| Assigning the values of a matrix to a larger matrix [message #88467] |
Wed, 30 April 2014 13:52  |
Fabrice Lambert
Messages: 6
Registered: March 2012
|
Junior Member |
|
|
Hello,
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
for example something like this (which unfortunately does not work:
Base[*,*,0:90]=Input[*,*,0]
Thanks,
Fabrice
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| Re: Assigning the values of a matrix to a larger matrix [message #88468 is a reply to message #88467] |
Wed, 30 April 2014 14:21  |
Heinz Stege
Messages: 189
Registered: January 2003
|
Senior Member |
|
|
Hello Fabrice.
On Wed, 30 Apr 2014 13:52:42 -0700 (PDT), fabrice.lambert@gmail.com
wrote:
> 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
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| Re: Assigning the values of a matrix to a larger matrix [message #88470 is a reply to message #88468] |
Wed, 30 April 2014 16:05 |
Fabrice Lambert
Messages: 6
Registered: March 2012
|
Junior Member |
|
|
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
|
|
|
|
| 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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| Re: Assigning the values of a matrix to a larger matrix [message #88498 is a reply to message #88497] |
Fri, 02 May 2014 16:51 |
Fabrice Lambert
Messages: 6
Registered: March 2012
|
Junior Member |
|
|
Hello Heinz,
Indeed, your initial code did produce an error. What I meant with "that works" was that I read up on rebin and tested it on some simple examples.
You understood my problem well now. I have many variables with plenty of dimensions and I've been looking for a way to avoid endless for loops to assign one dimension of one variable to another dimension of another variable. It works well enough with for loops, but I have a feeling there must be a more elegant solution.
I'll study your new solution, thanks a lot for the follow up.
-Fabrice
On Friday, May 2, 2014 2:16:45 PM UTC-4, Heinz Stege wrote:
> On Wed, 30 Apr 2014 16:05:11 -0700 (PDT),
>
> 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
|
|
|
|
| Re: Assigning the values of a matrix to a larger matrix [message #88525 is a reply to message #88497] |
Wed, 07 May 2014 09:35 |
Fabrice Lambert
Messages: 6
Registered: March 2012
|
Junior Member |
|
|
Hello,
Thank you very much for the help with rebin. That works great to work on the last dimension. Unfortunately, rebin doesn't support changes to other dimensions. Example:
for i=0,127 do begin
for j=0,63 do begin
for t=0,11 do begin
for b=0,5 do begin
for p=0,1 do begin
AtmConc[i,j,*,t,b,p]=SurfConc[i,j,t,b,p]*AtmFactor[i,j,*,t,b ,p]
endfor
endfor
endfor
endfor
endfor
Is there a way to avoid the for-loops in that case?
Thanks,
Fabrice
On Friday, May 2, 2014 2:16:45 PM UTC-4, Heinz Stege wrote:
> On Wed, 30 Apr 2014 16:05:11 -0700 (PDT),
>
> 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
|
|
|
|
|
|
| Re: Assigning the values of a matrix to a larger matrix [message #88528 is a reply to message #88527] |
Wed, 07 May 2014 16:40 |
Fabrice Lambert
Messages: 6
Registered: March 2012
|
Junior Member |
|
|
Just what I needed.
Thanks a lot Heinz!
-Fabrice
On Wednesday, May 7, 2014 1:51:19 PM UTC-4, Heinz Stege wrote:
> On Wed, 7 May 2014 09:35:45 -0700 (PDT), Fabrice Lambert wrote:
>
>
>
>> Is there a way to avoid the for-loops in that case?
>
>
>
> Yes again. REBIN has a partner. It's name is REFORM. REFORM is able to
>
> add extra dimensions within an array. Example
>
> a=indgen(3,2)
>
> a=reform(a,3,1,2,/overwrite)
>
> After this REBIN can be used to increase the new dimension.
>
>
>
> For more information you may want to read
>
> http://www.idlcoyote.com/tips/rebin_magic.html.
>
>
>
> Cheers, Heinz
|
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