plotting - Mark an 2d-area on a 3dPlot (ListPlot3d)

I have a list {(IC,s,FK)}, which I use to generate a 3d-plot with the help of the command ListPlot3D. IC and s go from 0 to 1.

enter image description here

There is an area (lets say 0.11 < IC < 0.3, 0.1 < s < 0.4) which is specially important, because this is the physical range. For some reason I want to plot the whole parameter space, but highlight (for example with a red rectangle) this physical range. It would be great to frame the physical range with a red rectangle on the surface of the 3d-Plot. (In the end, this 2d-rectangle should look similar to these mesh lines, but it should frame the range, defined above).

Do you have any idea?

Answer

With

 data = Table[{x = RandomReal[{-1, 1}], y = RandomReal[{-1, 1}], x^2 - y^2}, {300}];

Three possible methods are

using ColorFunction

using a combination of Mesh, MeshFunctions and MeshShading, (or, and better yet, just Mesh and MeshShading as in @Brett's answer)

produce two plots using different RegionFunction settings and combine them using Show.

Using ColorFunction:

 ListPlot3D[data, BoxRatios -> Automatic, Mesh -> None, 
  ColorFunction -> Function[{x, y, z}, If[-.5 < x < .5 && -.3 < y < .1, Red, White]],
  ColorFunctionScaling -> False, BoxRatios -> Automatic, 
   MaxPlotPoints -> 100, Lighting -> "Neutral"]

enter image description here

or with a different setting for the ColorFunction, say:

 ColorFunction -> Function[{x, y, z}, If[-.5 < x < .5 && -.3 < y < .1,
    ColorData["DeepSeaColors", (1 + x)/2], Directive[Opacity[.7], Hue[(1 + z)/2]]]]

enter image description here

Using Mesh, MeshFunctions and MeshShading:

 ListPlot3D[data,
  MeshFunctions -> {Boole[-.5 < #1 < .5 && -.2 < #2 < .75] &}, 
  Mesh -> {{1}},  MeshShading -> {White, Red}, 

  BoxRatios -> Automatic,  MaxPlotPoints -> 100, Lighting -> "Neutral"]

enter image description here

Using RegionFunction and Show:

lp1 = ListPlot3D[data,
    RegionFunction -> (! (-.5 < #1 < .5 && -.2 < #2 < .75) &),
    Mesh -> None, BoxRatios -> 1,
    MaxPlotPoints -> 100, Lighting -> "Neutral",
    ColorFunction -> (Directive[Opacity[.9], Hue[#1]] &), 
    Lighting -> "Neutral",   ImageSize -> 300];

lp2 = ListPlot3D[data,
    RegionFunction -> ((-.5 < #1 < .5 && -.2 < #2 < .75) &),
    Mesh -> None, BoxRatios -> 1,
    MaxPlotPoints -> 100, Lighting -> "Neutral",
    ColorFunction -> (Red &), Lighting -> "Neutral",
    ImageSize -> 300];
    Panel@Row[{lp1, lp2, Show[{lp1, lp2}]}, Spacer[5]]

enter image description here

Comments

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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}]

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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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