| interpolate weirdness [message #87659] |
Fri, 21 February 2014 04:00  |
rogass
Messages: 200
Registered: April 2008
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Senior Member |
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Hi Folks,
I know that there is no 'correct' way for interpolation, but maybe a 'best' IDL way. The following example easily demonstrates what I mean:
IDL> d=dist(20) & print,interpolate(d,[.5],[.5],/grid)
0.853553
IDL> print,interpolate(d,[.5],[.5],/grid,/cubic)
0.622236
IDL> print,interpolate(d,[.400544],[.400544],/grid)
0.707107
IDL> print,interpolate(d,[.581946],[.581946],/grid,/cubic)
0.707107
IDL> print,'The result should be: ',sqrt(.5)
The result should be: 0.707107
It does not play a role which IDL version is used - for IDL 6.4 and for IDL 8.3 I got the same results. Is there any way to get 'closer' to a reasonable result?
Thanks and Cheers
Chris
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| Re: interpolate weirdness [message #87660 is a reply to message #87659] |
Fri, 21 February 2014 04:33  |
Helder Marchetto
Messages: 520
Registered: November 2011
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Senior Member |
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On Friday, February 21, 2014 1:00:36 PM UTC+1, CR wrote:
> Hi Folks,
>
>
>
> I know that there is no 'correct' way for interpolation, but maybe a 'best' IDL way. The following example easily demonstrates what I mean:
>
>
>
> IDL> d=dist(20) & print,interpolate(d,[.5],[.5],/grid)
>
> 0.853553
>
>
>
> IDL> print,interpolate(d,[.5],[.5],/grid,/cubic)
>
> 0.622236
>
>
>
> IDL> print,interpolate(d,[.400544],[.400544],/grid)
>
> 0.707107
>
>
>
> IDL> print,interpolate(d,[.581946],[.581946],/grid,/cubic)
>
> 0.707107
>
>
>
> IDL> print,'The result should be: ',sqrt(.5)
>
> The result should be: 0.707107
>
>
>
>
>
> It does not play a role which IDL version is used - for IDL 6.4 and for IDL 8.3 I got the same results. Is there any way to get 'closer' to a reasonable result?
>
>
>
> Thanks and Cheers
>
>
>
> Chris
I'm not sure what you're trying to do here, but interpolating near borders is kind of complicated because you can only have one side to rely on. However, if you're going to use bicubic interpolation, I would recommend using:
print,interpolate(d,[.5],[.5],/grid,cubic=-0.5)
then you get:
0.738791
Hope it helps,
h
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| Re: interpolate weirdness [message #87661 is a reply to message #87660] |
Fri, 21 February 2014 05:59 |
Helder Marchetto
Messages: 520
Registered: November 2011
|
Senior Member |
|
|
On Friday, February 21, 2014 1:33:45 PM UTC+1, Helder wrote:
> On Friday, February 21, 2014 1:00:36 PM UTC+1, CR wrote:
>
>> Hi Folks,
>
>>
>
>>
>
>>
>
>> I know that there is no 'correct' way for interpolation, but maybe a 'best' IDL way. The following example easily demonstrates what I mean:
>
>>
>
>>
>
>>
>
>> IDL> d=dist(20) & print,interpolate(d,[.5],[.5],/grid)
>
>>
>
>> 0.853553
>
>>
>
>>
>
>>
>
>> IDL> print,interpolate(d,[.5],[.5],/grid,/cubic)
>
>>
>
>> 0.622236
>
>>
>
>>
>
>>
>
>> IDL> print,interpolate(d,[.400544],[.400544],/grid)
>
>>
>
>> 0.707107
>
>>
>
>>
>
>>
>
>> IDL> print,interpolate(d,[.581946],[.581946],/grid,/cubic)
>
>>
>
>> 0.707107
>
>>
>
>>
>
>>
>
>> IDL> print,'The result should be: ',sqrt(.5)
>
>>
>
>> The result should be: 0.707107
>
>>
>
>>
>
>>
>
>>
>
>>
>
>> It does not play a role which IDL version is used - for IDL 6.4 and for IDL 8.3 I got the same results. Is there any way to get 'closer' to a reasonable result?
>
>>
>
>>
>
>>
>
>> Thanks and Cheers
>
>>
>
>>
>
>>
>
>> Chris
>
>
>
> I'm not sure what you're trying to do here, but interpolating near borders is kind of complicated because you can only have one side to rely on. However, if you're going to use bicubic interpolation, I would recommend using:
>
> print,interpolate(d,[.5],[.5],/grid,cubic=-0.5)
>
> then you get:
>
> 0.738791
>
> Hope it helps,
>
> h
Just for sake of comparison, if you do the same for a point not next to the boundary, you get:
print,interpolate(d,[5.5],[5.5],/grid, cubic=-0.5)
7.77795
print, sqrt(2)*5.5
7.77817
The difference is 0.00022220612, what is much better.
Hope it helps.
