mesh - How discretize a region placing vertices on a specific non-uniform grid

Given a generic region, for example:

Ω = 
  ImplicitRegion[
   2 x^2 + 3 y^2 + 2 x y - 2 <= 0 ∧ x^2 + y^2 > .1, {x, y}];

and a non-uniform grid, for example:

Ωb = RegionBounds[Ω];
{xg, yg} = 

  N@Map[bound \[Function] 
     Range[bound[[1]], 
      bound[[2]], (bound[[2]] - bound[[1]])/20], Ωb];
{xg, yg} = 
  Map[g \[Function] {g[[1]], 
     Sequence @@ (g[[2 ;; -2]] RandomReal[{1 - .04, 1 + .04}, 
         Length[g] - 2]), g[[-1]]}, {xg, yg}];

which, together, can be shown with:

Ωg = 

  RegionPlot[Ω, AspectRatio -> Automatic];

gg = Graphics[{LightGray,
    Table[
     Line[{{x, Ωb[[2, 1]]}, {x, Ωb[[2, 
         2]]}}], {x, xg}],
    Table[
     Line[{{Ωb[[1, 1]], y}, {Ωb[[1, 2]], 
        y}}], {y, yg}]
    }];


Show[gg, Ωg]

Mathematica graphics

I'm searching a way to discretize the region to a MeshRegion or ElementMesh in such a way that all vertices (MeshCoordinates or incidents) are placed on at least one gridline.

At present I'm working on a routine that discretize the region with BoundaryDiscretizeRegion with a reasonable MaxCellMeasure. Then I'm trying to split all Line mesh cells crossing any gridline into two or more Line such that at least one end is on a gridline. Then I plan to delete vertices not on a grid line and properly reconnect the sourronding vertices.

Mathematica graphics

This is the only way I could think to, but it appear difficult, because there are many branches tho examine.

A more elegant and simple approach would be helpful. For example, there is a way to use ToElementMesh and ToBoundaryMesh to accomplish this task? In the end, the routine should work with thousand of vertices and hundred of gridlines in a reasonable time.

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....

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