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ListPlot join points by nearest neighbor rather than order
FindCurvePath for lines (rather than points)How to deal with overlapping data points in ListPLotShow[List of plot] how to color them differently?Blank ListPlot with a tableListLinePlot not plotting full dataGetting the error out of a prediction and a measurementMore resolution on ListContourPlot?ListLinePlot/ListPlot: How to center the ticks on the steps in the plot?
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$begingroup$
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.
plotting order
edited Apr 15 at 21:48
Carl Woll
90.2k3 gold badges117 silver badges229 bronze badges
asked Apr 15 at 13:44
zackzack
1257 bronze badges
$endgroup$
|
show 1 more comment
$begingroup$
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.
plotting order
edited Apr 15 at 21:48
Carl Woll
90.2k3 gold badges117 silver badges229 bronze badges
asked Apr 15 at 13:44
zackzack
1257 bronze badges
$endgroup$
4
$begingroup$
TryFindShortestTour
$endgroup$
– C. E.
Apr 15 at 13:48
2
$begingroup$
Try something likeListLinePlot[data[[Last@FindShortestTour@data]]]but it is not perfect
$endgroup$
– J42161217
Apr 15 at 14:39
2
$begingroup$
Will your data always be in convex hulls?
$endgroup$
– MikeY
Apr 15 at 18:22
$begingroup$
@MikeY unfortunately not! They are the results of a Bayesian analysis and many have differing, strange, and non-analytical forms.
$endgroup$
– zack
Apr 15 at 21:04
$begingroup$
Related: (136181)
$endgroup$
– Mr.Wizard♦
Apr 15 at 22:33
|
show 1 more comment
$begingroup$
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.
plotting order
edited Apr 15 at 21:48
Carl Woll
90.2k3 gold badges117 silver badges229 bronze badges
asked Apr 15 at 13:44
zackzack
1257 bronze badges
$endgroup$
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.
plotting order
plotting order
edited Apr 15 at 21:48
Carl Woll
90.2k3 gold badges117 silver badges229 bronze badges
asked Apr 15 at 13:44
zackzack
1257 bronze badges
edited Apr 15 at 21:48
Carl Woll
90.2k3 gold badges117 silver badges229 bronze badges
asked Apr 15 at 13:44
zackzack
1257 bronze badges
edited Apr 15 at 21:48
Carl Woll
90.2k3 gold badges117 silver badges229 bronze badges
edited Apr 15 at 21:48
Carl Woll
90.2k3 gold badges117 silver badges229 bronze badges
edited Apr 15 at 21:48
Carl Woll
90.2k3 gold badges117 silver badges229 bronze badges
90.2k3 gold badges117 silver badges229 bronze badges
asked Apr 15 at 13:44
zackzack
1257 bronze badges
asked Apr 15 at 13:44
zackzack
1257 bronze badges
asked Apr 15 at 13:44
zackzack
1257 bronze badges
1257 bronze badges
4
$begingroup$
TryFindShortestTour
$endgroup$
– C. E.
Apr 15 at 13:48
2
$begingroup$
Try something likeListLinePlot[data[[Last@FindShortestTour@data]]]but it is not perfect
$endgroup$
– J42161217
Apr 15 at 14:39
2
$begingroup$
Will your data always be in convex hulls?
$endgroup$
– MikeY
Apr 15 at 18:22
$begingroup$
@MikeY unfortunately not! They are the results of a Bayesian analysis and many have differing, strange, and non-analytical forms.
$endgroup$
– zack
Apr 15 at 21:04
$begingroup$
Related: (136181)
$endgroup$
– Mr.Wizard♦
Apr 15 at 22:33
|
show 1 more comment
4
$begingroup$
TryFindShortestTour
$endgroup$
– C. E.
Apr 15 at 13:48
2
$begingroup$
Try something likeListLinePlot[data[[Last@FindShortestTour@data]]]but it is not perfect
$endgroup$
– J42161217
Apr 15 at 14:39
2
$begingroup$
Will your data always be in convex hulls?
$endgroup$
– MikeY
Apr 15 at 18:22
$begingroup$
@MikeY unfortunately not! They are the results of a Bayesian analysis and many have differing, strange, and non-analytical forms.
