Skip to main content

performance tuning - How do you determine the optimal autocompilation length on your system


When you pack lists there is an overhead therefore packing a list with, say, 2 elements is likely to cost more than you get back in efficiency. Mathematica has default list lengths for which functions creating those lists will pack the list (i.e. if the list length is less than the numbers shown below the list will not be packed):



SystemOptions["CompileOptions"]

{"CompileOptions" -> {"ApplyCompileLength" -> \[Infinity],
"ArrayCompileLength" -> 250, "AutoCompileAllowCoercion" -> False,
"AutoCompileProtectValues" -> False, "AutomaticCompile" -> False,
"BinaryTensorArithmetic" -> False, "CompileAllowCoercion" -> True,
"CompileConfirmInitializedVariables" -> True,
"CompiledFunctionArgumentCoercionTolerance" -> 2.10721,
"CompiledFunctionMaxFailures" -> 3,
"CompileDynamicScoping" -> False,

"CompileEvaluateConstants" -> True,
"CompileOptimizeRegisters" -> False,
"CompileReportCoercion" -> False, "CompileReportExternal" -> False,
"CompileReportFailure" -> False, "CompileValuesLast" -> True,
"FoldCompileLength" -> 100, "InternalCompileMessages" -> False,
"ListableFunctionCompileLength" -> 250, "MapCompileLength" -> 100,
"NestCompileLength" -> 100, "NumericalAllowExternal" -> False,
"ProductCompileLength" -> 250, "ReuseTensorRegisters" -> True,
"SumCompileLength" -> 250, "SystemCompileOptimizations" -> All,
"TableCompileLength" -> 250}}


So, for example, if you make a list using Table


Developer`PackedArrayQ[Table[i, {i, 1, 249}]]
False

Developer`PackedArrayQ[Table[i, {i, 1, 251}]]
True

I am assuming that if you plotted the time to make uncompiled lists using Table, vs making compiled lists, the lines would intersect at ~250, beyond which packed lists become more efficient. Is that a correct interpetation of what the autocompilation length represents?


I would expect that the optimal lengths for compilation (incl. packing) vary on system to system, therefore I want to know the best way to construct a set of tests to test that proposition, and to determine the optimal list length for packing for the functions listed above.



Edit


For clarity, as per Albert's comments, there are cases when the evaluations taking place prevent compilation so these discussions are redundant, i.e. compilation is prevented regardless of the default settings. But I am curious about the optimal list lengths in cases where compilation occurs.




Comments

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}]

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 - 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....