Does Mathematica reuse previous computations?












7












$begingroup$


I am doing an analysis of experimental results in which I need to repeat the same GaussianFilter hundred of times on different data. As explained in the documentation, GaussianFilter just convolves the data with a Gaussian kernel. Does it recompute the kernel every time I call the function, or will it somehow preserve and reuse the previous kernel? Would it be more efficient computationally for me to precompute the kernel (which I could do easily by applying GaussianFilter to a KroneckerDelta array), then do hundreds of ListConvolves instead of hundreds of GaussianFilters?










share|improve this question











$endgroup$








  • 2




    $begingroup$
    My first guess is that GaussianFilter employs FFT (FFT, multiply, inverse FFT). In this case, it would be a very bad idea to replace it by a convolution unless the convolution is also implemented by FFT: The naive way of convolution would require $O(n^2)$ flops for a vector of length $n$ while the FFT method would need only $O(n , log(n))$ flops.
    $endgroup$
    – Henrik Schumacher
    Mar 15 at 15:18








  • 2




    $begingroup$
    Best way would be to just try it and compare the RepeatedTimings.
    $endgroup$
    – Henrik Schumacher
    Mar 15 at 15:21






  • 3




    $begingroup$
    @HenrikSchumacher I'm pretty sure ListConvolve also uses FFT (based on its performance).
    $endgroup$
    – Szabolcs
    Mar 15 at 15:22






  • 1




    $begingroup$
    ListConvolve does indeed use FFT.
    $endgroup$
    – Leon Avery
    Mar 15 at 16:27








  • 4




    $begingroup$
    Indeed it does.
    $endgroup$
    – J. M. is slightly pensive
    Mar 15 at 16:36
















7












$begingroup$


I am doing an analysis of experimental results in which I need to repeat the same GaussianFilter hundred of times on different data. As explained in the documentation, GaussianFilter just convolves the data with a Gaussian kernel. Does it recompute the kernel every time I call the function, or will it somehow preserve and reuse the previous kernel? Would it be more efficient computationally for me to precompute the kernel (which I could do easily by applying GaussianFilter to a KroneckerDelta array), then do hundreds of ListConvolves instead of hundreds of GaussianFilters?










share|improve this question











$endgroup$








  • 2




    $begingroup$
    My first guess is that GaussianFilter employs FFT (FFT, multiply, inverse FFT). In this case, it would be a very bad idea to replace it by a convolution unless the convolution is also implemented by FFT: The naive way of convolution would require $O(n^2)$ flops for a vector of length $n$ while the FFT method would need only $O(n , log(n))$ flops.
    $endgroup$
    – Henrik Schumacher
    Mar 15 at 15:18








  • 2




    $begingroup$
    Best way would be to just try it and compare the RepeatedTimings.
    $endgroup$
    – Henrik Schumacher
    Mar 15 at 15:21






  • 3




    $begingroup$
    @HenrikSchumacher I'm pretty sure ListConvolve also uses FFT (based on its performance).
    $endgroup$
    – Szabolcs
    Mar 15 at 15:22






  • 1




    $begingroup$
    ListConvolve does indeed use FFT.
    $endgroup$
    – Leon Avery
    Mar 15 at 16:27








  • 4




    $begingroup$
    Indeed it does.
    $endgroup$
    – J. M. is slightly pensive
    Mar 15 at 16:36














7












7








7


2



$begingroup$


I am doing an analysis of experimental results in which I need to repeat the same GaussianFilter hundred of times on different data. As explained in the documentation, GaussianFilter just convolves the data with a Gaussian kernel. Does it recompute the kernel every time I call the function, or will it somehow preserve and reuse the previous kernel? Would it be more efficient computationally for me to precompute the kernel (which I could do easily by applying GaussianFilter to a KroneckerDelta array), then do hundreds of ListConvolves instead of hundreds of GaussianFilters?










share|improve this question











$endgroup$




I am doing an analysis of experimental results in which I need to repeat the same GaussianFilter hundred of times on different data. As explained in the documentation, GaussianFilter just convolves the data with a Gaussian kernel. Does it recompute the kernel every time I call the function, or will it somehow preserve and reuse the previous kernel? Would it be more efficient computationally for me to precompute the kernel (which I could do easily by applying GaussianFilter to a KroneckerDelta array), then do hundreds of ListConvolves instead of hundreds of GaussianFilters?







numerics convolution caching






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Mar 17 at 12:29









J. M. is slightly pensive

98.4k10307466




98.4k10307466










asked Mar 15 at 15:09









Leon AveryLeon Avery

700318




700318








  • 2




    $begingroup$
    My first guess is that GaussianFilter employs FFT (FFT, multiply, inverse FFT). In this case, it would be a very bad idea to replace it by a convolution unless the convolution is also implemented by FFT: The naive way of convolution would require $O(n^2)$ flops for a vector of length $n$ while the FFT method would need only $O(n , log(n))$ flops.
    $endgroup$
    – Henrik Schumacher
    Mar 15 at 15:18








  • 2




    $begingroup$
    Best way would be to just try it and compare the RepeatedTimings.
    $endgroup$
    – Henrik Schumacher
    Mar 15 at 15:21






  • 3




    $begingroup$
    @HenrikSchumacher I'm pretty sure ListConvolve also uses FFT (based on its performance).
    $endgroup$
    – Szabolcs
    Mar 15 at 15:22






  • 1




    $begingroup$
    ListConvolve does indeed use FFT.
    $endgroup$
    – Leon Avery
    Mar 15 at 16:27








  • 4




    $begingroup$
    Indeed it does.
    $endgroup$
    – J. M. is slightly pensive
    Mar 15 at 16:36














  • 2




    $begingroup$
    My first guess is that GaussianFilter employs FFT (FFT, multiply, inverse FFT). In this case, it would be a very bad idea to replace it by a convolution unless the convolution is also implemented by FFT: The naive way of convolution would require $O(n^2)$ flops for a vector of length $n$ while the FFT method would need only $O(n , log(n))$ flops.
    $endgroup$
    – Henrik Schumacher
    Mar 15 at 15:18








  • 2




    $begingroup$
    Best way would be to just try it and compare the RepeatedTimings.
    $endgroup$
    – Henrik Schumacher
    Mar 15 at 15:21






  • 3




    $begingroup$
    @HenrikSchumacher I'm pretty sure ListConvolve also uses FFT (based on its performance).
    $endgroup$
    – Szabolcs
    Mar 15 at 15:22






  • 1




    $begingroup$
    ListConvolve does indeed use FFT.
    $endgroup$
    – Leon Avery
    Mar 15 at 16:27








  • 4




    $begingroup$
    Indeed it does.
    $endgroup$
    – J. M. is slightly pensive
    Mar 15 at 16:36








2




2




$begingroup$
My first guess is that GaussianFilter employs FFT (FFT, multiply, inverse FFT). In this case, it would be a very bad idea to replace it by a convolution unless the convolution is also implemented by FFT: The naive way of convolution would require $O(n^2)$ flops for a vector of length $n$ while the FFT method would need only $O(n , log(n))$ flops.
$endgroup$
– Henrik Schumacher
Mar 15 at 15:18






$begingroup$
My first guess is that GaussianFilter employs FFT (FFT, multiply, inverse FFT). In this case, it would be a very bad idea to replace it by a convolution unless the convolution is also implemented by FFT: The naive way of convolution would require $O(n^2)$ flops for a vector of length $n$ while the FFT method would need only $O(n , log(n))$ flops.
$endgroup$
– Henrik Schumacher
Mar 15 at 15:18






2




2




$begingroup$
Best way would be to just try it and compare the RepeatedTimings.
$endgroup$
– Henrik Schumacher
Mar 15 at 15:21




$begingroup$
Best way would be to just try it and compare the RepeatedTimings.
$endgroup$
– Henrik Schumacher
Mar 15 at 15:21




3




3




$begingroup$
@HenrikSchumacher I'm pretty sure ListConvolve also uses FFT (based on its performance).
$endgroup$
– Szabolcs
Mar 15 at 15:22




$begingroup$
@HenrikSchumacher I'm pretty sure ListConvolve also uses FFT (based on its performance).
$endgroup$
– Szabolcs
Mar 15 at 15:22




1




1




$begingroup$
ListConvolve does indeed use FFT.
$endgroup$
– Leon Avery
Mar 15 at 16:27






$begingroup$
ListConvolve does indeed use FFT.
$endgroup$
– Leon Avery
Mar 15 at 16:27






4




4




$begingroup$
Indeed it does.
$endgroup$
– J. M. is slightly pensive
Mar 15 at 16:36




$begingroup$
Indeed it does.
$endgroup$
– J. M. is slightly pensive
Mar 15 at 16:36










2 Answers
2






active

oldest

votes


















14












$begingroup$

Here I implemented three different versions of Gaussian filtering (for periodic data). It took me a while to adjust the constants and still some of them might be wrong.



Prepare the Gaussian kernel



n = 200000;
σ = .1;
t = Subdivide[-1. Pi, 1. Pi, n - 1];
ker = 1/Sqrt[2 Pi]/ σ Exp[-(t/σ)^2/2];
ker = Join[ker[[Quotient[n,2] + 1 ;;]], ker[[;; Quotient[n,2]]]];


Generate noisy function



u = Sin[t] + Cos[2 t] + 1.5 Cos[3 t] + .5 RandomReal[{-1, 1}, Length[t]];


The three methods with their timings. As Niki Estner pointed out, GaussianFilter with the option Method -> "Gaussian" performs much batter than GaussianFilter with the default emthod.



kerhat = 2 Pi/Sqrt[N@n] Fourier[ker];
vConvolve = (2. Pi/n) ListConvolve[ker, u, {-1, -1}]; // RepeatedTiming // First
vFFT = Re[Fourier[InverseFourier[u] kerhat]]; // RepeatedTiming // First
vFilter = GaussianFilter[u, 1./(Pi) σ n, Padding -> "Periodic"]; // RepeatedTiming // First
vGaussian = GaussianFilter[u, 1./(Pi) σ n, Padding -> "Periodic", Method -> "Gaussian"]; // RepeatedTiming // First



0.0038



0.0058



0.055



0.0072




ListLinePlot[{u, vFFT, vFilter, vConvolve}]


enter image description here



From further experiments with different values for n, GaussianFilter seems to be slower by a factor 10-20 over a wide range of n (from n = 1000 to n = 1000000). So it seems that it does use some FFT-based method (because it has the same speed asymptotics) but maybe some crucial part of the algorithm is not compiled (the factor 10 is somewhat an indicator for that) or does not use the fastest FFT implementation possible. A bit weird.



So, to my own surprise, your idea of computing the kernel once does help but for quite unexpected reasons.






share|improve this answer











$endgroup$









  • 1




    $begingroup$
    IIRC, it's using a discrete Gaussian kernel, which involves non-compilable modified Bessel functions, so that might be the reason for the added computational effort you observe.
    $endgroup$
    – J. M. is slightly pensive
    Mar 15 at 16:35










  • $begingroup$
    Yes, that is one reason I proposed using GaussianFilter itself to generate the kernel.
    $endgroup$
    – Leon Avery
    Mar 15 at 16:36






  • 1




    $begingroup$
    If I add Method -> "Gaussian" to the GaussianFilter call, it's about as fast as the other two
    $endgroup$
    – Niki Estner
    Mar 16 at 12:32






  • 2




    $begingroup$
    @NikiEstner Ah, that's good to know! Somewhat unexpected that we have to call out twice for a "Gaussian" in order to get one, isn't it?
    $endgroup$
    – Henrik Schumacher
    Mar 16 at 13:01










  • $begingroup$
    @Henrik Schumacher: Your results are about 6 times faster than mine. My platform is Xeon E3-1270, V3, 3.695 MHz; 4 cores, 8 threads; 32 GB ECC memory; Windows 8.1 Prof, 64-bit. I'm not used to such shenanigans. Can you suggest a reason? In any case, thanks for the good work.
    $endgroup$
    – CElliott
    Mar 20 at 16:39



















7












$begingroup$

It's hard to know for sure, but one way to test for caching is to apply a single command to lots of data sets, or to apply the command separately to each set. For instance:



n = 5000;
data = RandomReal[{-1, 1}, {n, 10000}];
GaussianFilter[#, 100] & /@ data; // AbsoluteTiming
Do[GaussianFilter[data[[i]], 100], {i, n}] // AbsoluteTiming
Do[GaussianFilter[data[[i]], 100 + RandomInteger[{-15, 15}]], {i, n}] // AbsoluteTiming


The second line generates 5000 different sets of data, each 10000 length. The third applies one Gaussian filter to all the data sets. The third line applies a separate GaussianFilter to each set. The final line forces the GaussianFilter to recompute its kernel. The timings are pretty much the same. This suggests that whatever is happening, the time needed to calculate the Gaussian filter parameters is pretty negligeable.






share|improve this answer









$endgroup$













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    2 Answers
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    2 Answers
    2






    active

    oldest

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    active

    oldest

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    active

    oldest

    votes









    14












    $begingroup$

    Here I implemented three different versions of Gaussian filtering (for periodic data). It took me a while to adjust the constants and still some of them might be wrong.