$endgroup$
– zack
Apr 15 at 21:04
$begingroup$
Related: (136181)
$endgroup$
– Mr.Wizard♦
Apr 15 at 22:33
4
4
$begingroup$
Try
FindShortestTour$endgroup$
– C. E.
Apr 15 at 13:48
$begingroup$
Try
FindShortestTour$endgroup$
– C. E.
Apr 15 at 13:48
2
2
$begingroup$
Try something like
ListLinePlot[data[[Last@FindShortestTour@data]]] but it is not perfect$endgroup$
– J42161217
Apr 15 at 14:39
$begingroup$
Try something like
ListLinePlot[data[[Last@FindShortestTour@data]]] but it is not perfect$endgroup$
– J42161217
Apr 15 at 14:39
2
2
$begingroup$
Will your data always be in convex hulls?
$endgroup$
– MikeY
Apr 15 at 18:22
$begingroup$
Will your data always be in convex hulls?
$endgroup$
– MikeY
Apr 15 at 18:22
$begingroup$
@MikeY unfortunately not! They are the results of a Bayesian analysis and many have differing, strange, and non-analytical forms.
$endgroup$
– zack
Apr 15 at 21:04
$begingroup$
@MikeY unfortunately not! They are the results of a Bayesian analysis and many have differing, strange, and non-analytical forms.
$endgroup$
– zack
Apr 15 at 21:04
$begingroup$
Related: (136181)
$endgroup$
– Mr.Wizard♦
Apr 15 at 22:33
$begingroup$
Related: (136181)
$endgroup$
– Mr.Wizard♦
Apr 15 at 22:33
|
show 1 more comment
4 Answers
4
active
oldest
votes
$begingroup$
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.
answered Apr 15 at 15:17
Carl WollCarl Woll
90.2k3 gold badges117 silver badges229 bronze badges
$endgroup$
1
$begingroup$
Why not just the closely relatedListCurvePathPlot?
$endgroup$
– Roman
Apr 15 at 15:32
1
$begingroup$
@Roman Did you try usingListCurvePathPlot? Because the data has such a small variation in theycoordinate,ListCurvePathPlotdoesn't work well. That's why I scaled the data and usedFindCurvePathto reorder the data, and then plotted the reordered data.
$endgroup$
– Carl Woll
Apr 15 at 16:25
1
$begingroup$
Ah yes, brilliant! Maybe even easier for automation would be a hands-free rescaling by the covariance matrix of the data, something likepath = First[FindCurvePath[data.(Transpose[#[[2]]]/Sqrt[#[[1]]] &@ Eigensystem[Covariance[data]])]], which tries to map the given data onto a unit circle before applyingFindCurvePath. What do you think?
$endgroup$
– Roman
Apr 15 at 17:34
1
$begingroup$
@Roman Adding automatic rescaling is a good idea. I added a simple version based onRescalingTransform. You can add an answer usingEigensystem/Covarianceif you want.
$endgroup$
– Carl Woll
Apr 15 at 17:58
$begingroup$
Thank you very much for your multiple solutions @CarlWoll! These worked perfectly for all my datasets other than the ones with kinks, those of which I can manually edit.
$endgroup$
– zack
Apr 15 at 21:03
add a comment
|
$begingroup$
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"]]
edited Apr 15 at 20:37
Bob Hanlon
65.7k3 gold badges37 silver badges100 bronze badges
answered Apr 15 at 18:01
Chip HurstChip Hurst
25.5k1 gold badge61 silver badges100 bronze badges
$endgroup$
add a comment
|
$begingroup$
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]
answered Apr 15 at 18:25
RomanRoman
15.9k1 gold badge21 silver badges54 bronze badges
$endgroup$
1
$begingroup$
You're missing the plot command for your first image and the command shown for it should be with the second image.
$endgroup$
– Bob Hanlon
Apr 15 at 20:35
1
$begingroup$
Thanks @BobHanlon , for some reason the formatting got scrambled when I added the second image.
$endgroup$
– Roman
Apr 15 at 20:50
$begingroup$
Thank you for this solution @Roman! It also works excellently.
$endgroup$
– zack
Apr 15 at 21:05
add a comment
|
$begingroup$
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]]
answered Apr 15 at 23:08
Daniel LichtblauDaniel Lichtblau
48.3k2 gold badges80 silver badges167 bronze badges
$endgroup$
add a comment
|
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4 Answers
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active
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4 Answers
4
active
oldest
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votes
active
oldest
votes
$begingroup$
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.
answered Apr 15 at 15:17
Carl WollCarl Woll
90.2k3 gold badges117 silver badges229 bronze badges
$endgroup$
1
$begingroup$
Why not just the closely relatedListCurvePathPlot?