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| Re: interpolate weirdness [message #87662 is a reply to message #87661] |
Fri, 21 February 2014 06:06 |
Helder Marchetto
Messages: 520
Registered: November 2011
|
Senior Member |
|
|
On Friday, February 21, 2014 2:59:46 PM UTC+1, Helder wrote:
> On Friday, February 21, 2014 1:33:45 PM UTC+1, Helder wrote:
>
>> On Friday, February 21, 2014 1:00:36 PM UTC+1, CR wrote:
>
>>
>
>>> Hi Folks,
>
>>
>
>>>
>
>>
>
>>>
>
>>
>
>>>
>
>>
>
>>> I know that there is no 'correct' way for interpolation, but maybe a 'best' IDL way. The following example easily demonstrates what I mean:
>
>>
>
>>>
>
>>
>
>>>
>
>>
>
>>>
>
>>
>
>>> IDL> d=dist(20) & print,interpolate(d,[.5],[.5],/grid)
>
>>
>
>>>
>
>>
>
>>> 0.853553
>
>>
>
>>>
>
>>
>
>>>
>
>>
>
>>>
>
>>
>
>>> IDL> print,interpolate(d,[.5],[.5],/grid,/cubic)
>
>>
>
>>>
>
>>
>
>>> 0.622236
>
>>
>
>>>
>
>>
>
>>>
>
>>
>
>>>
>
>>
>
>>> IDL> print,interpolate(d,[.400544],[.400544],/grid)
>
>>
>
>>>
>
>>
>
>>> 0.707107
>
>>
>
>>>
>
>>
>
>>>
>
>>
>
>>>
>
>>
>
>>> IDL> print,interpolate(d,[.581946],[.581946],/grid,/cubic)
>
>>
>
>>>
>
>>
>
>>> 0.707107
>
>>
>
>>>
>
>>
>
>>>
>
>>
>
>>>
>
>>
>
>>> IDL> print,'The result should be: ',sqrt(.5)
>
>>
>
>>>
>
>>
>
>>> The result should be: 0.707107
>
>>
>
>>>
>
>>
>
>>>
>
>>
>
>>>
>
>>
>
>>>
>
>>
>
>>>
>
>>
>
>>> It does not play a role which IDL version is used - for IDL 6.4 and for IDL 8.3 I got the same results. Is there any way to get 'closer' to a reasonable result?
>
>>
>
>>>
>
>>
>
>>>
>
>>
>
>>>
>
>>
>
>>> Thanks and Cheers
>
>>
>
>>>
>
>>
>
>>>
>
>>
>
>>>
>
>>
>
>>> Chris
>
>>
>
>>
>
>>
>
>> I'm not sure what you're trying to do here, but interpolating near borders is kind of complicated because you can only have one side to rely on. However, if you're going to use bicubic interpolation, I would recommend using:
>
>>
>
>> print,interpolate(d,[.5],[.5],/grid,cubic=-0.5)
>
>>
>
>> then you get:
>
>>
>
>> 0.738791
>
>>
>
>> Hope it helps,
>
>>
>
>> h
>
>
>
> Just for sake of comparison, if you do the same for a point not next to the boundary, you get:
>
> print,interpolate(d,[5.5],[5.5],/grid, cubic=-0.5)
>
> 7.77795
>
> print, sqrt(2)*5.5
>
> 7.77817
>
> The difference is 0.00022220612, what is much better.
>
>
>
> Hope it helps.
One more commment about the bilinear interpolation:
IDL> d=dist(20) & print,interpolate(d,[.5],[.5],/grid)
0.853553
The value 0.853553 is excately what is expected!
Linear interpolation on x gives 0.5 and (1+sqrt(2))/2=2.414, then take the average of those two values and you get:
IDL> (0.5+(1.0+sqrt(2))/2.0)/2.0
0.85355341
Correct answer. See http://en.wikipedia.org/wiki/Bilinear_interpolation
Cheers,
Helder
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| Re: interpolate weirdness [message #87863 is a reply to message #87662] |
Fri, 28 February 2014 08:31 |
rogass
Messages: 200
Registered: April 2008
|
Senior Member |
|
|
Dear Helder,
thank you for clarification. The situation has changed. Desired values are now far away from boundaries and I can use something like this:
function cr_interpol,dat,mat_old,mat_new,cubic=cubic
ndim = size(dat,/ndimensions)
dims = size(dat,/dimensions)
case (1b) of
ndim eq 1 : begin
newx = (interpol(ulindgen(dims[0]),(mat_old[0,*])[*],(mat_new[0,*]) [*]))[*]
result = interpolate(dat[*],newx,/grid,cubic=cubic)
end
ndim eq 2 : begin
newx = (interpol(ulindgen(dims[0]),(mat_old[0,*])[*],(mat_new[0,*]) [*]))[*]
newy = (interpol(ulindgen(dims[1]),(mat_old[1,*])[*],(mat_new[1,*]) [*]))[*]
result = interpolate(dat,newx,newy,/grid,cubic=cubic)
end
ndim eq 3 : begin
newx = (interpol(ulindgen(dims[0]),(mat_old[0,*])[*],(mat_new[0,*]) [*]))[*]
newy = (interpol(ulindgen(dims[1]),(mat_old[1,*])[*],(mat_new[1,*]) [*]))[*]
newz = (interpol(ulindgen(dims[2]),(mat_old[2,*])[*],(mat_new[2,*]) [*]))[*]
result = interpolate(dat,newx,newy,newz,/grid,cubic=cubic)
end
endcase
return,result
end
Cheers
Chris
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