ]]])
# Triangle 0 is the only neighbor of triangle 1, and it's opposite to
# vertex 1 of triangle 1:
tri.neighbors[1]
# array([-1, 0, -1], dtype=int32)
points[tri.simplices[1,1]]
# array([ 0. , 1.1])
# We can find out which triangle points are in:
p = np.array([(0.1, 0.2), (1.5, 0.5), (0.5, 1.05)])
tri.find_simplex(p)
# array([ 1, -1, 1], dtype=int32)
# We can also compute barycentric coordinates in triangle 1 for
# these points:
b = tri.transform[1,:2].dot(np.transpose(p - tri.transform[1,2]))
np.c_[np.transpose(b), 1 - b.sum(axis=0)]
# array([[ 0.1 , 0.09090909, 0.80909091],
# [ 1.5 , -0.90909091, 0.40909091],
# [ 0.5 , 0.5 , 0. ]])
# The coordinates for the first point are all positive, meaning it
# is indeed inside the triangle. The third point is on a vertex,
# hence its null third coordinate.