On Thu, 2008-09-18 at 07:57 -0500, Tim Chase wrote:
> > Code: Select all
> > for i in range(len(IN)): #scan all elements of the list IN
> > for j in range(len(IN)):
> > if i <> j:
> > if IN[i].coordinates[0] == IN[j].coordinates[0]:
> > if IN[i].coordinates[1] == IN[j].coordinates[1]:
> > SN.append(IN[i].label)
> >
> >
> Such changes might look something like
>
> 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 you aren't checking j values less than i, you might want to do both
SN.append(IN[i].label)
SN.append(IN[j].label)
on the same pass.
> 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.
>
> -tkc
>
>
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>
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