improving a huge double-for cycle

[Tim Chase]

> Tim Chase:
>>    for i in xrange(len(IN)):
>>      for j in xrange(i+1, len(IN)):
>>        if IN[i].coordinates == IN[j].coordinates:
>>          SN.append(IN[i].label)
>>
>> If my college algorithms memory serves me sufficiently, this
>> reduces your O(N^2) to O(N log N) which will garner you some
>> decent time savings.
> 
> That's O(n^2) still, it's just half matrix, a triangle.

Ah, good catch as I'm thinking about it more, you're right...it's 
O((N^2)/2) which is just O(N^2).  Sigh.  I'm not awake enough 
yet.  However, dividing the time by 2 (from 15hr to 7.5hr) is 
still a significant savings in my book :)

However, if list-comprehensions are faster as well, you might be 
able to do something like

   SN = [d.label
         for (i,d) in enumerate(IN)
         for j in xrange(i+1, len(IN))
         if d.coordinates == IN[j].coordinates
         ]

or the slightly more verbose (but closer to my original code) version

   SN = [IN[i].label
         for i in xrange(len(IN))
         for j in xrange(i+1, len(IN))
         if IN[i].coordinates == IN[j].coordinates
         ]

To the OP:  As always, throw some timing code on the various 
samples you get back from the list and see what works for you.

-tkc