Skip to main content

performance tuning - Adding three integer sparse matrices is very slow. Adding only two is fast


Bug introduced in 9.0.0 and fixed in 10.0.0




Adding more than two sparse matrices in one step in Mathematica 9 is very slow (in fact I couldn't even wait for it to finish).


Here's an example. Let's generate a large sparse matrix:


am = AdjacencyMatrix@RandomGraph[{25000, 50000}];

am + am; // Timing (* very fast *)

(am + am) + am; // Timing (* very fast *)


am + am + am; // Timing (* so slow I didn't wait for it to finish *)

Can anyone reproduce this? I used Mathematica 9.0.0 on OS X. Mathematica 8 does not have this problem.


Does anyone have any ideas what may be going wrong here?



Answer



I have reported this as a bug. Having said that, here is some more info:


Note that when you use:


am = AdjacencyMatrix@RandomGraph[{25000, 50000}];//N


you do not see the issue. So this is only an integer sparse array issue. What happened until V8 internally is that there is a loop for adding each sparse array as a binary operation. E.g. something like:


res = Plus[Plus[sp1,sp2],sp3]

In version 9 this has been changed to use var args. In essence what you do with using parentheses is to enforce the old behavior. The reason this was changed is that the var args form of the code is faster and uses less memory:


am = (AdjacencyMatrix@RandomGraph[{2500, 5000}]) // N;
MaxMemoryUsed[am + am; // Timing] (*very fast*)
MaxMemoryUsed[(am + am) + am; // Timing] (*very fast*)
MaxMemoryUsed[
am + am + am; // Timing] (*so slow I didn't wait for it to finish*)


So for now you can either convert with N if that is possible, or use the parentheses.


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....