http://imagej.273.s1.nabble.com/Curve-Fitting-issue-question-tp5018961p5018983.html
thank you very much for your exceedingly informative reply. I do not
I tested your suggestions for both. The resulting data array is checked
against min/max of the source data. So if there might be a peak like
The interpolation via image is also useful. Maybe I'll incorporate that
> Hi Rainer,
>
>> I tend to use Gaussian equation, which works well so far. But it seems
>> that sometimes, I get undesired results, like the one from the
>> attached macro, where -995468 as curve point is calculated.
>
> After analyzing your case, I am quite confident that this is not a bug.
> The reason for the "strange" fit result is the following:
> Your data set is essentially a flat curve with a few outliers at
> significantly more negative values than the rest. There are many
> equivalent solutions for a best fit, all of them are very narrow
> Gaussians that are essentially constant everywhere except for one
> outlier point, which is fitted very well by the Gaussian.
> With just one outlier point defining the Gaussian, the curve fitter can
> put the peak of the Gaussian essentially anywhere in between the
> surrounding data points; the peak does not necessarily have to be
> exactly at the outlier point. If the peak position is a bit off the
> outlier point, that point will be somewhere on the slope of the
> Gaussian. If the point happens to rather far from the peak of the
> Gaussian, the Gaussian has to be very high to accurately fit the outlier
> point.
>
> With your data set, the outlier is at 23, and the CurveFitter happens to
> put the peak of the Gaussian at 23.65, which is still far from the
> adjacent points. I have attached a high-resolution plot. It shows that
> the outlier point at 23 is exactly on fitting curve, and it makes no
> difference for the adjacent points whether one shifts the Gaussian a bit
> to the left or right (but this would make a huge difference for the
> Gaussian's height).
>
> By the way, there is also another local solution to your fitting
> problem, which is rather broad peak at c=29.5, with a width of d=14.5,
> but that one is worse than fitting one of the outliers by putting a very
> sharp Gaussian there (sum of residuals squared = 1763.3 vs. 1559.8,
> correlation coefficient 0.043 vs. 0.117).
>
> To avoid such problems, one could think of doing a fit with a 'penalty
> function' for such high values, e.g., something in the sense of maximum
> entropy methods, but this is not implemented in ImageJ and it would be
> rather difficult to do so.
>
https://en.wikipedia.org/wiki/Principle_of_maximum_entropy>
> By the way, the fitting process is based on minimizing the least-squares
> deviation with the Nelder-Mead method, which is rather robust even for
> badly conditioned minimization problems
>
https://en.wikipedia.org/wiki/Nelder%E2%80%93Mead_method> In addition, for more robust fitting, it eliminates up to two parameters
> by linear regression (in case of the Gaussian, the baseline and height
> of the Gaussian, so only two fit parameters remain to be determined by
> the Nelder-Mead method).
> ---
>
> In your case, maybe you could detect whether the r-squared of the
> Gaussian fit is too bad (say, below 0.4), or if the width ('d'
> parameter) of the Gaussian is too narrow, and if so, use a different fit
> function such as a fourth-order polynomial? It won't give you a much
> better fit, but it can't run into the problem of a Gaussian trying to
> catch a single outlier point.
>
> ---
>
> The second of your questions is easier to answer:
>> Just out of curiosity. Is there an equation or function to interpolate
>> only missing positions of a curve, where each given point is kept?
>
> You can do linear interpolation
>
https://en.wikipedia.org/wiki/Linear_interpolation> cubic interpolation, splines, etc.
>
https://en.wikipedia.org/wiki/Spline_(mathematics)
>
> In ImageJ, if you have equidistant points simply create an image with a
> height of one pixel and the data as pixel values, then you can use the
> macro function
> getPixel(x, 0)
> to get the interpolated value (linear by default; setOption("bicubic,
> true) for cubic interpolation.
>
> ImageJ uses splines for 'rounding' segmented line rois, but I am not
> aware of a macro interface for it's built-in SplineFitter. One could
> probably access it via JavaScript.
>
>
> Michael
> ________________________________________________________________
>
>
> On 24/06/2017 16:37, Rainer M. Engel wrote:
>> Hello everyone,
>>
>> I'm working with curve fitting, to smooth data but in the first place to
>> fill invalid positions in a curve.
>>
>> I tend to use Gaussian equation, which works well so far. But it seems
>> that sometimes, I get undesired results, like the one from the attached
>> macro, where -995468 as curve point is calculated.
>>
>> Is this an error/bug?
>>
>> If this is correct, what would you suggest to avoid/remove such
>> deflections?
>>
>> Just out of curiosity. Is there an equation or function to interpolate
>> only missing positions of a curve, where each given point is kept?
>>
>> Any help/hint is much appreciated.
>>
>> Kind regards,
>> Rainer
>>
>>
>
> --
> ImageJ mailing list:
http://imagej.nih.gov/ij/list.htmlRainer M. Engel, Dipl. Digital Artist