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I have data of closed shape form

ListPlot[data, AspectRatio -> 1]

enter image description here

I would like to Interpolate this data to get the missing points so I am using BSplineFunction like this

interdata=BSplineFunction[Flatten[points, 1]];
ParametricPlot[bsF[t], {t, 0, 1}]

enter image description here

as you can see, the data needs to be sorted in a specific way to make all the points on the circumference of the closed shape and get something like this

enter image description here

Update

The last thing I would like to do is to get points on the circumference that are equally spaced. my idea was that after interpolation I will be able to do that. According to @kglr answer, it can be done as follows

equallyspaceddata = Table[{bsF[t]}, {t, 0, 1, 0.01}];
ListPlot[equallyspaceddata, AspectRatio -> 1, PlotStyle -> Black]

enter image description here

as you can see the points in the top half are denser compared to the bottom one. How can I get equally spaced points?

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2 Answers 2

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Update: " to get points on the circumference that are equally spaced":

Use MeshFunctions -> {ArcLength} and Mesh -> m to get m equally-spaced points on the parametric curve:

m = 50;

ParametricPlot[bsF[t], {t, 0, 1}, 
 MeshFunctions -> {ArcLength}, 
 Mesh -> m, 
 MeshStyle -> Directive[PointSize @ Medium, Red], 
 Epilog -> {Directive[PointSize @ Medium, Green], Point @ bsF[0]}]

enter image description here

Original answer:

Use FindCurvePath:

fcp = FindCurvePath[data];

sorteddata = data[[fcp[[1]]]];

ListLinePlot[sorteddata, AspectRatio -> 1]

enter image description here

bsF = BSplineFunction[sorteddata];

ParametricPlot[bsF[t], {t, 0, 1}]

enter image description here

Try also ListCurvePathPlot:

ListCurvePathPlot[data]

enter image description here

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  • $\begingroup$ AWESOME, thanks a lot! $\endgroup$
    – valar morghulis
    Dec 28, 2021 at 9:25
  • $\begingroup$ may you please see my update? $\endgroup$
    – valar morghulis
    Dec 28, 2021 at 10:29
  • $\begingroup$ Thanks, but how can I get the red pint as a list to use them in a calculation? $\endgroup$
    – valar morghulis
    Dec 28, 2021 at 11:55
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    $\begingroup$ @valarmorghulis, one way is to extract the point coordinates from ParametricPlot: e.g,, pp=ParametricPlot[..., MeshFunctions -> {ArcLength}, Mesh ->50]; pointcoords= Cases[Normal@pp, Point[x_] :> x, All] $\endgroup$
    – kglr
    Dec 28, 2021 at 12:24
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data = Import["/Users/roberthanlon/Downloads/data.dat"];

Since you are dealing with a circle:

EDIT 2: Using minimization to determine the center and radius

{center, radius} = {{x0, y0}, r} /. Minimize[{Total[
      (EuclideanDistance[#, {x0, y0}] - r)^2 & /@ data],
     r > 0}, {x0, y0, r}][[2]]

(* {{-4.22506*10^-14, -4.04645*10^-14}, 0.693182} *)

Manipulate[
 Graphics[{
   AbsoluteThickness[1], Red,
   Circle[center, radius], Blue,
   Point[CirclePoints[center, radius, n]]}],
 {{n, 51}, 3, 300, 3, Appearance -> "Labeled"}]

enter image description here

EDIT: Version 13 has a new function RegionFit

{center, radius} = List @@ RegionFit[data, "Circle"]

(* {{-0.00115749, -9.88079*10^-16}, 0.694887} *)
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  • $\begingroup$ that is nice, but it is not always a circle, it can be any closed shape such as a rectangle or triangle. for example, your approach cant work with this data $\endgroup$
    – valar morghulis
    Dec 29, 2021 at 6:27

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