    Prepare the Gaussian kernel



    n = 200000;
    σ = .1;
    t = Subdivide[-1. Pi, 1. Pi, n - 1];
    ker = 1/Sqrt[2 Pi]/ σ Exp[-(t/σ)^2/2];
    ker = Join[ker[[Quotient[n,2] + 1 ;;]], ker[[;; Quotient[n,2]]]];


    Generate noisy function



    u = Sin[t] + Cos[2 t] + 1.5 Cos[3 t] + .5 RandomReal[{-1, 1}, Length[t]];


    The three methods with their timings. As Niki Estner pointed out, GaussianFilter with the option Method -> "Gaussian" performs much batter than GaussianFilter with the default emthod.



    kerhat = 2 Pi/Sqrt[N@n] Fourier[ker];
    vConvolve = (2. Pi/n) ListConvolve[ker, u, {-1, -1}]; // RepeatedTiming // First
    vFFT = Re[Fourier[InverseFourier[u] kerhat]]; // RepeatedTiming // First
    vFilter = GaussianFilter[u, 1./(Pi) σ n, Padding -> "Periodic"]; // RepeatedTiming // First
    vGaussian = GaussianFilter[u, 1./(Pi) σ n, Padding -> "Periodic", Method -> "Gaussian"]; // RepeatedTiming // First



    0.0038



    0.0058



    0.055



    0.0072




    ListLinePlot[{u, vFFT, vFilter, vConvolve}]


    enter image description here



    From further experiments with different values for n, GaussianFilter seems to be slower by a factor 10-20 over a wide range of n (from n = 1000 to n = 1000000). So it seems that it does use some FFT-based method (because it has the same speed asymptotics) but maybe some crucial part of the algorithm is not compiled (the factor 10 is somewhat an indicator for that) or does not use the fastest FFT implementation possible. A bit weird.



    So, to my own surprise, your idea of computing the kernel once does help but for quite unexpected reasons.






    share|improve this answer











    $endgroup$









    • 1




      $begingroup$
      IIRC, it's using a discrete Gaussian kernel, which involves non-compilable modified Bessel functions, so that might be the reason for the added computational effort you observe.
      $endgroup$
      – J. M. is slightly pensive
      Mar 15 at 16:35










    • $begingroup$
      Yes, that is one reason I proposed using GaussianFilter itself to generate the kernel.
      $endgroup$
      – Leon Avery
      Mar 15 at 16:36






    • 1




      $begingroup$
      If I add Method -> "Gaussian" to the GaussianFilter call, it's about as fast as the other two
      $endgroup$
      – Niki Estner
      Mar 16 at 12:32






    • 2




      $begingroup$
      @NikiEstner Ah, that's good to know! Somewhat unexpected that we have to call out twice for a "Gaussian" in order to get one, isn't it?
      $endgroup$
      – Henrik Schumacher
      Mar 16 at 13:01










    • $begingroup$
      @Henrik Schumacher: Your results are about 6 times faster than mine. My platform is Xeon E3-1270, V3, 3.695 MHz; 4 cores, 8 threads; 32 GB ECC memory; Windows 8.1 Prof, 64-bit. I'm not used to such shenanigans. Can you suggest a reason? In any case, thanks for the good work.
      $endgroup$
      – CElliott
      Mar 20 at 16:39
















    14












    $begingroup$

    Here I implemented three different versions of Gaussian filtering (for periodic data). It took me a while to adjust the constants and still some of them might be wrong.



    Prepare the Gaussian kernel



    n = 200000;
    σ = .1;
    t = Subdivide[-1. Pi, 1. Pi, n - 1];
    ker = 1/Sqrt[2 Pi]/ σ Exp[-(t/σ)^2/2];
    ker = Join[ker[[Quotient[n,2] + 1 ;;]], ker[[;; Quotient[n,2]]]];


    Generate noisy function



    u = Sin[t] + Cos[2 t] + 1.5 Cos[3 t] + .5 RandomReal[{-1, 1}, Length[t]];


    The three methods with their timings. As Niki Estner pointed out, GaussianFilter with the option Method -> "Gaussian" performs much batter than GaussianFilter with the default emthod.



