suicidaleggroll@gmail.com writes:
>
> On Monday, September 30, 2013 2:09:10 PM UTC-6, David Fanning wrote:
>> Gus writes:
>>
>>
>>
>>> I've read a few of the older posts on this topic, but their solution didn't really help me solve the problem that I am currently having with the poly_fit function. The set of coefficients generated by the function (a 4th degree polynomial) produces some rather absurd results. Here is a short version of the problem I am having.
>>
>>>
>>
>>> X = [0.000000, 11.6667, 822.914, 3458.85, 27703.4, 133928.]
>>
>>> Y = [15.9000, 16.0000, 17.0000, 18.0000, 19.0000, 20.0000]
>>
>>>
>>
>>> C = poly_fit(X, Y, /double, yfit=D)
>>
>>>
>>
>>> IDL generates the following coefficients (for C)
>>
>>>
>>
>>> 15.940691
>>
>>> 0.0015355228
>>
>>> -3.0965110e-007
>>
>>> 1.1170193e-011
>>
>>> -6.6767399e-017
>>
>>
>>
>> This code doesn't seem to work for me:
>>
>>
>>
>> IDL> X = [0.000000, 11.6667, 822.914, 3458.85, 27703.4, 133928.]
>>
>> IDL> Y = [15.9000, 16.0000, 17.0000, 18.0000, 19.0000, 20.0000]
>>
>> IDL> C = poly_fit(X, Y, /double, yfit=D)
>>
>> % Compiled module: POLY_FIT.
>>
>> % Variable is undefined: NDEGREE.
>>
>>
>>
>> Are you sure you are using the right POLY_FIT?
>>
>>
>>
>> Cheers,
>>
>>
>>
>> David
>>
>>
>>
>> --
>>
>> David Fanning, Ph.D.
>>
>> Fanning Software Consulting, Inc.
>>
>> Coyote's Guide to IDL Programming: http://www.idlcoyote.com/
>>
>> Sepore ma de ni thue. ("Perhaps thou speakest truth.")
>
>
> He just missed the "4" in the call (for a 4th order polynomial).
>
> Gus - actually Excel gives the EXACT same answer as IDL, which, as you said, is completely ridiculous. The problem is you're fitting a 4th order polynomial to 5 data points. Because of this, the solution will be mathematically perfect (R^2 = 1), because the solution is not overdetermined and no least squares fitting can be performed.
>
> You need more points in order to generate a "valid" 4th order poly fit so the "fit" can actually do some good, rather than just reproduce your 5 values exactly (with god knows what in between).
>
> I've run into this in the past, and in that application it was reasonable to linearly interpolate my 5 points to, say, 1000 points, and then perform the poly fit on that.
>
> For example:
> X = [0.000000, 11.6667, 822.914, 3458.85, 27703.4, 133928.]
> Y = [15.9000, 16.0000, 17.0000, 18.0000, 19.0000, 20.0000]
>
> newX = dindgen(1000)/999 * (max(X)-min(X)) + min(X)
> newY = interpol(Y, X, newX)
> C = poly_fit(newX, newY, 4, /double, yfit=D)
> print, C
> 17.356485
> 0.00010074819
> -2.0171981e-09
> 1.8082251e-14
> -5.6591322e-20
It is probably worth pointing out that the order of the coefficients in
the variable C are e, d, c, b, and a. That sometimes (nearly every time
with me) gets missed.
Cheers,
David
--
David Fanning, Ph.D.
Fanning Software Consulting, Inc.
Coyote's Guide to IDL Programming: http://www.idlcoyote.com/
Sepore ma de ni thue. ("Perhaps thou speakest truth.")
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