$endgroup$
– Roman
Apr 15 at 15:32
1
$begingroup$
@Roman Did you try usingListCurvePathPlot? Because the data has such a small variation in theycoordinate,ListCurvePathPlotdoesn't work well. That's why I scaled the data and usedFindCurvePathto reorder the data, and then plotted the reordered data.
$endgroup$
– Carl Woll
Apr 15 at 16:25
1
$begingroup$
Ah yes, brilliant! Maybe even easier for automation would be a hands-free rescaling by the covariance matrix of the data, something likepath = First[FindCurvePath[data.(Transpose[#[[2]]]/Sqrt[#[[1]]] &@ Eigensystem[Covariance[data]])]], which tries to map the given data onto a unit circle before applyingFindCurvePath. What do you think?
$endgroup$
– Roman
Apr 15 at 17:34
1
$begingroup$
@Roman Adding automatic rescaling is a good idea. I added a simple version based onRescalingTransform. You can add an answer usingEigensystem/Covarianceif you want.
$endgroup$
– Carl Woll
Apr 15 at 17:58
$begingroup$
Thank you very much for your multiple solutions @CarlWoll! These worked perfectly for all my datasets other than the ones with kinks, those of which I can manually edit.
$endgroup$
– zack
Apr 15 at 21:03
add a comment
|
$begingroup$
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.
answered Apr 15 at 15:17
Carl WollCarl Woll
90.2k3 gold badges117 silver badges229 bronze badges
$endgroup$
1
$begingroup$
Why not just the closely relatedListCurvePathPlot?
$endgroup$
– Roman
Apr 15 at 15:32
1
$begingroup$
@Roman Did you try usingListCurvePathPlot? Because the data has such a small variation in theycoordinate,ListCurvePathPlotdoesn't work well. That's why I scaled the data and usedFindCurvePathto reorder the data, and then plotted the reordered data.
$endgroup$
– Carl Woll
Apr 15 at 16:25
1
$begingroup$
Ah yes, brilliant! Maybe even easier for automation would be a hands-free rescaling by the covariance matrix of the data, something likepath = First[FindCurvePath[data.(Transpose[#[[2]]]/Sqrt[#[[1]]] &@ Eigensystem[Covariance[data]])]], which tries to map the given data onto a unit circle before applyingFindCurvePath. What do you think?
$endgroup$
– Roman
Apr 15 at 17:34
1
$begingroup$
@Roman Adding automatic rescaling is a good idea. I added a simple version based onRescalingTransform. You can add an answer usingEigensystem/Covarianceif you want.
$endgroup$
– Carl Woll
Apr 15 at 17:58
$begingroup$
Thank you very much for your multiple solutions @CarlWoll! These worked perfectly for all my datasets other than the ones with kinks, those of which I can manually edit.
$endgroup$
– zack
Apr 15 at 21:03
add a comment
|
$begingroup$
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.
answered Apr 15 at 15:17
Carl WollCarl Woll
90.2k3 gold badges117 silver badges229 bronze badges
$endgroup$
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.
answered Apr 15 at 15:17
Carl WollCarl Woll
90.2k3 gold badges117 silver badges229 bronze badges
edited Apr 15 at 17:57
answered Apr 15 at 15:17
Carl WollCarl Woll
90.2k3 gold badges117 silver badges229 bronze badges
answered Apr 15 at 15:17
Carl WollCarl Woll
90.2k3 gold badges117 silver badges229 bronze badges
answered Apr 15 at 15:17
Carl WollCarl Woll
90.2k3 gold badges117 silver badges229 bronze badges
90.2k3 gold badges117 silver badges229 bronze badges
1
$begingroup$
Why not just the closely relatedListCurvePathPlot?
$endgroup$
– Roman
Apr 15 at 15:32
1
$begingroup$
@Roman Did you try usingListCurvePathPlot? Because the data has such a small variation in theycoordinate,ListCurvePathPlotdoesn't work well. That's why I scaled the data and usedFindCurvePathto reorder the data, and then plotted the reordered data.
