15
$\begingroup$

This seems like a simple problem, but I can't find any questions like it here so I'm making a new one. Apologies if I missed one.

I have a list of points contained in a file which can be visualized like so:

data = Import["http://www.sharecsv.com/dl/782ae64cbc8003d67f4ab98d784015d6/data.csv"];
{a, b} = {data[[1 ;; 335]], data[[336 ;; -1]]};
ListPointPlot3D[{a, b}, PlotStyle -> {Red, Blue}]

enter image description here

The problem is connecting/interpolating these points into a surface, ListSurfacePlot3D[] yields a garbled mess for any MaxPlotPoints.

ListSurfacePlot3D[data]

ListSurfacePlot3D of data.csv

How can I transform this set of points into a (smooth) surface?

Update:

Alright fellas I'm nearly there. The best way I have come up with is to use DelaunayMesh with RunnyKine's code to produce concave meshes here. First I split the data points into non-overlapping sections by a plane between them. This is a naff hack but what can you do. Then I construct a 2D mesh for the top and bottom halves, replacing the third coordinate with Dispatch.

alphaShapes2D[points_,crit_]:=Module[{alphacriteria,del=Quiet@DelaunayMesh@points,tetras,tetcoords,tetradii,selectExternalFaces},alphacriteria[tetrahedra_,radii_,rmax_]:=Pick[tetrahedra,UnitStep@Subtract[rmax,radii],1];
selectExternalFaces[facets_]:=MeshRegion[points,facets];
If[Head[del]===EmptyRegion,del,tetras=MeshCells[del,2];
tetcoords=MeshPrimitives[del,2][[All,1]];
tetradii=Quiet@Thread[Circumsphere[tetcoords]][[All,2]];
selectExternalFaces@alphacriteria[tetras,tetradii,crit]]];

data=Standardize@Import["http://www.sharecsv.com/dl/782ae64cbc8003d67f4ab98d784015d6/data.csv"];
{a,b}={data[[1;;335]],data[[336;;-1]]};
{bot, top} = {True, False} /. GroupBy[data, Last@# < .25 First@#&];
bottri=MeshPrimitives[alphaShapes2D[bot[[All,{1,2}]],.1],2][[All,1]]/.Dispatch[#[[{1,2}]]->#&/@bot];
toptri=MeshPrimitives[alphaShapes2D[top[[All,{1,2}]],.1],2][[All,1]]/.Dispatch[#[[{1,2}]]->#&/@top];
out=Graphics3D[{Polygon/@bottri,Polygon/@toptri}]

Joining DelaunayMesh objects

Now I'm just missing a way to replace the missing polygons. It would have been much better to cut the data along vertices belonging to both sets. Is there a simple way you can envisage doing this?

$\endgroup$
8
  • 1
    $\begingroup$ Thank you very much for providing the points at the outset; a lot of people don't do that. If I may, where'd these points come from? $\endgroup$ Jun 6, 2016 at 7:47
  • $\begingroup$ Some very messy and inelegant code. I'm looking at fixed points in random 2D vector fields. Here the data[[All,3]] axis is one coordinate of a particular fixed point and the data[[All,{1,2}]] axes map out the domain that fixed point exists. I think Mathematica is getting tripped up by the bifurcation resulting in a fold in the surface. $\endgroup$
    – Crêpo
    Jun 6, 2016 at 7:55
  • 1
    $\begingroup$ do you have access to the algorithm that generated the data, or just the raw points? It looks like its got a regular grid structure to it and if you had that grid info it would be useful. $\endgroup$
    – george2079
    Jun 6, 2016 at 12:15
  • $\begingroup$ I generated the data by first creating the boundary points, and then sampling uniformly in the data[[All,{1,2}]] plane. In principle I can generate as many points in whatever arrangement I want, although it doesn't seem to improve the result. $\endgroup$
    – Crêpo
    Jun 6, 2016 at 12:34
  • $\begingroup$ I added a figure from a different viewpoint to show this is a single simple smooth surface. $\endgroup$
    – george2079
    Jun 6, 2016 at 14:15

2 Answers 2

7
$\begingroup$

I think your best bet is to generate a triangualted polygon surface. Here is an initial stab at it:

p2d = b[[All, 1 ;; 2]];
tri1[i_, j_] := 
 If[EvenQ[j], {{i, j}, {i + 1, j}, {i + 1/2, j + 1}} .05, {{i + 1/2, 
     j}, {i + 1 + 1/2, j}, {i + 1, j + 1}} .05]
tri2[i_, j_] := 
 If[EvenQ[j], {{i, j}, {i + 1, j}, {i + 1/2, j - 1}} .05, {{i + 1/2, 
     j}, {i + 1 + 1/2, j}, {i + 1, j - 1}} .05]
triangles = 
  Flatten[Table[ tri2[i, j], {i, -45, 35}, {j, -55, 20}], 1]~Join~
   Flatten[Table[ tri1[i, j], {i, -45, 35}, {j, -55, 20}], 1];
near = Nearest[p2d];
good = Select[triangles, 
   Total[Norm[(near[#, 1][[1]] - #)] & /@ #] < .001 &];
p3d[x_] := SortBy[b, Norm[#[[1 ;; 2]] - x] &][[1]]
sp3d = (p3d /@ #) & /@ good;
Graphics3D[{EdgeForm[None], 
  Polygon /@ (Select[sp3d, Variance[#[[All, 3]]] < .0001 &])}, 
 BoxRatios -> {1, 1, 1}]

enter image description here

At this point I havn't dealt with the boundaries or the overlapping portions of the surface, but It should get you started.

$\endgroup$
1
  • $\begingroup$ Yes, this is the kick in the arse I needed. I can actually get my bulk triangles at runtime with a little jiggery-pokery, although it won't shed much light on how to perform this procedure more generally though. I've accepted this answer and will update the question with the full result when I get it. $\endgroup$
    – Crêpo
    Jun 6, 2016 at 17:11
10
$\begingroup$

Not sure how useful will this be. Usually Standardize-d data is giving better results.

We can also cut off parts of the surface which are distant from our data.

rf = Nearest[Standardize@data]

sr = ListSurfacePlot3D[Standardize @ data, 
  BoxRatios -> 1, 
  MaxPlotPoints -> 100, 
  RegionFunction -> Function[{x, y, z}, 
    Norm[{x, y, z} - rf[{x, y, z}][[1]]] < .2]
]

enter image description here

m = Mean @ data
sd = StandardDeviation @ data


Show[
 Graphics3D[
  GeometricTransformation[
   First@sr,
   TranslationTransform[m]@*ScalingTransform[sd]
   ]
  ]
 ,
 ListPointPlot3D[{a, b}, PlotStyle -> {Red, Blue}  ]
 ,
 SphericalRegion -> True, BoxRatios -> 1
 ]

enter image description here

$\endgroup$
2
  • $\begingroup$ I think the main problem for the sr in your answer is smoothness. $\endgroup$
    – xyz
    Jun 6, 2016 at 14:31
  • $\begingroup$ @ShutaoTANG don't claim it is best approach, but it is 1 line and looks way better than original approach. $\endgroup$
    – Kuba
    Jun 6, 2016 at 14:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.