Skip to main content

distributions - How to generate random points in a region?


The Mathematica 10 documentation was updated for FindInstance adding support for regions.


In my use case, I'm trying to sample points in a set of disks:



region = DiscretizeRegion@RegionUnion@Table[Disk[RandomReal[4, {2}], RandomReal[1]], {10}]
FindInstance[{x, y} ∈ region, {x, y}, Reals, 2] // N

However the above code fails and generates the following error:


FindInstance::elemc: "Unable to resolve the domain or region membership condition {x,y}∈"

What's going wrong here?



Answer



There are already good answers, but I'm going to improve the performance, generalize to any region in any dimensions and make the function more convenient. The main idea is to use DirichletDistribution (the uniform distribution on a simplex, e.g. triangle or tetrahedron). This idea was implemented by PlatoManiac and me in the related question obtaining random element of a set given by multiple inequalities (there is also Metropolis algorithm, but it is not suitable here).


The code is relatively short:



RegionDistribution /: 
 Random`DistributionVector[RegionDistribution[reg_MeshRegion], n_Integer, prec_?Positive] :=
  Module[{d = RegionDimension@reg, cells, measures, s, m},
   cells = Developer`ToPackedArray@MeshPrimitives[reg, d][[All, 1]];
   s = RandomVariate[DirichletDistribution@ConstantArray[1, d + 1], n];
   measures = PropertyValue[{reg, d}, MeshCellMeasure];
   m = RandomVariate[#, n] &@EmpiricalDistribution[measures -> Range@Length@cells];
   #[[All, 1]] (1 - Total[s, {2}]) + Total[#[[All, 2 ;;]] s, {2}] &@
   cells[[m]]]


Examples


Random disks (2D in 2D)


SeedRandom[0];
region = DiscretizeRegion@RegionUnion@Table[Disk[RandomReal[4, {2}], RandomReal[1]], {10}];
pts = RandomVariate[RegionDistribution[region], 10000]; // AbsoluteTiming
ListPlot[pts, AspectRatio -> Automatic]


{0.004473, Null}




enter image description here


Precise test


pts = RandomVariate[RegionDistribution[region], 200000000]; // AbsoluteTiming


{85.835022, Null}



Histogram3D[pts, 50, "PDF", BoxRatios -> {Automatic, Automatic, 1.5}]

enter image description here



It is fast for $2\cdot10^8$ points and the distribution is really flat!


Intervals (1D in 1D)


region = DiscretizeRegion[Interval[{0, 1}, {2, 4}]];
pts = RandomVariate[RegionDistribution[region], 100000]; // AbsoluteTiming
Histogram[Flatten@pts]


{0.062430, Null}



enter image description here



Random circles (1D in 2D)


region = DiscretizeRegion@RegionUnion[Circle /@ RandomReal[10, {100, 2}]];
pts = RandomVariate[RegionDistribution[region], 10000]; // AbsoluteTiming
ListPlot[pts, AspectRatio -> Automatic]


{0.006216, Null}



enter image description here


Balls (3D in 3D)



region = DiscretizeRegion@RegionUnion[Ball[{0, 0, 0}], Ball[{1.5, 0, 0}], Ball[{3, 0, 0}]];
pts = RandomVariate[RegionDistribution[region], 10000]; // AbsoluteTiming
ListPointPlot3D[pts, BoxRatios -> Automatic]


{0.082202, Null}



enter image description here


Surface cow disctribution (2D in 3D)


region = DiscretizeGraphics@ExampleData[{"Geometry3D", "Cow"}];

pts = RandomVariate[RegionDistribution[region], 2000]; // AbsoluteTiming
ListPointPlot3D[pts, BoxRatios -> Automatic]


{0.026357, Null}



enter image description here


Line in space (1D in 3D)


region = DiscretizeGraphics@ParametricPlot3D[{Sin[2 t], Cos[3 t], Cos[5 t]}, {t, 0, 2 π}];
pts = RandomVariate[RegionDistribution[region], 1000]; // AbsoluteTiming

ListPointPlot3D[pts, BoxRatios -> Automatic]


{0.005056, Null}



enter image description here


Comments

Post a Comment

Popular posts from this blog

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}]
Post a Comment
Read more

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}]
Post a Comment
Read more

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....
Post a Comment
Read more