| Re: 2d min [message #74362] |
Thu, 13 January 2011 09:07  |
Gray
Messages: 253
Registered: February 2010
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Senior Member |
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On Jan 13, 8:57 am, chris <rog...@googlemail.com> wrote:
> On 13 Jan., 14:11, Gray <grayliketheco...@gmail.com> wrote:
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>> On Jan 13, 7:26 am, Gray <grayliketheco...@gmail.com> wrote:
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>>> On Jan 13, 2:18 am, chris <rog...@googlemail.com> wrote:
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>>>> On 12 Jan., 23:21, Gray <grayliketheco...@gmail.com> wrote:
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>>>> > Hi all,
>
>>>> > I have a 3d array, NxNxM. What I would like is to find the minimum of
>>>> > each NxN slice, and note the index of the minimum in the slice. I can
>>>> > find my minimum by doing min(min(array,ind1,dim=1),dim=1,ind2), but
>>>> > I'm not sure how to turn those two index arrays into the indices that
>>>> > I need. Help...?
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>>>> > Thanks!
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>>>> > --Gray
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>>>> Hi,
>>>> maybe I missed something, but why don't you use something like this:
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>>>> IDL> a=randomn(seed,10,10,5)
>>>> IDL> min=min(a,dimension=3,ind)
>>>> IDL> help,min,ind
>>>> MIN FLOAT = Array[10, 10]
>>>> IND LONG64 = Array[10, 10]
>>>> IDL> ind2=array_indices(size(a,/dimensions),ind,/dimensions)
>>>> IDL> help,ind2
>>>> IND2 LONG64 = Array[3, 100]
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>>>> Is array_indices really to slow with the dimension keyword?
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>>>> Cheers
>
>>>> CR
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>>> This gets me a NxN array of minima... I want a vector of minima of
>>> length M (so the minimum in each plane).
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>> OK, for anyone else, here's what you have to do.
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>> IDL> a = randomu(seed,10,10,100)
>> IDL> minima = min(min(a,ind1,dim=1),ind2,dim=2)
>> IDL> ind1 -= rebin(indgen(1,100)*10^2.,10,100)
>> IDL> ind2 -= (indgen(100)*10)
>> IDL> xind = ind1[ind2,indgen(100)] mod 10
>> IDL> yind = ind2
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>> You have to use ind2 to find the right elements of ind1. Since you
>> get 1d indices from min, you need to subtract off N^2 or N to talk
>> about each plane individually. The reason I wanted to vectorize is
>> that my actual M is ~20k.
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> Ok, and this one?
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> t0=systime(1)
> a=randomu(seed,100l,100l,10000l)
> mins=min(reform(a,100l*100l,10000l,/over),ind,dimension=1)
> a=reform(a,100l,100l,10000l,/over) ;if you need this...
> help,ind,a &print,systime(1)-t0
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> IND LONG64 = Array[10000]
> A FLOAT = Array[100, 100, 10000]
> 1.9180000
> Maybe you can also use Thread Pool Keywords together with min.
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> Cheers
> CR
Ooh, reforming so that the first 2 dimensions become 1 that I can min
over... I didn't think about that possibility! Either way works, I
guess.
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