# Source code for pygeo.geo_utils.remove_duplicates

import numpy as np
from .norm import eDist

# --------------------------------------------------------------
#  Functions that removes duplicate entries from a list
# --------------------------------------------------------------

[docs]def unique(s):
r"""Return a list of the elements in s, but without duplicates.

For example, unique([1,2,3,1,2,3]) is some permutation of [1,2,3],
unique("abcabc") some permutation of ["a", "b", "c"], and
unique(([1, 2], [2, 3], [1, 2])) some permutation of
[[2, 3], [1, 2]].

For best speed, all sequence elements should be hashable.  Then
unique() will usually work in linear time.

If not possible, the sequence elements should enjoy a total
ordering, and if list(s).sort() doesn't raise TypeError it's
assumed that they do enjoy a total ordering.  Then unique() will
usually work in :math:\mathcal{O}(N\log_2(N)) time.

If that's not possible either, the sequence elements must support
equality-testing.  Then unique() will usually work in quadratic
time.
"""

n = len(s)
if n == 0:
return []

# Try using a dict first, as that's the fastest and will usually
# work.  If it doesn't work, it will usually fail quickly, so it
# usually doesn't np.cost much to *try* it.  It requires that all the
# sequence elements be hashable, and support equality comparison.
u = {}
try:
for x in s:
u[x] = 1
except TypeError:
pass
else:
return sorted(u.keys())

# We can't hash all the elements.  Second fastest is to sort,
# which brings the equal elements together; then duplicates are
# easy to weed out in a single pass.
# NOTE:  Python's list.sort() was designed to be efficient in the
# presence of many duplicate elements.  This isn't true of all
# sort functions in all languages or libraries, so this approach
# is more effective in Python than it may be elsewhere.

try:
t = list(s)
t.sort()
except TypeError:
pass
else:
assert n > 0
last = t[0]
lasti = i = 1
while i < n:
if t[i] != last:
t[lasti] = last = t[i]
lasti += 1
i += 1
return t[:lasti]

# Brute force is all that's left.

u = []
for x in s:
if x not in u:
u.append(x)
return u

[docs]def uniqueIndex(s, sHash=None):
"""
This function is based on :meth:unique.
The idea is to take a list s, and reduce it as per unique.

the same size as the original s, and points to where it ends up in
the the reduced list

if sHash is not specified for sorting, s is used

"""
if sHash is not None:
ind = np.argsort(np.argsort(sHash))
else:
ind = np.argsort(np.argsort(s))

n = len(s)
t = list(s)
t.sort()

diff = np.zeros(n, "bool")

last = t[0]
lasti = i = 1
while i < n:
if t[i] != last:
t[lasti] = last = t[i]
lasti += 1
else:
diff[i] = True
i += 1

b = np.where(diff)[0]
for i in range(n):
ind[i] -= b.searchsorted(ind[i], side="right")

return t[:lasti], ind

[docs]def pointReduce(points, nodeTol=1e-4):
"""Given a list of N points in ndim space, with possible
duplicates, return a list of the unique points AND a pointer list
for the original points to the reduced set"""

# First
points = np.array(points)
N = len(points)
if N == 0:
return points, None
dists = []
for ipt in range(N):
dists.append(np.sqrt(np.dot(points[ipt], points[ipt])))

# we need to round the distances to 8 decimals before sorting
# because 2 points might have "identical" distances to the origin,
# but they might differ on the 16 significant figure. As a result
# the argsort might flip their order even though the elements
# should not take over each other. By rounding them to 8
# significant figures, we somewhat guarantee that nodes that
# have similar distances to the origin dont get shuffled
# because of floating point errors
dists_rounded = np.around(dists, decimals=8)

# the "stable" sorting algorithm guarantees that entries
# with the same values dont overtake each other.
# The entries with identical distances are fully checked
# in the brute force check below.
ind = np.argsort(dists_rounded, kind="stable")

i = 0
cont = True
newPoints = []

while cont:
cont2 = True
tempInd = []
j = i
while cont2:
if abs(dists[ind[i]] - dists[ind[j]]) < nodeTol:
tempInd.append(ind[j])
j = j + 1
if j == N:  # Overrun check
cont2 = False
else:
cont2 = False

subPoints = []  # Copy of the list of sub points with the dists
for ii in range(len(tempInd)):
subPoints.append(points[tempInd[ii]])

# Brute Force Search them
newPoints.extend(subUniquePts)

for ii in range(len(tempInd)):

i = j - 1 + 1
if i == N:
cont = False

[docs]def pointReduceBruteForce(points, nodeTol=1e-4):
"""Given a list of N points in ndim space, with possible
duplicates, return a list of the unique points AND a pointer list
for the original points to the reduced set

Warnings
--------
This is the brute force version of :func:pointReduce.
"""
N = len(points)
if N == 0:
return points, None
uniquePoints = [points[0]]
for i in range(1, N):
foundIt = False
for j in range(len(uniquePoints)):
if eDist(points[i], uniquePoints[j]) < nodeTol: