ListPlot join points by nearest neighbor rather than order Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?FindCurvePath for lines (rather than points)How to specify DataRange in a ListPlot with an imported set of dataListPlot will not color markers in accordance with my color functionHow to deal with overlapping data points in ListPLotHow do I make ListPlot join the points in order?Show[List of plot] how to color them differently?Blank ListPlot with a tableListContourPlot not plotting all data pointsListLinePlot not plotting full dataGetting the error out of a prediction and a measurementMore resolution on ListContourPlot?
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ListPlot join points by nearest neighbor rather than order
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?FindCurvePath for lines (rather than points)How to specify DataRange in a ListPlot with an imported set of dataListPlot will not color markers in accordance with my color functionHow to deal with overlapping data points in ListPLotHow do I make ListPlot join the points in order?Show[List of plot] how to color them differently?Blank ListPlot with a tableListContourPlot not plotting all data pointsListLinePlot not plotting full dataGetting the error out of a prediction and a measurementMore resolution on ListContourPlot?
$begingroup$
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
$endgroup$
|
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
$endgroup$
4
$begingroup$
TryFindShortestTour
$endgroup$
– C. E.
2 days ago
2
$begingroup$
Try something likeListLinePlot[data[[Last@FindShortestTour@data]]]but it is not perfect
$endgroup$
– J42161217
2 days ago
2
$begingroup$
Will your data always be in convex hulls?
$endgroup$
– MikeY
2 days ago
$begingroup$
@MikeY unfortunately not! They are the results of a Bayesian analysis and many have differing, strange, and non-analytical forms.
$endgroup$
– zack
2 days ago
$begingroup$
Related: (136181)
$endgroup$
– Mr.Wizard♦
2 days ago
|
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
$endgroup$
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 2 days ago
Carl Woll
74.5k399194
74.5k399194
asked 2 days ago
zackzack
1036
1036
4
$begingroup$
TryFindShortestTour
$endgroup$
– C. E.
2 days ago
2
$begingroup$
Try something likeListLinePlot[data[[Last@FindShortestTour@data]]]but it is not perfect
$endgroup$
– J42161217
2 days ago
2
$begingroup$
Will your data always be in convex hulls?
$endgroup$
– MikeY
2 days ago
$begingroup$
@MikeY unfortunately not! They are the results of a Bayesian analysis and many have differing, strange, and non-analytical forms.
$endgroup$
– zack
2 days ago
$begingroup$
Related: (136181)
$endgroup$
– Mr.Wizard♦
2 days ago
|
show 1 more comment
4
$begingroup$
TryFindShortestTour
$endgroup$
– C. E.
2 days ago
2
$begingroup$
Try something likeListLinePlot[data[[Last@FindShortestTour@data]]]but it is not perfect
$endgroup$
– J42161217
2 days ago
2
$begingroup$
Will your data always be in convex hulls?
$endgroup$
– MikeY
2 days ago
$begingroup$
@MikeY unfortunately not! They are the results of a Bayesian analysis and many have differing, strange, and non-analytical forms.
$endgroup$
– zack
2 days ago
$begingroup$
Related: (136181)
$endgroup$
– Mr.Wizard♦
2 days ago
4
4
$begingroup$
Try
FindShortestTour$endgroup$
– C. E.
2 days ago
$begingroup$
Try
FindShortestTour$endgroup$
– C. E.
2 days ago
2
2
$begingroup$
Try something like
ListLinePlot[data[[Last@FindShortestTour@data]]] but it is not perfect$endgroup$
– J42161217
2 days ago
$begingroup$
Try something like
ListLinePlot[data[[Last@FindShortestTour@data]]] but it is not perfect$endgroup$
– J42161217
2 days ago
2
2
$begingroup$
Will your data always be in convex hulls?
$endgroup$
– MikeY
2 days ago
$begingroup$
Will your data always be in convex hulls?
$endgroup$
– MikeY
2 days ago
$begingroup$
@MikeY unfortunately not! They are the results of a Bayesian analysis and many have differing, strange, and non-analytical forms.
$endgroup$
– zack
2 days ago
$begingroup$
@MikeY unfortunately not! They are the results of a Bayesian analysis and many have differing, strange, and non-analytical forms.
$endgroup$
– zack
2 days ago
$begingroup$
Related: (136181)
$endgroup$
– Mr.Wizard♦
2 days ago
$begingroup$
Related: (136181)
$endgroup$
– Mr.Wizard♦
2 days ago
|
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.
$endgroup$
1
$begingroup$
Why not just the closely relatedListCurvePathPlot?
$endgroup$
– Roman
2 days ago
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
2 days ago
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
2 days ago
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
2 days ago
$begingroup$
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
2 days ago
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"]]

$endgroup$
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]

