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Wrapping a list density plot on a sphere

lei baida
Posted 12 years ago
dear all

I have a data set in the form of a list: {{Theta1,Phi1,f1},{Theta1,Phi1,f1} ...} which i would like to represent in the form of a density/Contour plot on the surface of a sphere

i tried using
SphericalPlot3D[1, {t, 0, 2 Pi}, {p, 0, Pi},
ColorFunction -> Function[{x, y, z, t, p, r}, Hue[f[t, p]]],
Mesh -> 10, MeshFunctions -> {Function[{x, y, z, t, p, r}, f[t, p]]}]

but it doesn't seem to work in my case

thanks in advance for your help
POSTED BY: lei baida
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Dear Lei,

first you said that you have "data", but then your own example makes use of a "function" - an analytic expression. Plotting data and functions require different graphical tools. So which is your case?
POSTED BY: EDITORIAL BOARD
lei baida
lei baida
Posted 12 years ago
first thank you for your reply. yes, i have a data set in the form of a list having function values depending on theta and phi, i.e. { {0,0,1}, {Pi,0,2}, {0,Pi,3} .... } which i would like to wrap on a sphere surface as a density plot. SphericalPlot3D was the only way i could find to do something similar but it doesn't work with a data set. is there another way to do it?
POSTED BY: lei baida
I will make a noisy data set to imitate the format you have given:
data = Flatten[Table[{{x, y}, Cos[4 x] + Sin[2 y] + .1 RandomReal[]}, {x, 0, Pi + .1, .01}, {y, 0, 2 Pi + .1, .01}], 1];

Now there are two ways I see how we can put it as density on a sphere surface. For both we need Interpolation function:
f = Interpolation[data];

Now I will use ContourPlot. I could use ListDensityPlot but it would basically do the same thing but slower. So:
plt = ContourPlot[f[x, y], {x, 0, Pi}, {y, 0, 2 Pi},
  PlotRangePadding -> 0, Frame -> False, ColorFunction -> "Rainbow", Contours -> 15, ContourStyle -> Opacity[.2]]



Now the 1st method is to use Texture:
SphericalPlot3D[1 , {u, 0, Pi}, {v, 0, 2 Pi}, Mesh -> None, TextureCoordinateFunction -> ({#5, 1 - #4} &),
PlotStyle -> Directive[Texture[plt]], Lighting -> "Neutral", SphericalRegion -> True, RotationAction -> "Clip"]



2nd method is to use custom ColorFunction and MeshFunctions based on the Interpolation you got before:
SphericalPlot3D[1, {u, 0, ?}, {v, 0, 2 ?}, ColorFunction ->
Function[{x, y, z, u, v, r}, ColorData["Rainbow"][Rescale[f[u, v], {-2, 2}]]],
ColorFunctionScaling -> False,
MeshFunctions -> {Function[{x, y, z, u, v, r}, f[u, v]]}, SphericalRegion -> True, RotationAction -> "Clip",
MeshStyle -> Opacity[.3]]



They don't come out exactly the same, but you can figure out which you like better and how you can improve them further.
POSTED BY: Vitaliy Kaurov
lei baida
lei baida
Posted 12 years ago
Dear Vitaliy,

Thank you for your reply.

i tried your suggested solution but unfortunately it didn't work in my case...

I am attaching a part of my data here, the format is {Theta,Phi,f}
 indata= {
 {0.000Pi, 0.000Pi, 0.600},
 {0.000Pi, 0.033Pi, 0.320},
 {0.500Pi, 0.033Pi, 0.405},
 {1.000Pi, 0.033Pi, 0.328},
 {1.500Pi, 0.033Pi, 0.406},
 {0.150Pi, 0.144Pi, 0.093},
 {0.350Pi, 0.144Pi, 0.115},
 {0.650Pi, 0.144Pi, 0.095},
{0.850Pi, 0.144Pi, 0.151},
{1.150Pi, 0.144Pi, 0.150},
{1.350Pi, 0.144Pi, 0.102},
{1.650Pi, 0.144Pi, 0.115},
{1.850Pi, 0.144Pi, 0.084},
{0.000Pi, 0.206Pi, 1.541},
{0.050Pi, 0.206Pi, 1.165},
{0.450Pi, 0.206Pi, -1.403},
{0.500Pi, 0.206Pi, -1.505},
{0.550Pi, 0.206Pi, -1.449},
{1.450Pi, 0.206Pi, -1.415},
{1.500Pi, 0.206Pi, -1.468},
{1.950Pi, 0.206Pi, 1.153},
{0.189Pi, 0.256Pi, -1.197},
{0.311Pi, 0.256Pi, -0.757}
}

As you can see the internal step i am using isn't necessarily regular since i am generating the data outside of mathematica.

for interpolation:
f = Interpolation;
i get the following error: "The coordinates do not lie on a structured tensor product grid"

and for:
f = Interpolation;
it just doesn't do anything..

I wonder if the data structure is the reason? If so, I'm sorry i didn't make it perfectly clear.
Is there a way to get around it?

Thank you again for your help.

Dan (Bai DaLei)
POSTED BY: lei baida
Yes, your data structure is wrong for Interpolation function. I would recommend first read documentation on functions you use and looking through examples given - you could literally use them to guide you. Documentation is very extensive. So in your case the ptoblem is fixed as
f = Interpolation[{{#[[1]], #[[2]]}, #[[3]]} & /@ indata, InterpolationOrder -> 1];

Then you plot becomes:
plt = ContourPlot[f[x, y], {x, 0, 6}, {y, 0, 0.8}, PlotRangePadding -> 0,
Frame -> False, ColorFunction -> "Rainbow", Contours -> 15, ContourStyle -> Opacity[.2]]



The same plot you would get simply using:
plt = ListContourPlot[indata, InterpolationOrder -> All, PlotRangePadding -> 0,
Frame -> False, ColorFunction -> "Rainbow", Contours -> 15, ContourStyle -> Opacity[.2]]

But now you have to be more careful with the rest, because you angle variables do not run complete spherical coordinate ranges {0, Pi} and {0, 2Pi}.
POSTED BY: Vitaliy Kaurov
lei baida
lei baida
Posted 12 years ago
Dear Vitaliy,

Thank you again for your reply.

i guess i will have to re-generate my data and try to put it on a regular grid.

hopefully if will work after that.
POSTED BY: lei baida
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