functions - Probable Bug in HighlightMesh with 2D Convex Hulls created via ConvexHullMesh

Bug introduced in 10.0 and fixed in 10.2

One nice way to view the new and exciting Mesh objects introduced in Version 10 is by utilizing the function HighlightMesh, so I've been using it heavily. Problem is, when trying to visualize 2D convex hull regions (3D case works fine) created using ConvexHullMesh, HighlightMesh essentially ignores the Options for styling the face of the 2D polygon region. Here is an example:

SeedRandom[0];
pts = RandomReal[4, {200, 2}];

chull = ConvexHullMesh[pts]

Mathematica graphics

Now let's Style it using HighlightMesh

HighlightMesh[chull, {Style[0, Directive[PointSize[0.02], Red]], 
  Style[1, Thin, Green], Style[2, Directive[Yellow]]}]

Mathematica graphics

Notice the Yellow Color under Style[2, ..], that's for the Polygon face. This is obviously ignored for the Default color used above. Can anyone reproduce this on Windows 8.1 and other operating systems and is there an easy workaround for this?

Answer

An alternative workaround is to convert the BoundaryMeshRegion into a MeshRegion from the MeshCoordinates and MeshCells. This lets you use HighlightMesh as desired:

SeedRandom[0];
pts = RandomReal[4, {200, 2}];
chull = ConvexHullMesh[pts];
styles = MapThread[Style, {{0, 1, 2}, {Red, Green, Yellow}}];

fullmesh[bm_] := MeshRegion[MeshCoordinates[bm], MeshCells[bm, All]]

HighlightMesh[chull, styles]
HighlightMesh[fullmesh @ chull, styles]

enter image description here

Note that this won't work for higher dimensional regions:

MeshCells[ConvexHullMesh[RandomReal[1, {4, 3}]], 3]

MeshCells::cnorep: There is no simple cell representation for the specified cells of the BoundaryMeshRegion in dimension 3.

Comments

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]

plotting - Mathematica: 3D plot based on combined 2D graphs

I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1....

Blog

Search This Blog