| About INTERPOLATE function. [message #91564] |
Wed, 29 July 2015 18:21  |
Dae-Kyu Shin
Messages: 25
Registered: February 2015
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Junior Member |
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Example 1.
a = findgen(4)
a[1] = !values.f_nan
print, a
0.000000 NaN 2.00000 3.00000
print, interpolate(a, [0, 1, 2, 3])
NaN NaN 2.00000 3.00000
Example 2.
a = transpose(a)
print, interpolate(a, 0, [0, 1, 2, 3], /grid)
0.000000
NaN
2.00000
3.00000
different result~!!
is this right output?
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| Re: About INTERPOLATE function. [message #91565 is a reply to message #91564] |
Wed, 29 July 2015 20:55  |
wlandsman
Messages: 743
Registered: June 2000
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Senior Member |
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I think this a case of the location values being interpreted as floating point, and you are on a knife-edge when next to an NaN value
if a = [0,!values.f_nan,2,3]
then one expects interpolate(a,0) = 0 and interpolate(a,2) = 2
because the zeroth element has value of 0 and the second element has a value of 2.
But one also expects interpolate(a,0.00000000001)=NaN
and interpolate(a,1.99999999999(=NaN
because any type of interpolation/calculation involving an NaN value -- no matter how tiny -- should give NaN as an answer. If the location values are interpreted as floating point then inputing a 0 could mean a value very slightly smaller or larger. Thus IDL gives
IDL> print,interpolate(a,0),interpolate(a,2)
NaN 2.00000
though evidently with the /grid keyword the rounding works differently
My advice is -- don't use INTERPOLATE() when you have NaN values.
On Wednesday, July 29, 2015 at 9:21:04 PM UTC-4, Dae-Kyu Shin wrote:
> Example 1.
>
> a = findgen(4)
> a[1] = !values.f_nan
> print, a
> 0.000000 NaN 2.00000 3.00000
>
> print, interpolate(a, [0, 1, 2, 3])
> NaN NaN 2.00000 3.00000
>
> Example 2.
> a = transpose(a)
> print, interpolate(a, 0, [0, 1, 2, 3], /grid)
> 0.000000
> NaN
> 2.00000
> 3.00000
>
> different result~!!
> is this right output?
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