Skip to main content

Export[] performance


Consider this sequence of commands:


foo10 = RandomReal[{0, 1}, {5000, 10}];
Export["D:\\Dropper\\Dropbox\\footest.csv", foo10] // AbsoluteTiming
{0.898086, "D:\\Dropper\\Dropbox\\footest.csv"}

foo100 = RandomReal[{0, 1}, {5000, 100}];
Export["D:\\Dropper\\Dropbox\\footest.csv", foo100] // AbsoluteTiming

{7.428425, "D:\\Dropper\\Dropbox\\footest.csv"}

foo1000 = RandomReal[{0, 1}, {5000, 1000}];
Export["D:\\Dropper\\Dropbox\\footest.csv", foo1000] // AbsoluteTiming
{65.063763, "D:\\Dropper\\Dropbox\\footest.csv"}

Two (at least) things leap to mind:


The first is that the performance is truly atrocious (this is on a very fast machine, on a local file system.


The second is that the times grow sublinearly in the size of the input, which indicates a whole'nother kind of weirdness.


Does anyone understand what is going on, and if there is a reasonable workaround?



EDIT Following up on @belisarius' suggestion, here are the Timing[] results for the three commands:


0.671875 4.65250 47.968750


From which we see that both the OS interaction and the internal computation are ridiculously slow. But it gets worse. In the middle of doing this experiment, when I was running the commands exactly as above with AbsoluteTiming[] replaced by Timing[], between the first and the second command Mathematica informed me that it could not do the Export[] because the "footest.csv" file was already open (which means that it did not do the necessary fclose(), or whatever Windows calls it, of course changing the file name fixed it.


For posterity, this is Mathematica 10.0.1 on Windows 8.1


ANOTHER EDIT Just for completeness: I tried the same experiment with ".dat" instead of ".csv", and the results were essentially identical (including the fclose() bug). Which makes it surprising (though gratifying) that there is an actual format that works quickly, as pointed out by @Jens.



Answer



The speed issue is similar on Mac OS X. I can only speculate what's going on, so I better not even start.


But let me suggest a work-around. Since CSV is not in any way optimized for exporting numerical data, I would instead suggest to use a different format that has similar flexibility but is particularly designed for numerical data.


Of course, this is a workaround only if your intention is to save numerical data for later use by an application that's also able to read the alternate format. Here is what I tried, and it's very fast by comparison:


foo1000 = RandomReal[{0, 1}, {5000, 1000}];


Export["footest.h5", foo1000] // AbsoluteTiming

(* ==> {0.521861, "footest.h5"} *)

foo = Import["footest.h5", "Data"]; // AbsoluteTiming

(* ==> {0.145216, Null} *)

foo[[1]] == foo1000


(* ==> True *)

The format I used is HDF5.


If you're able to work with packed arrays, then things get even faster:


foo1000 = 
Developer`ToPackedArray@RandomReal[{0, 1}, {5000, 1000}];

Export["footest.h5", foo1000] // AbsoluteTiming


(* ==> {0.358825, "footest.h5"} *)

foo = Import["footest.h5", "Data"]; // AbsoluteTiming

(* ==> {0.084686, Null} *)

foo[[1]] == foo1000

(* ==> True *)

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....