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Re: Nearest Neighbor ... again! [message #71629 is a reply to message #71627]

Thu, 08 July 2010 09:06

jeanh
Messages: 79
Registered: November 2009

Member

> The stupid solution is (sorry for the very very ugly code):
>
> ---
> for i = 0l, n2 - 1 do begin
> quad = (x2[i] - x1)^2 + (y2[i]- y1)^2
> for j=0, 3 do begin
> minquad = min(quad, p)
> if N_ELEMENTS(p) gt 1 then p = p[0] ; it happens.....
> out[j,i] = p
> quad[p] = max(quad) * 2. ;dummy large distance
> endfor
> endfor
> ---
>
> But I just cannot find a cleverer solution with triangulation...
> Someone clever than me to help ?
>
> Thanks a lot!
>
> Fabz

While there is certainly a better solution, you can start by removing
the j loop, and use sort(quad) instead of min() ... you can then take
the first 4 index.
Jean

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[Message index]

		Re: Nearest Neighbor ... again! By: Wout De Nolf on Fri, 09 July 2010 01:56
		Re: Nearest Neighbor ... again! By: penteado on Thu, 08 July 2010 09:27
		Re: Nearest Neighbor ... again! By: jeanh on Thu, 08 July 2010 09:06
		Re: Nearest Neighbor ... again! By: Fabzi on Thu, 08 July 2010 06:40
		Re: Nearest Neighbor ... again! By: David Fanning on Thu, 08 July 2010 06:21
		Re: Nearest Neighbor ... again! By: Fabzi on Mon, 12 July 2010 01:14

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