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Home/ Questions/Q 8636405
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Editorial Team
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Editorial Team
Asked: June 12, 20262026-06-12T10:13:29+00:00 2026-06-12T10:13:29+00:00

I’m using python and numpy to compare two arrays or equal shape with coordinates

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I’m using python and numpy to compare two arrays or equal shape with coordinates (x,y,z) in order to match them, which look like that:

coordsCFS
array([[ 0.02      ,  0.02      ,  0.        ],
       [ 0.03      ,  0.02      ,  0.        ],
       [ 0.02      ,  0.025     ,  0.        ],
        ..., 
       [ 0.02958333,  0.029375  ,  0.        ],
       [ 0.02958333,  0.0290625 ,  0.        ],
       [ 0.02958333,  0.0296875 ,  0.        ]])

and

coordsRMED
array([[ 0.02      ,  0.02      ,  0.        ],
       [ 0.02083333,  0.02      ,  0.        ],
       [ 0.02083333,  0.020625  ,  0.        ],
       ..., 
       [ 0.03      ,  0.0296875 ,  0.        ],
       [ 0.02958333,  0.03      ,  0.        ],
       [ 0.02958333,  0.0296875 ,  0.        ]]) 

The data are read from two hdf5 files with h5py.
For the comparison I use allclose, which tests for “almost equality”. The coordinates do not match within python’s regular floating point precision. This is the reason I used the for loops, otherwise it would have worked with numpy.where. I usually try to avoid for loops, but in this context I couldn’t figure out how. So I came up with this surprisingly slow snippet:

mapList = []
for cfsXYZ in coordsCFS:
    # print cfsXYZ
    indexMatch = 0
    match = []
    for asterXYZ in coordRMED:
        if numpy.allclose(asterXYZ,cfsXYZ):
            match.append(indexMatch)
            # print "Found match at index " + str(indexMatch)
            # print asterXYZ
        indexMatch += 1

    # check: must only find one match. 
    if len(match) != 1:
        print "ERROR matching"
        print match
        print cfsXYZ
        return 1

    # save to list
    mapList.append(match[0])

if len(mapList) != coordsRMED.shape[0]:
    print "ERROR: matching consistency check"
    print mapList
    return 1

This is very slow for my test sample size (800 rows). I plan to compare much larger sets. I could remove the consistency check and use break in the inner for loop for some speed benefit. Is there still a better way?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-12T10:13:30+00:00Added an answer on June 12, 2026 at 10:13 am

    You could get rid of the inner loop with something like this:

    for cfsXYZ in coordsCFS:
        match = numpy.nonzero(
            numpy.max(numpy.abs(coordRMED - cfsXYZ), axis=1) < TOLERANCE)
    
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