Skip to main content

export - Exporting extremely large images; arrays generated via TensorProduct use 5 times as much RAM as expected?



I am generating plots of the roots of certain classes of polynomials, in the spirit of the multicolored visuals on the Wiki page on algebraic numbers. I've thus far been able to make some extremely spectacular plots, such as this 3600 x 3600 image of the roots of the monic quadratics and cubics (very low-res preview below)


enter image description here


and this 4600 x 4600 image of the roots of the cubics with lead coefficient 5 (low-res preview below)


enter image description here


and this 4600 x 4600 image of the roots of the cubics with lead coefficient 1 or 5 (low-res preview below).


enter image description here


I have been running into a RAM usage bottleneck during the Export step as I try to make more detailed visuals. To make the images, I essentially execute


Export["Imagename.PNG",Image[A]]

where $A$ is roughly a $4600\times 4600\times 3$ rank-3 array specifying the RGB color channels. I was having issues with the MathKernel swallowing many gigabytes of RAM and initially thought it was from the Image Export step, but it turns out it's due to the TensorProduct I used to generate $A$.



I've begun to observe that arrays generated via TensorProduct occupy 5 times as much RAM as I would naively expect. For example,


$HistoryLength = 0;
L = 2000;
a = RandomReal[{-10^6, 10^6}, {L, L}];
b = RandomReal[{-10^6, 10^6}, 3];
c = TensorProduct[a, b];
d = RandomReal[{-10^6, 10^6}, {L, L, 3}];
ByteCount[a]
ByteCount[b]
ByteCount[c]

ByteCount[d]
Quit[]

indicates that $c$ occupies 480 MB RAM, while $d$ occupies 96MB RAM. Choosing $L=4600$ shows that $c$ occupies 2.6 GB RAM, while $d$ occupies 500 MB RAM. Changing the ranges of RandomReal from $10^6$ to other numbers does not significantly alter the result. This is a bit unexpected, since both $c$ and $d$ are of the same dimension and same MachinePrecision entry type, and yet the manner of array generation seems to strongly impact the space it occupies in RAM.


Question: Does anyone know why using TensorProduct to generate an array creates an array object which occupies 5 times the normal RAM amount as comparable arrays generated via other means?


EDIT: After applying the packed array fixes, things work well. Here are links to the two notebooks that can be used to generate the images I used. Root Generator is used to compute and export the polynomial roots of one's choice to sparse matrix formats, and Root Visualizer imports the sparse matrices and exports the images. I use packed arrays in the RAM-intensive parts, and Compile the convolution kernel used to blur the image. If anyone sees any obvious improvements that could be made, I'd love to hear it.



Mathematica 10 has updated TensorProduct so that it returns packed arrays. So if you are running version 10 or higher, you should not run into this problem.



Answer



For unknown reasons, TensorProduct produces unpacked array (see packed arrays here). You can use Outer[Times, a, b] instead:



$HistoryLength = 0;
L = 2000;
a = RandomReal[{-10^6, 10^6}, {L, L}];
b = RandomReal[{-10^6, 10^6}, 3];
c = TensorProduct[a, b];
c2 = Outer[Times, a, b];
d = RandomReal[{-10^6, 10^6}, {L, L, 3}];
ByteCount /@ {a, b, c, c2, d} // Column
Developer`PackedArrayQ /@ {c, c2} // Column
Max@Abs[c - c2]



32000152
128
480800392
96000160
96000160

False
True


0.

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....