| Re: yet another 2d matching question [message #72001 is a reply to message #71998] |
Fri, 30 July 2010 08:25   |
Gray
Messages: 253
Registered: February 2010
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Senior Member |
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On Jul 30, 11:23 am, Gray <grayliketheco...@gmail.com> wrote:
> On Jul 30, 11:15 am, Paolo <pgri...@gmail.com> wrote:
>
>
>
>
>
>> On Jul 30, 10:01 am, Gray <grayliketheco...@gmail.com> wrote:
>
>>> Hi all,
>
>>> For quite a while I've been using JD Smith's match_2d routine to match
>>> xy coords between lists. However, this and all the other matching
>>> codes I've seen out there suffer from a variation of the uniqueness of
>>> matches problem.
>
>>> Codes like SRCOR in the NASA IDL library let you specify a one-to-one
>>> match, i.e. enforcing that each element in list 2 only be matched to
>>> one element in list 1; using match_2d's match_distance keyword one
>>> could implement the same effect oneself. However, while that excludes
>>> multiple matches to the same element, it's all done after the fact,
>>> after the original match was determined.
>
>>> What I'm looking for is an algorithm that matches 2 lists, identifies
>>> multiple-matches, and then looks for additional matches within the
>>> search radius for elements which would become unmatched after
>>> enforcing a one-to-one relationship. What I mean is, say element 0 in
>>> list 2 is matched to both element 3 and element 5 in list 1, and that
>>> the distance between 2_0 and 1_3 is smaller than the distance between
>>> 2_0 and 1_5. In that case, 1_5 would become unmatched; but what if
>>> there is element 2_1 which is also within the search radius of 1_5?
>>> Then, 1_5 should be re-matched with 2_1.
>
>>> My best idea thus far is to run match_2d once, identify multiple-
>>> matches, keep the matches with minimum distance using match_distance,
>>> then iterate with the remaining elements until match_2d returns no
>>> matches. Can anyone come up with a better solution?
>
>> Hmmm... what about starting with first point (a) in list 1, finding
>> the nearest
>> point (b) to (a) in list 2, removing (b) from list 2 and repeat for
>> all points
>> in list 1? [this assumes list 1 and list 2 have the same number of
>> elements N,
>> which is a necessary condition for a one-to-one matching].
>
>> With some smart partitioning of list 1 it will take ~log(N) to find
>> the nearest
>> point, so we are looking at ~ N log(N) operations...
>
>> Ciao,
>> Paolo
>
>>> --Gray
>
> I'm fine with having there be points which don't match at all w/in the
> search radius, I'm just looking to force any matches that exist to be
> recognized.
>
> The straight FOR-loop method is certainly serviceable, but I had hoped
> there was a more efficient way to do it... but it's certainly possible
> (or even likely) that anything fancier I try to do is LESS efficient.
>
> --Gray
Though I have trouble believing that FOR is the way to go when I have
~50k elements in each list.
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