Skip to main content

list manipulation - Why does this nested sum appear to sleep between iterations?


I have a weird performance problem in a nested sum, which I've reduced to the following test case:


testTab = Table[1.`20, {i, 188}, {j, 301}, {k, 20}];
test[] := Sum[testTab[[1, 1, 1]] Sum[testTab[[1, 1, 1]], {n, 1, 250}], {m, 1, 10}]
Monitor[test[], {n, m}]

Here the output from Monitor is {n, 1} for a second, then it changes to {n, 2}, and so on until I get the result of 2500.0000000000000000. Obviously, table accesses shouldn't be that slow.



Even more interestingly, if I change the summation limit of the inner sum from 250 to 249 or smaller, I don't get this slowdown, the result appears almost instantly. I can even make table dimensions way larger, but this 250→249 transition still results in drastic performance difference.


What's happening here? Is it a bug?


I'm using Mathematica "11.1.0 for Linux x86 (32-bit) (March 13, 2017)", but this problem also happens on "9.0 for Linux x86 (32-bit) (November 20, 2012)" and on "11.0.1 for Linux ARM (32-bit) (January 17, 2017)" (Raspberry Pi 3).



Answer



The reason is that Mathematica tries to compile the code for sufficiently long Tables, Sums, and Products. The outer sum has 10 summands. That's too few. But starting the inner sum with 250 or more summands implies a compilation. 250 is the default value of the suboption "SumCompileLength" of the system option "CompileOptions":


"SumCompileLength" /. ("CompileOptions" /. SystemOptions[])


250




This compilation is done for each iteration of the outer Sum and induces some overhead. This is why it is a good idea to merge multiple sums into a single instance of Sum like this:


test2[] := 
Sum[testTab[[1, 1, 1]] testTab[[1, 1, 1]], {n, 1, 250}, {m, 1, 10}]

Moreover, trying to use matrix-vector producs or Total may be even more efficient. Needless to say that nothing will beat testTab[[1, 1, 1]] 250 10 in this case.


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....