$endgroup$
– Carl Woll
Apr 15 at 16:25
1
$begingroup$
Ah yes, brilliant! Maybe even easier for automation would be a hands-free rescaling by the covariance matrix of the data, something likepath = First[FindCurvePath[data.(Transpose[#[[2]]]/Sqrt[#[[1]]] &@ Eigensystem[Covariance[data]])]], which tries to map the given data onto a unit circle before applyingFindCurvePath. What do you think?
$endgroup$
– Roman
Apr 15 at 17:34
1
$begingroup$
@Roman Adding automatic rescaling is a good idea. I added a simple version based onRescalingTransform. You can add an answer usingEigensystem/Covarianceif you want.
$endgroup$
– Carl Woll
Apr 15 at 17:58
$begingroup$
Thank you very much for your multiple solutions @CarlWoll! These worked perfectly for all my datasets other than the ones with kinks, those of which I can manually edit.
$endgroup$
– zack
Apr 15 at 21:03
add a comment
|
1
$begingroup$
Why not just the closely relatedListCurvePathPlot?
$endgroup$
– Roman
Apr 15 at 15:32
1
$begingroup$
@Roman Did you try usingListCurvePathPlot? Because the data has such a small variation in theycoordinate,ListCurvePathPlotdoesn't work well. That's why I scaled the data and usedFindCurvePathto reorder the data, and then plotted the reordered data.
$endgroup$
– Carl Woll
Apr 15 at 16:25
1
$begingroup$
Ah yes, brilliant! Maybe even easier for automation would be a hands-free rescaling by the covariance matrix of the data, something likepath = First[FindCurvePath[data.(Transpose[#[[2]]]/Sqrt[#[[1]]] &@ Eigensystem[Covariance[data]])]], which tries to map the given data onto a unit circle before applyingFindCurvePath. What do you think?
$endgroup$
– Roman
Apr 15 at 17:34
1
$begingroup$
@Roman Adding automatic rescaling is a good idea. I added a simple version based onRescalingTransform. You can add an answer usingEigensystem/Covarianceif you want.
$endgroup$
– Carl Woll
Apr 15 at 17:58
$begingroup$
Thank you very much for your multiple solutions @CarlWoll! These worked perfectly for all my datasets other than the ones with kinks, those of which I can manually edit.
$endgroup$
– zack
Apr 15 at 21:03
1
1
$begingroup$
Why not just the closely related
ListCurvePathPlot?$endgroup$
– Roman
Apr 15 at 15:32
$begingroup$
Why not just the closely related
ListCurvePathPlot?$endgroup$
– Roman
Apr 15 at 15:32
1
1
$begingroup$
@Roman Did you try using
ListCurvePathPlot? Because the data has such a small variation in the y coordinate, ListCurvePathPlot doesn't work well. That's why I scaled the data and used FindCurvePath to reorder the data, and then plotted the reordered data.$endgroup$
– Carl Woll
Apr 15 at 16:25
$begingroup$
@Roman Did you try using
ListCurvePathPlot? Because the data has such a small variation in the y coordinate, ListCurvePathPlot doesn't work well. That's why I scaled the data and used FindCurvePath to reorder the data, and then plotted the reordered data.$endgroup$
– Carl Woll
Apr 15 at 16:25
1
1
$begingroup$
Ah yes, brilliant! Maybe even easier for automation would be a hands-free rescaling by the covariance matrix of the data, something like
path = First[FindCurvePath[data.(Transpose[#[[2]]]/Sqrt[#[[1]]] &@ Eigensystem[Covariance[data]])]], which tries to map the given data onto a unit circle before applying FindCurvePath. What do you think?$endgroup$
– Roman
Apr 15 at 17:34
$begingroup$
Ah yes, brilliant! Maybe even easier for automation would be a hands-free rescaling by the covariance matrix of the data, something like
path = First[FindCurvePath[data.(Transpose[#[[2]]]/Sqrt[#[[1]]] &@ Eigensystem[Covariance[data]])]], which tries to map the given data onto a unit circle before applying FindCurvePath. What do you think?$endgroup$
– Roman
Apr 15 at 17:34
1
1
$begingroup$
@Roman Adding automatic rescaling is a good idea. I added a simple version based on
RescalingTransform. You can add an answer using Eigensystem/Covariance if you want.$endgroup$
– Carl Woll
Apr 15 at 17:58
$begingroup$
@Roman Adding automatic rescaling is a good idea. I added a simple version based on
RescalingTransform. You can add an answer using Eigensystem/Covariance if you want.$endgroup$
– Carl Woll
Apr 15 at 17:58
$begingroup$
Thank you very much for your multiple solutions @CarlWoll! These worked perfectly for all my datasets other than the ones with kinks, those of which I can manually edit.
