Removing duplicates (within a given tolerance) from a Numpy array of vectors

I have an Nx5 array containing N vectors of form 'id', 'x', 'y', 'z' and 'energy'. I need to remove duplicate points (i.e. where x, y, z all match) within a tolerance of say 0.1. Ideally I could create a function where I pass in the array, columns that need to match and a tolerance on the match.

Following this thread on Scipy-user, I can remove duplicates based on a full array using record arrays, but I need to just match part of an array. Moreover this will not match within a certain tolerance.

I could laboriously iterate through wit开发者_开发问答h a for loop in Python but is there a better Numponic way?

You might look at scipy.spatial.KDTree. How big is N ?
Added: oops, tree.query_pairs is not in scipy 0.7.1 .

When in doubt, use brute force: split the space (here side^3) into little cells, one point per cell:

""" 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]

I have finally got a solution that I am happy with, this is a slightly cleaned up cut and paste from my own code. There may yet be some bugs.

Note: that it still uses a 'for' loop. I could use Denis's idea of KDTree above coupled with the rounding to get the full solution.

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: http://stackoverflow.com/questions/2433882/
    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()

Have not tested this but if you sort your array along x then y then z this should get you the list of duplicates. You then need to choose which to keep.

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

Also just found this http://mail.scipy.org/pipermail/scipy-user/2008-April/016504.html