Bsrgvty

I have found some software that allows me to "data mine" the values from publication figures. I have a bunch of contours from papers that I've mined using this software, and am having some trouble plotting the points with the Joined command.

Unfortunately, the downloaded points are sorted by increasing x values, which makes the plotting of Gaussian-esque contours very difficult. I've searched around the forums and haven't found anyone mentioning this problem.

Here's an example on a very small, simpler distribution (note my other sets are much larger so brute force definitely won't work.)

data=62.0774, 0.598737, 62.2377, 0.619119, 62.4048, 
0.580509, 62.5466, 0.637818, 62.9276, 0.654518, 62.9668, 
 0.566973, 63.3095, 0.671261, 63.8137, 0.688518, 63.8913, 
 0.565805, 64.4067, 0.703821, 64.8157, 0.568541, 65.1005, 
 0.718671, 65.7401, 0.573603, 65.9282, 0.732056, 66.6646, 
 0.580678, 66.7973, 0.743456, 67.6058, 0.589303, 67.7571, 
 0.755602, 68.5512, 0.599853, 68.6815, 0.761419, 69.4, 
 0.614478, 69.6059, 0.76384, 70.1679, 0.631668, 70.5117, 
 0.759937, 70.5514, 0.759266, 70.7216, 0.649606, 71.3609, 
 0.666955, 71.3764, 0.751005, 71.7909, 0.736308, 71.8078, 
 0.687055, 71.947, 0.702022, 72.0491, 0.717738

Using ListPlot gives me this:

ListPlot[data]

While using ListLinePlot gives me this

ListLinePlot[data]

because the points are ordered with increasing x-value.

So, is there any way to either join the points by nearest neighbor, or re-order the list such that the joined command will give me a neat line? This seems like a traveling-salesman type problem, which could quickly get slow as I increase the number of points too much.

You can use FindCurvePath to reorder your data. However, FindCurvePath expects the scale of the two coordinates to be close, so you need to rescale first:

new = FindCurvePath[data . 1, 0, 0, 100]
ListLinePlot[data[[#]]& /@ new]

2, 1, 3, 6, 9, 11, 13, 15, 17, 19, 21, 23, 26, 27, 30, 31, 32, 29,
28, 25, 24, 22, 20, 18, 16, 14, 12, 10, 8, 7, 5, 4, 2

Update

Roman suggested automating the scaling of the data. Here is one possibility for rescaling the data:

rescale = RescalingTransform[CoordinateBounds[data]] @ data;

Then, using FindCurvePath on the rescaled data:

new = FindCurvePath @ rescale

2, 1, 3, 6, 9, 11, 13, 15, 17, 19, 21, 23, 26, 27, 30, 31, 32, 29, 28, 25,
24, 22, 20, 18, 16, 14, 12, 10, 8, 7, 5, 4, 2

produces the same result.

Since your data can form a star convex polygon, we can sort by the angle with respect to a certain point:

center = Mean[data];
ListLinePlot[ArrayPad[SortBy[data, ArcTan @@ (# - center) &], 0, 1, "Periodic"]]

By scaling the data into the covariance ellipsoid, we can achieve hands-free auto-scaling before calculating a FindCurvePath along @CarlWoll 's solution:

path = First@FindCurvePath[
 data.Transpose[#[[2]]/Sqrt[#[[1]]]&@Eigensystem[Covariance[data]]]]

2, 1, 3, 6, 9, 11, 13, 15, 17, 19, 21, 23, 26, 27, 30, 31, 32, 29, 28, 25, 24, 22, 20, 18, 16, 14, 12, 10, 8, 7, 5, 4, 2

ListPlot[data[[path]]]

Alternatively, if the data points are meant to describe a closed loop, the path can be found with

path = Last@FindShortestTour[
 data.Transpose[#[[2]]/Sqrt[#[[1]]]&@Eigensystem[Covariance[data]]]]

1, 2, 4, 5, 7, 8, 10, 12, 14, 16, 18, 20, 22, 24, 25, 28, 29, 32, 31, 30, 27, 26, 23, 21, 19, 17, 15, 13, 11, 9, 6, 3, 1

The transformed data that are fed into FindCurvePath or FindShortestTour have a unit covariance matrix, which makes it easier to find a good path:

Sdata = data.Transpose[#[[2]]/Sqrt[#[[1]]]&@Eigensystem[Covariance[data]]];
Chop@Covariance[Sdata]

1., 0, 0, 1.

We can see that these scaled points nearly lie on a circle:

ListPlot[Sdata, AspectRatio -> Automatic]

Sorta lame, but rescaling and Nearest can be used to get triples, with Line to connect the triples (each has a point and its two closest neighbors which in this case will do what you want).

data2 = Map[1, 100*# &, data];
nf = Nearest[data2];
triples0 = Map[RotateRight, nf[data2, 3]];
triples = Map[Line, Map[1, 1/100*# &, triples0, 2]];

Show[ListPlot[data, ColorFunction -> (Black &)], 
 Graphics[Green, triples]]