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InterpolationOrder -> 2 does not work (error) in Manipulate

Posted 9 years ago

Hi,

Can someone tell me why InterpolationOrder ->2 or 3 etc. does not work in Manipulate?

See file attached.

DateListPlot[ {Sdata2014[[1 ;; Test]] , Hdata2014[[1 ;; Test]]} , Joined -> True , Filling -> {1 -> {{2}, Red}} , PlotStyle -> White , InterpolationOrder -> 2] , {Test, 1, Length[Sdata2014], 1} ]

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POSTED BY: Jos Klaps
3 Replies
Posted 9 years ago

Hi,

InterpolationOrder is indeed an option for DateListPlot, set by default to None. This can be checked by evaluating

Options[DateListPlot]

InterpolationOrder works fine in Manipulate. The error comes from the minimum value given to Test. For instance for an InterpolationOrder of 3, one should set it to 4 at least.

Manipulate[
 DateListPlot[
    {Sdata2014[[1 ;; Test]], Hdata2014[[1 ;; Test]]}, 
    Joined -> True, Filling -> {1 -> {{2}, Red}}, PlotStyle -> White, InterpolationOrder -> 3], 
{Test, 4, Length[Sdata2014], 1}]

This allows the option to have enough data (here from the tables Sdata2014 and Hdata2014) for the interpolation to be done. InterpolationOrder - > n uses a polynomial of degree n, so I think it expects a set of at least n points to make the interpolation, otherwise it returns an error.

If the minimal value to Test in the above example was set to a smaller value, say 1, you would get an error for every value up to 3 (even though you will still be able to manipulate the output in any case).

POSTED BY: Xavier Roy

Hi Xavier,

Thank you very much for your help and professional explanations. This was really what I needed.

Kind regards,......Jos

POSTED BY: Jos Klaps

The documentation for DateListPlot does not list InterpolationOrder as an option.

POSTED BY: Frank Kampas
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