Skip to main content

code review - Performance on Binomial Deviate is slow


I hope I won't get spanked for reposting. This is related to a previous question discussed here: Performance on Multinomial Deviate is slow.


but this version is simpler because it is a binomial rather than multinomial which is my justification for the new question.


I've kluged together a real solution for this problem that someone might find useful. I'm hoping someone will improve on the readability and/or performance. The performance is acceptable but still a bit slow compared to C. I think there might be another factor of 2 or 3 speedup to be had. My question is: does anyone have suggestions that will improve on my solution?


The problem is that I need a bunch of deviates where the parameters of the distribution are different for each deviate. Here is the simple way to do it which is unacceptably slow.


Nboxes = 8000;

p = RandomReal[{0, .01}, Nboxes];
rt = RandomInteger[{0, 10}, Nboxes];
rp = RandomInteger[{0, 10}, Nboxes];
probs = 1 - p rp;
out = MapThread[
RandomVariate[BinomialDistribution[#1, #2]] &, {rt, probs}]; // Timing

The timing result was over 4.7 seconds on my laptop. I create one below that is about 25 times faster.


Now to generate a binomial deviate with parameters n and p you generate n uniform deviates on (0,1) and count how many are < p. The total number of uniform deviates needed then is Total[rt]. Those can be generated in a small fraction of a second. The following code implements a binomial deviate generator and runs in a total of under 0.2 seconds. The deviates are stored in "counts". If anyone can improve on this code for performance, readability, etc, I'm all ears. First I generate all the uniform deviates I need. Then I "box" those up according to what rt is and count the number of deviates in each partition that are less than that boxes p value.


unitdeviates = RandomReal[1, Total[rt]]; // Timing

nt = ConstantArray[0, Nboxes + 1];
Table[nt[[i]] = nt[[i - 1]] + rt[[i - 1]], {i, 2, Length[rt] + 1}]; // Timing

counts = ConstantArray[0, Nboxes];
Do[counts[[nboxes]] = Count[unitdeviates[[nt[[nboxes]] + 1 ;;
nt[[nboxes + 1]]]], _?(# < probs[[nboxes]] &)], {nboxes, 1,Nboxes}] // Timing

Here I show that the mean and variance of the two sets of deviates are consistent


N[Mean[out]]
N[Mean[counts]]

N[Variance[out]]
N[Variance[counts]]

Answer



The main thing that I am trying to show is that you can use Accumulate and that almost all these functions are compilable. I hope it also shows when to use Table rather than Do, to avoid making unnecessary ConstantArrays. I personally find the use of Table in your code confusing. Of course it is nice to localise variables from time to time, which is also done in this code. All in all it is pretty boring code and maybe it looks a lot like what you have written in C, but I hope you learn something anyway.


In the code below, there is quite a lot of code to define a single function. There are multiple ways to avoid this, for example calling other CompiledFunctions inside the CompiledFunction. Another way is to inline definitions, which can also be done in multiple ways. For simplicity, I have just dumped everything in one definition.


cfu =
Compile[{{nBoxes, _Integer, 0}},
Block[
{unifs, rt, min, max, acc, count, p, rp, probs}
,

rt = RandomInteger[{0, 10}, nBoxes];

acc = Accumulate[rt];

unifs = RandomReal[1., acc[[-1]]];

p = RandomReal[{0., .01}, nBoxes];
rp = RandomInteger[{0, 10}, nBoxes];
probs = 1 - p rp;


min = 1;

Table[
max = acc[[nb]];

count = 0;
Do[
If[
unifs[[j]] < probs[[nb]]
,

count++
]
,
{j, min, max}
];

min = max + 1;
count
,
{nb, 1, nBoxes}

]

]
,
CompilationTarget -> "C"
]

Let's also make definitions to compare this with your code


yourInit :=
(

Nboxes = 8000;
p = RandomReal[{0, .01}, Nboxes];
rt = RandomInteger[{0, 10}, Nboxes];
rp = RandomInteger[{0, 10}, Nboxes];
probs = 1 - p rp;
)

yours1 :=
(
yourInit;

out = MapThread[
RandomVariate[BinomialDistribution[#1, #2]] &, {rt, probs}];
);

yours2 :=
(
yourInit;
unitdeviates = RandomReal[1, Total[rt]];
nt = ConstantArray[0, Nboxes + 1];
Table[nt[[i]] = nt[[i - 1]] + rt[[i - 1]], {i, 2, Length[rt] + 1}];


counts = ConstantArray[0, Nboxes];
Do[counts[[nboxes]] =
Count[unitdeviates[[nt[[nboxes]] + 1 ;;
nt[[nboxes + 1]]]], _?(# < probs[[nboxes]] &)], {nboxes, 1,
Nboxes}]
)

We then have


yours1 // Timing // First

Mean[out] // N
Variance[out] // N


3.770396  
4.85013
9.70163

yours2 // Timing // First
Mean[counts] // N

Variance[counts] // N


0.132868  
4.8515
9.53689

(res = cfu[8000]) // Timing // First
Mean@res // N
Variance@res // N



0.002822  
4.87888
9.66966

So the speedup is similar to your own C code. Using LibraryLink we should be able to do a little bit better.


Comments

Popular posts from this blog

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 - 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 - 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.