[Numpy-discussion] Robust Sorting of Points

Pierre Haessig pierre.haessig at crans.org
Tue Oct 29 06:37:24 EDT 2013

Hi Freddie,

Le 29/10/2013 10:21, Freddie Witherden a écrit :
> The order itself does not need to satisfy any specific properties.
I can't agree with you : if there is no specific property, then keeping
the list *unchanged* would be a fine solution (and very fast and very
very robust) ;-)

what about defining a comparison function (in the sense of the old cmp
keyword of list.sort) such as :

def compare(point, other):
    delta = point - other
    argmax = np.abs(delta).argmax()
    delta_max = delta[argmax]
    if delta_max > 0:
        return 1
    elif delta_max < 0:
        return -1
        return 0

This function returns a comparison of the coordinates with the  biggest
absolute difference. Of course this doesn't define an *absolute order*
(since it doesn't exist). But I think it defines a *relative order* (if
this notion exists mathematically !!!!) which is indeed robust.

To apply this comparison function, the simple solution I see (for Python
<3) is, using your first example ;

>>> my_array = np.array([[-0.5, 0, 2**0.5],
                               [0.5, 0, 2**0.5 - 1e-15]])

>>> l = list(my_array)

>>>  l.sort(cmp = compare)

>>> l
[array([-0.5       ,  0.        ,  1.41421356]),
 array([ 0.5       ,  0.        ,  1.41421356])]

Now, if this comparison function indeed answers your question, the next
step is to plug it either in numpy sorting machinery or in the new
Python sorting which uses a "key function" (cf.


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