Skip to main content

graphics - How to compile Heike's winding number function?


Heike gave the following function for winding number:


winding[poly_, pt_] :=  
Round[(Total@Mod[(# - RotateRight[#]) &@(ArcTan @@ (pt - #) & /@ poly),
2 Pi, -Pi]/2./Pi)]

I attempted to compile it as follows:


winding2 := Compile[{{poly, _Real, 2}, {pt, _Real, 1}},   
Round[(Total@Mod[(# - RotateRight[#]) &@(ArcTan @@ (pt - #) & /@ poly),

2 Pi, -Pi]/2./Pi)]]

Applied to the following simple problem, the compiled version gives error messages:


poly = {{0., 0.}, {10., 0.}, {10., 6.}, {0, 6}, {0., 0.}};
pt = {5., 3.};
winding2[poly, pt]

The error messages include:


Compile::cpapot: Compilation of ArcTan@@(ptCompile`GetElement[poly,System`Private`CompileSymbol[0]]) 
is not supported for the function argument ArcTan. The only function arguments supported are

Times, Plus, or List. Evaluation will use the uncompiled function. >>
CompiledFunction::cfse: Compiled expression 6.283185307179586` should be a machine-size integer. >>
CompiledFunction::cfex: Could not complete external evaluation at instruction 1;
proceeding with uncompiled evaluation. >>

Where am I going wrong?



Answer



First, if you use := in your assignment, then the compilation is not done instantly but every time you call winding2. That's btw the reason why you get the error message when you try to call the function because it is not compiled until then and the error is a compilation error.


Secondly, as the error messages sais, @@ can only be used with Times, Plus or List, so you have to expand this part:


winding2 = 

Compile[{{poly, _Real, 2}, {pt, _Real, 1}},
Round[Total@Mod[# - RotateRight[#] &@(ArcTan[#[[1]], #[[2]]] &@
(Transpose@poly - pt)), 2 Pi, -Pi]/(2. Pi)]
]

Seems to work pretty smoothly


enter image description here


And here the code:


winding2 = 
Compile[{{poly, _Real, 2}, {pt, _Real, 1}},

0 != Round[
Total@Mod[# -
RotateRight[#] &@(ArcTan[#[[1]], #[[2]]] &@(Transpose@
poly - pt)), 2 Pi, -Pi]/(2. Pi)],
CompilationTarget -> "C", RuntimeOptions -> "Speed"];

With[{gr =
RegionPlot[x^2 + y^3 < 2, {x, -2, 2}, {y, -2, 2}, Mesh -> All,
FrameTicks -> None, PlotPoints -> 3, MaxRecursion -> 4,
PlotStyle -> RGBColor[14/15, 232/255, 71/85],

MeshStyle -> RGBColor[88/255, 22/51, 39/85]]},
DynamicModule[{pt = {0, 1}, polyPts},
polyPts =
gr[[1, 1, #]] & /@
Flatten[Cases[gr, Polygon[pts__] :> Sequence[pts], Infinity], 1];
LocatorPane[Dynamic[pt],
Dynamic@Show[gr,
Graphics[{Opacity[.5], RGBColor[38/255, 139/255, 14/17],
Polygon[Pick[polyPts, winding2[#, pt] & /@ polyPts]]}]]]]]

Comments

Popular posts from this blog

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}]

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}]

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.