$endgroup$
– zack
Apr 15 at 21:03
$begingroup$
Thank you very much for your multiple solutions @CarlWoll! These worked perfectly for all my datasets other than the ones with kinks, those of which I can manually edit.
$endgroup$
– zack
Apr 15 at 21:03
add a comment
|
$begingroup$
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"]]
edited Apr 15 at 20:37
Bob Hanlon
65.7k3 gold badges37 silver badges100 bronze badges
answered Apr 15 at 18:01
Chip HurstChip Hurst
25.5k1 gold badge61 silver badges100 bronze badges
$endgroup$
add a comment
|
$begingroup$
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"]]
edited Apr 15 at 20:37
Bob Hanlon
65.7k3 gold badges37 silver badges100 bronze badges
answered Apr 15 at 18:01
Chip HurstChip Hurst
25.5k1 gold badge61 silver badges100 bronze badges
$endgroup$
add a comment
|
$begingroup$
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"]]
edited Apr 15 at 20:37
Bob Hanlon
65.7k3 gold badges37 silver badges100 bronze badges
answered Apr 15 at 18:01
Chip HurstChip Hurst
25.5k1 gold badge61 silver badges100 bronze badges
$endgroup$
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"]]
edited Apr 15 at 20:37
Bob Hanlon
65.7k3 gold badges37 silver badges100 bronze badges
answered Apr 15 at 18:01
Chip HurstChip Hurst
25.5k1 gold badge61 silver badges100 bronze badges
edited Apr 15 at 20:37
Bob Hanlon
65.7k3 gold badges37 silver badges100 bronze badges
edited Apr 15 at 20:37
Bob Hanlon
65.7k3 gold badges37 silver badges100 bronze badges
edited Apr 15 at 20:37
Bob Hanlon
65.7k3 gold badges37 silver badges100 bronze badges
65.7k3 gold badges37 silver badges100 bronze badges
answered Apr 15 at 18:01
Chip HurstChip Hurst
25.5k1 gold badge61 silver badges100 bronze badges
answered Apr 15 at 18:01
Chip HurstChip Hurst
25.5k1 gold badge61 silver badges100 bronze badges
answered Apr 15 at 18:01
Chip HurstChip Hurst
25.5k1 gold badge61 silver badges100 bronze badges
25.5k1 gold badge61 silver badges100 bronze badges
add a comment
|
add a comment
|
$begingroup$
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]
answered Apr 15 at 18:25
RomanRoman
15.9k1 gold badge21 silver badges54 bronze badges
$endgroup$
1
$begingroup$
You're missing the plot command for your first image and the command shown for it should be with the second image.
$endgroup$
– Bob Hanlon
Apr 15 at 20:35
1
$begingroup$
Thanks @BobHanlon , for some reason the formatting got scrambled when I added the second image.
$endgroup$
– Roman
Apr 15 at 20:50
$begingroup$
Thank you for this solution @Roman! It also works excellently.
$endgroup$
– zack
Apr 15 at 21:05
add a comment
|
$begingroup$
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]
answered Apr 15 at 18:25
RomanRoman
15.9k1 gold badge21 silver badges54 bronze badges
$endgroup$
1
$begingroup$
You're missing the plot command for your first image and the command shown for it should be with the second image.
$endgroup$
– Bob Hanlon
Apr 15 at 20:35
1
$begingroup$
Thanks @BobHanlon , for some reason the formatting got scrambled when I added the second image.
$endgroup$
– Roman
Apr 15 at 20:50
$begingroup$
Thank you for this solution @Roman! It also works excellently.
