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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:
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>> Hi Folks,
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>> I know that there is no 'correct' way for interpolation, but maybe a 'best' IDL way. The following example easily demonstrates what I mean:
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>> IDL> d=dist(20) & print,interpolate(d,[.5],[.5],/grid)
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>> 0.853553
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>> IDL> print,interpolate(d,[.5],[.5],/grid,/cubic)
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>> 0.622236
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>> IDL> print,interpolate(d,[.400544],[.400544],/grid)
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>> 0.707107
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>> IDL> print,interpolate(d,[.581946],[.581946],/grid,/cubic)
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>> 0.707107
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>> IDL> print,'The result should be: ',sqrt(.5)
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>> The result should be: 0.707107
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>> 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?
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>> Thanks and Cheers
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>> Chris
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> 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:
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> print,interpolate(d,[.5],[.5],/grid,cubic=-0.5)
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> then you get:
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> 0.738791
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> Hope it helps,
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> 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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[Message index]

		interpolate weirdness By: rogass on Fri, 21 February 2014 04:00
		Re: interpolate weirdness By: Helder Marchetto on Fri, 21 February 2014 04:33
		Re: interpolate weirdness By: Helder Marchetto on Fri, 21 February 2014 05:59
		Re: interpolate weirdness By: Helder Marchetto on Fri, 21 February 2014 06:06
		Re: interpolate weirdness By: rogass on Fri, 28 February 2014 08:31

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