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I'm using ContourPlot to extract data points from a rather non-trivial curve in the $(x,y)$-plane. However, the points are spit out in the order of increasing $x$ value. This curve has many turning points, and I want to order the points in such a way that they correspond to a parameter moving along the length of the curve. I'm wondering if there's a way to reorder these data points in a compatible way for parameterizing with respect to length along the curve.

Since my code is messy and involves, maybe an example would be best. 

data2 = Table[{x, Cos[x]}, {x, -2*Pi, 2*Pi, \[Pi]/16}] // N
dataa = RandomSample[data2]
ListPlot[dataa]

So here we have a small number of data points from a cosine curve, and I've placed the points in a random order. Can one reorder the points in a way that corresponds to moving smoothly from the left to the right along the curve? Clearly in this case, there's a trivial solution where you order the points from smallest to largest $x$ value. In my case, the curve is not a function of $x$, so this won't work! So maybe is there another way, other than this trivial solution? Thanks in advance.

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3 Answers 3

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datab = FindCurvePath[dataa];
dataa = dataa[[Flatten[datab]]];
ListPlot[dataa]

enter image description here

If the list is split, just run 

datab = FindCurvePath[dataa];
dataa = dataa[[Flatten[datab]]];

again, or go to the next step. If it still doesn't work, run this code

datad = Table[
   dataa[[i]] - dataa[[i + 1]], {i, 1, Length[dataa] - 1}];
splitp = Position[datad[[All, 2]], Max[datad[[All, 2]]]][[1]][[1]]
dataa = Catenate[{Reverse[dataa[[1 ;; splitp]]], 
    dataa[[splitp+1 ;; Length[dataa]]]}];

As the code doesn't always produce the same result, you may have to run the entire code more than once, but eventually it sorts. You can run:

Manipulate[ListPlot[dataa[[1 ;; a]],
PlotRange -> {{-2 Pi, 2 Pi}, {-1, 1}}], {a,1, Length[dataa], 1}]

To determine if the list is properly arranged.

If the list is reversed, use Reverse[dataa] to fix.

enter image description here

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  • $\begingroup$ Caveat for @spietro, the smooth curve here is found in two pieces, as can be seen from ListPlot[dataa[[#]] & /@ datab, Joined -> True]. $\endgroup$
    – BoLe
    Commented Aug 7, 2016 at 9:31
  • $\begingroup$ @BoLe This happens occasionally, but not always, it's something that can be fixed, but the exact code depends on the situation. $\endgroup$
    – Feyre
    Commented Aug 7, 2016 at 9:47
  • $\begingroup$ @BoLe Actually, just rerunning fixes this. $\endgroup$
    – Feyre
    Commented Aug 7, 2016 at 9:55
  • $\begingroup$ I'm not so sure. The number of FindCurvePath pieces can't be determined since dataa is in random order. However, the pieces returned by first FCP aren't in any particular order, so I don't think Flatten or Join reduce the disorder really. In addition to that, every two neighboring pieces share an index, so the indexing dataa[[Flatten[datab]]] should run in trouble. Did you try and run the code a couple of times and get a fixed-point FCP result of length one? $\endgroup$
    – BoLe
    Commented Aug 7, 2016 at 10:36
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    $\begingroup$ I have copied your first two blocks of code, placed then in a single Do loop, and iterated 200 times. The result never is entirely sorted. In fact, the Manipulate plot in your answer does not show the points in order either. Watch carefully the far left of the plot, and you will see another point appear there about three-quarters the way through the animation. Use ListLinePlot instead of ListPlot to see this more clearly. $\endgroup$
    – bbgodfrey
    Commented Aug 7, 2016 at 15:24
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Your problem can likely be avoided entirely by using Jens's lovely routine contourRegionPlot:

Compare these results:

p1 = ContourPlot[Im[ArcSin[x + I y]], {x, -2, 2}, {y, -2, 2}, Exclusions -> None];

p2 = contourRegionPlot[Im[ArcSin[x + I y]], {x, -2, 2}, {y, -2, 2}];

GraphicsRow[{p1, p2}]

{pts1, pts2} = Cases[#, {_, _}, {-2}] & /@ {p1, p2};

ListLinePlot /@ {pts1, pts2} // GraphicsColumn

enter image description here

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Perhaps ListCurvePathPlot can help you do what you want?

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  • $\begingroup$ That looks potentially promising. Applying it, it certainly connects the data points (in the proper order) with a smooth curve, but I'm not sure it will reorder the data points in the original list. Or maybe I can just extract data points from the smooth curve it produces? Using something like Union which works for ContourPlot $\endgroup$
    – Benighted
    Commented Aug 7, 2016 at 5:26
  • $\begingroup$ Cases[ListCurvePathPlot[dataa], Line[z_] -> z, Infinity] extracts the curve. Unfortunately, it is in multiple parts that are not necessarily in order. In other words, it still requires some sorting. $\endgroup$
    – bbgodfrey
    Commented Aug 7, 2016 at 15:50

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