Skip: > indexes = range(len(IN)) > for i in indexes: #scan all elements of the list IN > for j in indexes:
Nope, use xrange in both situations, and save a list. 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. Bye, bearophile -- http://mail.python.org/mailman/listinfo/python-list
