tree.query_pairs 不在 scipy 0.7.1 中。""" scatter points to little cells, 1 per cell """
from __future__ import division
import sys
import numpy as np
side = 100
npercell = 1 # 1: ~ 1/e empty
exec "\n".join( sys.argv[1:] ) # side= ...
N = side**3 * npercell
print "side: %d npercell: %d N: %d" % (side, npercell, N)
np.random.seed( 1 )
points = np.random.uniform( 0, side, size=(N,3) )
cells = np.zeros( (side,side,side), dtype=np.uint )
id = 1
for p in points.astype(int):
cells[tuple(p)] = id
id += 1
cells = cells.flatten()
# A C, an E-flat, and a G walk into a bar.
# The bartender says, "Sorry, but we don't serve minors."
nz = np.nonzero(cells)[0]
print "%d cells have points" % len(nz)
print "first few ids:", cells[nz][:10]
我终于找到了一个令我满意的解决方案，这是从我的代码中稍微整理过的剪贴板。可能还存在一些错误。
请注意：它仍然使用“for”循环。我可以使用Denis上面提出的KDTree的想法加上四舍五入来获得完整的解决方案。
import numpy as np
def remove_duplicates(data, dp_tol=None, cols=None, sort_by=None):
'''
Removes duplicate vectors from a list of data points
Parameters:
data An MxN array of N vectors of dimension M
cols An iterable of the columns that must match
in order to constitute a duplicate
(default: [1,2,3] for typical Klist data array)
dp_tol An iterable of three tolerances or a single
tolerance for all dimensions. Uses this to round
the values to specified number of decimal places
before performing the removal.
(default: None)
sort_by An iterable of columns to sort by (default: [0])
Returns:
MxI Array An array of I vectors (minus the
duplicates)
EXAMPLES:
Remove a duplicate
>>> import wien2k.utils
>>> import numpy as np
>>> vecs1 = np.array([[1, 0, 0, 0],
... [2, 0, 0, 0],
... [3, 0, 0, 1]])
>>> remove_duplicates(vecs1)
array([[1, 0, 0, 0],
[3, 0, 0, 1]])
Remove duplicates with a tolerance
>>> vecs2 = np.array([[1, 0, 0, 0 ],
... [2, 0, 0, 0.001 ],
... [3, 0, 0, 0.02 ],
... [4, 0, 0, 1 ]])
>>> remove_duplicates(vecs2, dp_tol=2)
array([[ 1. , 0. , 0. , 0. ],
[ 3. , 0. , 0. , 0.02],
[ 4. , 0. , 0. , 1. ]])
Remove duplicates and sort by k values
>>> vecs3 = np.array([[1, 0, 0, 0],
... [2, 0, 0, 2],
... [3, 0, 0, 0],
... [4, 0, 0, 1]])
>>> remove_duplicates(vecs3, sort_by=[3])
array([[1, 0, 0, 0],
[4, 0, 0, 1],
[2, 0, 0, 2]])
Change the columns that constitute a duplicate
>>> vecs4 = np.array([[1, 0, 0, 0],
... [2, 0, 0, 2],
... [1, 0, 0, 0],
... [4, 0, 0, 1]])
>>> remove_duplicates(vecs4, cols=[0])
array([[1, 0, 0, 0],
[2, 0, 0, 2],
[4, 0, 0, 1]])
'''
# Deal with the parameters
if sort_by is None:
sort_by = [0]
if cols is None:
cols = [1,2,3]
if dp_tol is not None:
# test to see if already an iterable
try:
null = iter(dp_tol)
tols = np.array(dp_tol)
except TypeError:
tols = np.ones_like(cols) * dp_tol
# Convert to numbers of decimal places
# Find the 'order' of the axes
else:
tols = None
rnd_data = data.copy()
# set the tolerances
if tols is not None:
for col,tol in zip(cols, tols):
rnd_data[:,col] = np.around(rnd_data[:,col], decimals=tol)
# TODO: For now, use a slow Python 'for' loop, try to find a more
# numponic way later - see: https://dev59.com/YUzSa4cB1Zd3GeqPnYim
sorted_indexes = np.lexsort(tuple([rnd_data[:,col] for col in cols]))
rnd_data = rnd_data[sorted_indexes]
unique_kpts = []
for i in xrange(len(rnd_data)):
if i == 0:
unique_kpts.append(i)
else:
if (rnd_data[i, cols] == rnd_data[i-1, cols]).all():
continue
else:
unique_kpts.append(i)
rnd_data = rnd_data[unique_kpts]
# Now sort
sorted_indexes = np.lexsort(tuple([rnd_data[:,col] for col in sort_by]))
rnd_data = rnd_data[sorted_indexes]
return rnd_data
if __name__ == '__main__':
import doctest
doctest.testmod()我没有测试过，但如果你按照x、y、z的顺序对数组进行排序，这应该会给你重复项列表。然后你需要选择要保留哪些。
def find_dup_xyz(anarray, x, y, z): #for example in an data = array([id,x,y,z,energy]) x=1 y=2 z=3
dup_xyz=[]
for i, row in enumerated(sortedArray):
nx=1
while (abs(row[x] - sortedArray[i+nx[x])<0.1) and (abs(row[z] and sortedArray[i+nx[y])<0.1) and (abs(row[z] - sortedArray[i+nx[z])<0.1):
nx=+1
dup_xyz.append(row)
return dup_xyz
还发现了这个: http://mail.scipy.org/pipermail/scipy-user/2008-April/016504.html