Skip to main content

performance tuning - Faster Alternatives to DateDifference


I need a faster implementation of FractionOfYear and FractionOfMonth, which do the following:



Input: A time/date specified by {y_, m_, d_, h_, n_, s_}


Output: A real number from 0 to 1 representing the fraction of the year or month that the given time/date spec occurs in.


Leap days and leap seconds complicate things, so I thought I could just rely on DateDifference, but it is too slow:


RandomDateList[] := {RandomInteger[{1800, 2100}], RandomInteger[{1, 12}], RandomInteger[{1, 28}], RandomInteger[{0, 23}], RandomInteger[{0, 59}], RandomInteger[{0, 59}]};
RandomDates[n_] := Table[RandomDateList[],{n}]

secondOfYear[{y_, m_, d_, h_, n_, s_}] := 
   First[DateDifference[{y - 1, 12, 31, 24, 0, 0}, {y, m, d, h, n, s}, 
               "Second"]] / First[DateDifference[{y - 1, 12, 31, 24, 0, 0}, {y, 12, 31, 24, 0, 0}, "Second"]]


secondOfMonth[{y_, m_, d_, h_, n_, s_}] := First[DateDifference[{y, m, 1, 0, 0, 0}, 
{y, m, d, h, n, s}, "Second"]]/First[DateDifference[{y, m, 1, 0, 0, 0}, If[m==12, {y+1, 1, 1, 0, 0, 0}, {y, m+1, 1, 0, 0, 0}], "Second"]]

AbsoluteTiming [secondOfYear /@ RandomDates[1000]] takes 6 seconds.


There must be a faster easier way of doing this! I'll accept the first answer that takes under a second for 100,000 elements.



Answer



Ok, Java solution, by popular demand.


Solution





  1. Load the Java reloader




  2. Compile this class:


    JCompileLoad@
      "import java.util.Calendar;

       public class SecondOfYearVectorized{  
         public static double[] secondOfYear(int[][] dates){
            Calendar calendar = Calendar.getInstance();

            double[] result = new double[dates.length];
            for(int i=0;i            calendar.set(dates[i][0],dates[i][1]-1,dates[i][2],
                      dates[i][3], dates[i][4],dates[i][5]);
                long time = (long) (calendar.getTimeInMillis()/1000);
                calendar.set(dates[i][0]-1,11,31,24,0,0);
                long timepy = (long) (calendar.getTimeInMillis()/1000);
                calendar.set(dates[i][0],11,31,24,0,0);
                long timey = (long) (calendar.getTimeInMillis()/1000);
                result[i]= ((double)(timepy-time))/(timepy-timey);

            }   
            return result;      
        }
    }"


Usage and comparisons


dates = RandomDates[100000];

SecondOfYearVectorized`secondOfYear[dates]//Short//AbsoluteTiming



  {0.2460938,{0.310529,0.0296395,<<99996>>,0.0393697,0.0470913}}

N[secondOfYear2 /@ dates]//Short//AbsoluteTiming


  {1.5937500,{0.310643,0.0296395,<<99996>>,0.0393697,0.0470913}}

Remarks



The Java solution seems about 6 times faster than Brett's one, and about 4 times faster than the fastest vectorized solution I could cook up with AbsoluteTime. The AbsoluteTime itself is pretty fast though, so I don't know to what should I attribute the speed-up (my guess is that I save on data transfer, which causes a performance hit, since I only transfer data once. And probably Java Calendar-based functionality is still faster by itself). Note that results obtained with Java and Mathematica solution are only approximately the same, but the differences are quite small, when any. Note also that Java solution reaches full speed after several runs, likely due to a warm-up of JVM HotSpot JIT compiler.


Comments

Post a Comment

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}]
Post a Comment
Read more

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}]
Post a Comment
Read more

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.
Post a Comment
Read more