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Re: Assigning the values of a matrix to a larger matrix [message #88497 is a reply to message #88470]

Fri, 02 May 2014 11:16

Heinz Stege
Messages: 189
Registered: January 2003

Senior Member

On Wed, 30 Apr 2014 16:05:11 -0700 (PDT), fabrice.lambert@gmail.com
wrote:

> Hello Heinz,
>
> That works, thanks a lot!
>
> I find it very counter-intuitive, though: In my example to fill in the values we want in the 3rd dimension of the base vector, we have to create a 4th dimension where every index contains a copy of the original 3d-input vector.
>
> Is there an easy way to understand that logic?
>
> Thanks,
> Fabrice
>
> On Wednesday, April 30, 2014 5:21:35 PM UTC-4, Heinz Stege wrote:
>> Hello Fabrice.
>>
>>
>>
>>
>>
>>
>>> Is there a way to write the following code in IDL/GDL without a for loop?
>>
>>>
>>
>>> for i=0,90 do begin
>>
>>> Base[*,*,i]=Input[*,*,0]
>>
>>> endfor
>>
>>
>>
>> Yes, you can write:
>>
>>
>>
>> n=size(input,/dimensions)
>>
>> base[0,0,0]=rebin(input,[n,91],/sample)
>>
>>
>>
>> If the base array does not have to be bigger than the 91x input
>>
>> array, you don't need to create it explicitly and can more simply
>>
>> write:
>>
>>
>>
>> n=size(input,/dimensions)
>>
>> base=rebin(input,[n,91],/sample)
>>
>>
>>
>> Cheers, Heinz

Hello Fabrice,

I believe, something went absolutely wrong here. Are you really sure,
that you get a correct result with my solution?

You say, that rebin(input,[n,91],/sample) has 4 dimensions. This
means, that input itself has 3 dimensions (and the size of the 3rd
dimension is greater or equal 2). It was my failure, not to think of
this possibility. Something made me thinking, that it was an "extra
dimension" you addressed with the zero in "Input[*,*,0]". Sorry for
this.

I wonder, why IDL didn't throw an error message. Does your base array
have more than 3 dimensions, which you don't want to touch with this
part of code?

Anyway, you want to write 3 dimensions into the base array. And as I
understand now, your input array has 3 dimensions too, but you want to
repeatedly use the first element of the third dimension. Here is a
detailed example, how to do this without a loop:

input=indgen(5,4,2)
base=intarr(5,4,3)
base[0,0,0]=rebin(input[*,*,0],5,4,3,/sample)

A print displays the following result:
IDL> print,base
0 1 2 3 4
5 6 7 8 9
10 11 12 13 14
15 16 17 18 19

0 1 2 3 4
5 6 7 8 9
10 11 12 13 14
15 16 17 18 19

0 1 2 3 4
5 6 7 8 9
10 11 12 13 14
15 16 17 18 19

Another note: It is not mandatory, that for big arrays, this solution
is faster than a loop. If your routine is time-consuming, you should
test it.

I hope, this makes the things more clear. And sorry again for my
mistake.

Cheers, Heinz

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[Message index]

		Assigning the values of a matrix to a larger matrix By: Fabrice Lambert on Wed, 30 April 2014 13:52
		Re: Assigning the values of a matrix to a larger matrix By: Heinz Stege on Wed, 30 April 2014 14:21
		Re: Assigning the values of a matrix to a larger matrix By: Fabrice Lambert on Wed, 30 April 2014 16:05
		Re: Assigning the values of a matrix to a larger matrix By: Heinz Stege on Fri, 02 May 2014 11:16
		Re: Assigning the values of a matrix to a larger matrix By: Fabrice Lambert on Fri, 02 May 2014 16:51
		Re: Assigning the values of a matrix to a larger matrix By: Fabrice Lambert on Wed, 07 May 2014 09:35
		Re: Assigning the values of a matrix to a larger matrix By: Heinz Stege on Wed, 07 May 2014 10:51
		Re: Assigning the values of a matrix to a larger matrix By: Fabrice Lambert on Wed, 07 May 2014 16:40

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