$endgroup$
– zack
Apr 15 at 21:05
add a comment
|
$begingroup$
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]
answered Apr 15 at 18:25
RomanRoman
15.9k1 gold badge21 silver badges54 bronze badges
$endgroup$
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]
answered Apr 15 at 18:25
RomanRoman
15.9k1 gold badge21 silver badges54 bronze badges
edited Apr 15 at 20:57
answered Apr 15 at 18:25
RomanRoman
15.9k1 gold badge21 silver badges54 bronze badges
answered Apr 15 at 18:25
RomanRoman
15.9k1 gold badge21 silver badges54 bronze badges
answered Apr 15 at 18:25
RomanRoman
15.9k1 gold badge21 silver badges54 bronze badges
15.9k1 gold badge21 silver badges54 bronze badges
1
$begingroup$
You're missing the plot command for your first image and the command shown for it should be with the second image.
$endgroup$
– Bob Hanlon
Apr 15 at 20:35
1
$begingroup$
Thanks @BobHanlon , for some reason the formatting got scrambled when I added the second image.
$endgroup$
– Roman
Apr 15 at 20:50
$begingroup$
Thank you for this solution @Roman! It also works excellently.
$endgroup$
– zack
Apr 15 at 21:05
add a comment
|
1
$begingroup$
You're missing the plot command for your first image and the command shown for it should be with the second image.
$endgroup$
– Bob Hanlon
Apr 15 at 20:35
1
$begingroup$
Thanks @BobHanlon , for some reason the formatting got scrambled when I added the second image.
$endgroup$
– Roman
Apr 15 at 20:50
$begingroup$
Thank you for this solution @Roman! It also works excellently.
$endgroup$
– zack
Apr 15 at 21:05
1
1
$begingroup$
You're missing the plot command for your first image and the command shown for it should be with the second image.
$endgroup$
– Bob Hanlon
Apr 15 at 20:35
$begingroup$
You're missing the plot command for your first image and the command shown for it should be with the second image.
$endgroup$
– Bob Hanlon
Apr 15 at 20:35
1
1
$begingroup$
Thanks @BobHanlon , for some reason the formatting got scrambled when I added the second image.
$endgroup$
– Roman
Apr 15 at 20:50
$begingroup$
Thanks @BobHanlon , for some reason the formatting got scrambled when I added the second image.
$endgroup$
– Roman
Apr 15 at 20:50
$begingroup$
Thank you for this solution @Roman! It also works excellently.
$endgroup$
– zack
Apr 15 at 21:05
$begingroup$
Thank you for this solution @Roman! It also works excellently.
$endgroup$
– zack
Apr 15 at 21:05
add a comment
|
$begingroup$
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]]
answered Apr 15 at 23:08
Daniel LichtblauDaniel Lichtblau
48.3k2 gold badges80 silver badges167 bronze badges
$endgroup$
add a comment
|
$begingroup$
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]]
answered Apr 15 at 23:08
Daniel LichtblauDaniel Lichtblau
48.3k2 gold badges80 silver badges167 bronze badges
$endgroup$
add a comment
|
$begingroup$
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]]
answered Apr 15 at 23:08
Daniel LichtblauDaniel Lichtblau
48.3k2 gold badges80 silver badges167 bronze badges
$endgroup$
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]]
answered Apr 15 at 23:08
Daniel LichtblauDaniel Lichtblau
48.3k2 gold badges80 silver badges167 bronze badges
answered Apr 15 at 23:08
Daniel LichtblauDaniel Lichtblau
48.3k2 gold badges80 silver badges167 bronze badges
answered Apr 15 at 23:08
Daniel LichtblauDaniel Lichtblau
48.3k2 gold badges80 silver badges167 bronze badges
answered Apr 15 at 23:08
Daniel LichtblauDaniel Lichtblau
48.3k2 gold badges80 silver badges167 bronze badges
48.3k2 gold badges80 silver badges167 bronze badges
add a comment
|
add a comment
|
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4
$begingroup$
Try
FindShortestTour$endgroup$
– C. E.
Apr 15 at 13:48
2
$begingroup$
Try something like
ListLinePlot[data[[Last@FindShortestTour@data]]]but it is not perfect$endgroup$
– J42161217
Apr 15 at 14:39
2
$begingroup$
Will your data always be in convex hulls?
$endgroup$
– MikeY
Apr 15 at 18:22
$begingroup$
@MikeY unfortunately not! They are the results of a Bayesian analysis and many have differing, strange, and non-analytical forms.
$endgroup$
– zack
Apr 15 at 21:04
$begingroup$
Related: (136181)
$endgroup$
– Mr.Wizard♦
Apr 15 at 22:33