$endgroup$
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
2 days ago
1
$begingroup$
Thanks @BobHanlon , for some reason the formatting got scrambled when I added the second image.
$endgroup$
– Roman
2 days ago
$begingroup$
Thank you for this solution @Roman! It also works excellently.
$endgroup$
– zack
2 days ago
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]]

$endgroup$
add a comment |
Your Answer
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
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.
$endgroup$
1
$begingroup$
Why not just the closely relatedListCurvePathPlot?
$endgroup$
– Roman
2 days ago
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
2 days ago
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
2 days ago
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
2 days ago
$begingroup$
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
2 days ago
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.
$endgroup$
1
$begingroup$
Why not just the closely relatedListCurvePathPlot?
$endgroup$
– Roman
2 days ago
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
2 days ago
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
2 days ago
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
2 days ago
$begingroup$
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
2 days ago
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.
$endgroup$
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.
edited 2 days ago
answered 2 days ago
Carl WollCarl Woll
74.5k399194
74.5k399194
1
$begingroup$
Why not just the closely relatedListCurvePathPlot?
$endgroup$
– Roman
2 days ago
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
2 days ago
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
2 days ago
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
2 days ago
$begingroup$
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
2 days ago
add a comment |
1
$begingroup$
Why not just the closely relatedListCurvePathPlot?
$endgroup$
– Roman
2 days ago
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
2 days ago
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
2 days ago
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
2 days ago
$begingroup$
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
2 days ago
1
1
$begingroup$
Why not just the closely related
ListCurvePathPlot?$endgroup$
– Roman
2 days ago
$begingroup$
Why not just the closely related
ListCurvePathPlot?$endgroup$
– Roman
2 days ago
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
2 days ago
$begingroup$
@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
2 days ago
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
2 days ago
$begingroup$
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
2 days ago
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
2 days ago
$begingroup$
@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
2 days ago
$begingroup$
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
2 days ago
$begingroup$
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
2 days ago
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"]]

$endgroup$
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"]]

$endgroup$
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"]]

$endgroup$
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 2 days ago
Bob Hanlon
61.7k33598
61.7k33598
answered 2 days ago
Chip HurstChip Hurst
23.5k15995
23.5k15995
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]

$endgroup$
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
2 days ago
1
$begingroup$
Thanks @BobHanlon , for some reason the formatting got scrambled when I added the second image.
$endgroup$
– Roman
2 days ago
$begingroup$
Thank you for this solution @Roman! It also works excellently.
$endgroup$
– zack
2 days ago
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]

$endgroup$
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
2 days ago
1
$begingroup$
Thanks @BobHanlon , for some reason the formatting got scrambled when I added the second image.
$endgroup$
– Roman
2 days ago
$begingroup$
Thank you for this solution @Roman! It also works excellently.
$endgroup$
– zack
2 days ago
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]

$endgroup$
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]

edited 2 days ago
answered 2 days ago
RomanRoman
5,56111131
5,56111131
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
2 days ago
1
$begingroup$
Thanks @BobHanlon , for some reason the formatting got scrambled when I added the second image.
$endgroup$
– Roman
2 days ago
$begingroup$
Thank you for this solution @Roman! It also works excellently.
$endgroup$
– zack
2 days ago
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
2 days ago
1
$begingroup$
Thanks @BobHanlon , for some reason the formatting got scrambled when I added the second image.
$endgroup$
– Roman
2 days ago
$begingroup$
Thank you for this solution @Roman! It also works excellently.
$endgroup$
– zack
2 days ago
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
2 days ago
$begingroup$
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
2 days ago
1
1
$begingroup$
Thanks @BobHanlon , for some reason the formatting got scrambled when I added the second image.
$endgroup$
– Roman
2 days ago
$begingroup$
Thanks @BobHanlon , for some reason the formatting got scrambled when I added the second image.
$endgroup$
– Roman
2 days ago
$begingroup$
Thank you for this solution @Roman! It also works excellently.
$endgroup$
– zack
2 days ago
$begingroup$
Thank you for this solution @Roman! It also works excellently.
$endgroup$
– zack
2 days ago
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]]

$endgroup$
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]]

$endgroup$
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]]

$endgroup$
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 2 days ago
Daniel LichtblauDaniel Lichtblau
47.6k277165
47.6k277165
add a comment |
add a comment |
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4
$begingroup$
Try
FindShortestTour$endgroup$
– C. E.
2 days ago
2
$begingroup$
Try something like
ListLinePlot[data[[Last@FindShortestTour@data]]]but it is not perfect$endgroup$
– J42161217
2 days ago
2
$begingroup$
Will your data always be in convex hulls?
$endgroup$
– MikeY
2 days ago
$begingroup$
@MikeY unfortunately not! They are the results of a Bayesian analysis and many have differing, strange, and non-analytical forms.
$endgroup$
– zack
2 days ago
$begingroup$
Related: (136181)
$endgroup$
– Mr.Wizard♦
2 days ago