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Similar to http://forums.wolfram.com/mathgroup/archive/2009/May/msg00133.html or Plot 4D data with color as 4th dimension, I would like to plot single points {x, y, z} in 3D with a 4th value Q as color function where each dataset has its own Q value. So i cannot use a function because each Q has to be connected to its {x, y, z}-set, but the z level is constant for each dataset. For example: 

DataA: {{xA1, yA2, zA, QA1}, {xA2, yA2, zA, QA2}, ...
DataB: {{xB1, yB2, zB, QB1}, {xB2, yB2, zB, QB2}, ...

How would this be implemented here?: 

circle3D[centre_: {0, 0, 0}, radius_: 20, normal_: {0, 0, 1}, angle_: {0, 2 Pi}] :=Composition[Line, Map[RotationTransform[{{0, 0, 1}, normal}, 
centre], #] &, Map[Append[#, Last@centre] &, #] &, 
Append[DeleteDuplicates[Most@#], Last@#] &, Level[#, {-2}] &,
MeshPrimitives[#, 1] &, DiscretizeRegion, If]
[First@Differences@angle >= 2 Pi, Circle[Most@centre, radius],
Circle[Most@centre, radius, angle]]

colors = 
  {{1, RGBColor[1, 0.11, 0.02, 0.75]}, 
   {2, RGBColor[1, 0.22, 0.23, 0.75]}, 
   {3, RGBColor[1, 0.52, 0, 0.75]}, 
   {4, RGBColor[0.93, 0.79, 0.05, 0.75]}, 
   {5, RGBColor[0.63, 0.84, 0.09, 0.75]}, 
   {6, RGBColor[0.4, 0.86, 0.15, 0.75]}, 
   {7, RGBColor[0.1, 0.86, 0.17, 0.75]}};

Data1 = 
  {{20, 0, z1, 3}, {10, 0, z1, 3}, {0, 0, z1, 3.67}, {-10, 0, z1, 3}, 
   {-20, 0, z1, 3}, {0, 10, z1, 3}, {0, 20, z1, 3}, {0, 10, z1, 3}, 
   {0, -20, z1, 3}, {z1, 100}};
Data2 = 
  {{20, 0, z2, 2.67}, {10, 0, z2, 3.33}, {0, 0, z2, 4}, {0, 10, z2, 3.67}, 
   {0, 20, z2, 2.33}, {0, -10, z2, 3.67}, {0, -20, z2, 2.33}, {-20, 0, z2, 2.67}, 
   {-10, 0, z2, 3.33}, {z2, 200}};

c1 = Graphics3D[{Dashed, circle3D[{0, 0, 100}, 20]}];
c2 = Graphics3D[{Dashed, circle3D[{0, 0, 200}, 20]}];

Show[
  ListPointPlot3D[{Data1, Data2},
    Axes -> True, 
    AxesOrigin -> {0, 0, 0},
    ColorFunction -> (Blend[colors, #1] &),
    ColorFunctionScaling -> False,
    Boxed -> False,
    PlotStyle -> PointSize[0.02],
    PlotRange -> Automatic,
    ImageSize -> Large,
    ViewPoint -> {0.8 Pi, -1.5 Pi, 0.20 Pi}],
  c1, c2]
Community
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StephanJ
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  • Possible duplicate of Plot 4D data with color as 4th dimension – chuy Feb 23 '16 at 21:03
  • Can you point out what does not work for you in the solutions proposed in the question "Plot 4D data with color as 4th dimension" that you link to? Also, why is the last entry in your Data lists completely different from the others? Can you give us the actual list of data you want to plot? Can you define circle3D, or link to its definition? – MarcoB Feb 23 '16 at 21:28
  • You can not plot Data1 and Data2 as you currently define them with ListPointPlot3D because they are not lists of 3D points. You must transform them into such lists and while doing so assign numeric values to z1 and z2. – m_goldberg Feb 23 '16 at 21:52
  • @StephanJ, in what way is this answer not the perfect solution to your problem? As your problem is stated, that's the answer, so if it doesn't work for you, let us know why. – Jason B. Feb 24 '16 at 10:36
  • Thanks for the hints. Sorry, i missed the circle3D definition, I added this one. My problem with the linked examples are, that there are always functions added, as with the flatten or table option. I here have points in x,y and z position with a certain value for the color (4th value), this value is unique for each data point and i want to link this 4th value to the color function. Only the z value is the same for each data list because all points of one list are located in one z level (Data2 --> z2, and so on). I currently do not now how to transform the data as m_goldberg suggested. – StephanJ Feb 24 '16 at 10:37
  • You still haven't fixed the fundamental problem that Data1 and Data2 are not suitable arguments for ListPointPlot3D because they not lists of 3D points. Until you do that, I see no reason to reopen this question. – m_goldberg Feb 24 '16 at 17:39
  • Ok, im afraid but exactly here is my problem. To bring these datasets of 4D points in a form to further use the transpose option like in the linked example is my bottleneck here. There a function is used. Up to now i could not combine my datasets with this method. The graph itself is already finished. – StephanJ Feb 25 '16 at 10:53
  • Update: Question might be deleted. I found my (unfortunatly) very basic mistake. The linked answer fits very well. Thanks though. – StephanJ Feb 25 '16 at 16:37

0 Answers0