| Re: faster then where possible? [message #66275] |
Thu, 07 May 2009 09:42  |
Jean H.
Messages: 472
Registered: July 2006
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Senior Member |
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rogass@googlemail.com wrote:
> Hi,
> i'm searching for some alternative approaches to compute the following
> "much" faster:
>
> -> matrix1 has m columns and n rows, matrix2 has 2 columns and n rows
> -> the values in matrix2 are NOT in matrix1, but within the min-max-
> range of matrix1
>
> szm1=size(matrix1,/dimensions)
> szm2=size(matrix2,/dimensions)
> index={ind:ptr_new()}
> indices=replicate(index,szm2[1])
>
> for j=0ull,szm1[1] do begin
> helpindex= where(matrix1[*,j] ge matrix2[0,j] and matrix1[*,j] le
> matrix2[1,j],c)
> if c gt 0 then begin
> indices[j] = ptr_new(uintarr(c))
> (*indices)[j]=helpindex
> endif else continue
> endfor
>
> It seems to be a typical Nearest-Neighbor-Problem, but all alternative
> approaches I tried were always slower. Maybe someone here has a good
> idea?
>
> Thank you and best regards
>
> Christian
>
Hi,
if you have enough memory, you could do something like this (not tested):
minArray = rebin(matrix2[0,*], n_elements(matrix1[*,0],
n_elements(matrix1[0,*]))
maxArray = rebin(matrix2[1,*], n_elements(matrix1[*,0],
n_elements(matrix1[0,*]))
goodIdx = where (matrix2 lt minArray and matrix2 gt minarray)
then you just have to transform the index so they match each row
Jean
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| Re: faster then where possible? [message #66464 is a reply to message #66275] |
Fri, 08 May 2009 02:12  |
rogass
Messages: 200
Registered: April 2008
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Senior Member |
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Dear Jean,
good idea, but unfortunately it is much slower (20 times) in my
specific case (spectral resampling routine). But thank you!
Any other ideas? Maybe some triangulation magic for getting nearest
neighbor? :)
Best regards
Christian
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