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list manipulation - How to punch a hole in some 3D distribution of points


Suppose we have a long list of 3D Cartesian coordinates, defining a distribution of random points in 3D space. How could we remove all the points inside a sphere of radius sphereRadius located at coordinates sphereLocation = {X, Y, Z}?


This Boolean subtraction is probably trivial, but I didn't found any useful info on how to do it with Mathematica 7.0. Maybe it isn't trivial, after all.




Generalisation : How can we do the same with an arbitrary closed surface, instead of a sphere, if the hole is defined as a deformed sphere ?


holeLocation = {X, Y, Z};
hole[theta_, phi_] = holeLocation  + 
    radius[theta, phi] {Sin[theta]Cos[phi], Sin[theta]Sin[phi], Cos[theta]};

where theta and phi are the usual spherical coordinates.



Answer



Using my solution to a similar question asked on StackOverflow some time ago,


Pick[dalist,UnitStep[criticalRadius^2-Total[(Transpose[dalist]-frameCenter)^2]],0]


which is for any number of (Euclidean) dimensions and should be quite fast.


EDIT


Ok, here is a generalization of the vectorized approach I proposed:


ClearAll[cutHole];
cutHole[relativeData_, holdRadiusF_] :=
  Module[{r, theta, phi, x, y, z},
    {x, y, z} = Transpose[relativeData];
    r = Sqrt[Total[relativeData^2, {2}]];
    theta = ArcCos[z/r];

    phi = ArcTan[x,y];
    Pick[relativeData, UnitStep[r - holdRadiusF[theta, phi]], 1]];

Here is an illustration:


data = RandomReal[6, {10^6, 3}];
holeLoc = {3, 3, 3};
relativeData = Transpose[ Transpose[data] - holeLoc];

Define some particular shape of the hole:


holeRad[theta_, phi_] := 1 + 4 Sqrt[Abs[Cos[theta]]]


Pick the points:


kept =cutHole[relativeData,holeRad];//AbsoluteTiming

(* {0.631836,Null}  *)

Visualize:


Show[{
  ListPointPlot3D[Cases[kept, {_, _, _?Positive}]], 
  SphericalPlot3D[

    holeRad[\[Theta], \[Phi]], {\[Theta], Pi/4, Pi/2}, {\[Phi], 0, 2 Pi}, 
    PlotStyle -> Directive[Orange, Specularity[White, 10]], Mesh -> None]}]

enter image description here


The main point here is that the filtering function is vectorized and therefore quite fast.


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