Re: yet another 2d matching question [message #72002 is a reply to message #72001] |
Fri, 30 July 2010 08:23   |
Gray
Messages: 253 Registered: February 2010
|
Senior Member |
|
|
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
|
|
|