    kerhat = 2 Pi/Sqrt[N@n] Fourier[ker];
    vConvolve = (2. Pi/n) ListConvolve[ker, u, {-1, -1}]; // RepeatedTiming // First
    vFFT = Re[Fourier[InverseFourier[u] kerhat]]; // RepeatedTiming // First
    vFilter = GaussianFilter[u, 1./(Pi) σ n, Padding -> "Periodic"]; // RepeatedTiming // First
    vGaussian = GaussianFilter[u, 1./(Pi) σ n, Padding -> "Periodic", Method -> "Gaussian"]; // RepeatedTiming // First



    0.0038



    0.0058



    0.055



    0.0072




    ListLinePlot[{u, vFFT, vFilter, vConvolve}]


    enter image description here



    From further experiments with different values for n, GaussianFilter seems to be slower by a factor 10-20 over a wide range of n (from n = 1000 to n = 1000000). So it seems that it does use some FFT-based method (because it has the same speed asymptotics) but maybe some crucial part of the algorithm is not compiled (the factor 10 is somewhat an indicator for that) or does not use the fastest FFT implementation possible. A bit weird.



    So, to my own surprise, your idea of computing the kernel once does help but for quite unexpected reasons.






    share|improve this answer











    $endgroup$









    • 1




      $begingroup$
      IIRC, it's using a discrete Gaussian kernel, which involves non-compilable modified Bessel functions, so that might be the reason for the added computational effort you observe.
      $endgroup$
      – J. M. is slightly pensive
      Mar 15 at 16:35










    • $begingroup$
      Yes, that is one reason I proposed using GaussianFilter itself to generate the kernel.
      $endgroup$
      – Leon Avery
      Mar 15 at 16:36






    • 1




      $begingroup$
      If I add Method -> "Gaussian" to the GaussianFilter call, it's about as fast as the other two
      $endgroup$
      – Niki Estner
      Mar 16 at 12:32






    • 2




      $begingroup$
      @NikiEstner Ah, that's good to know! Somewhat unexpected that we have to call out twice for a "Gaussian" in order to get one, isn't it?
      $endgroup$
      – Henrik Schumacher
      Mar 16 at 13:01










    • $begingroup$
      @Henrik Schumacher: Your results are about 6 times faster than mine. My platform is Xeon E3-1270, V3, 3.695 MHz; 4 cores, 8 threads; 32 GB ECC memory; Windows 8.1 Prof, 64-bit. I'm not used to such shenanigans. Can you suggest a reason? In any case, thanks for the good work.
      $endgroup$
      – CElliott
      Mar 20 at 16:39














    14












    14








    14





    $begingroup$

    Here I implemented three different versions of Gaussian filtering (for periodic data). It took me a while to adjust the constants and still some of them might be wrong.



    Prepare the Gaussian kernel



    n = 200000;
    σ = .1;
    t = Subdivide[-1. Pi, 1. Pi, n - 1];
    ker = 1/Sqrt[2 Pi]/ σ Exp[-(t/σ)^2/2];
    ker = Join[ker[[Quotient[n,2] + 1 ;;]], ker[[;; Quotient[n,2]]]];


    Generate noisy function



    u = Sin[t] + Cos[2 t] + 1.5 Cos[3 t] + .5 RandomReal[{-1, 1}, Length[t]];


    The three methods with their timings. As Niki Estner pointed out, GaussianFilter with the option Method -> "Gaussian" performs much batter than GaussianFilter with the default emthod.



    kerhat = 2 Pi/Sqrt[N@n] Fourier[ker];
    vConvolve = (2. Pi/n) ListConvolve[ker, u, {-1, -1}]; // RepeatedTiming // First
    vFFT = Re[Fourier[InverseFourier[u] kerhat]]; // RepeatedTiming // First
    vFilter = GaussianFilter[u, 1./(Pi) σ n, Padding -> "Periodic"]; // RepeatedTiming // First
    vGaussian = GaussianFilter[u, 1./(Pi) σ n, Padding -> "Periodic", Method -> "Gaussian"]; // RepeatedTiming // First



    0.0038



    0.0058



    0.055



    0.0072




    ListLinePlot[{u, vFFT, vFilter, vConvolve}]


    enter image description here



    From further experiments with different values for n, GaussianFilter seems to be slower by a factor 10-20 over a wide range of n (from n = 1000 to n = 1000000). So it seems that it does use some FFT-based method (because it has the same speed asymptotics) but maybe some crucial part of the algorithm is not compiled (the factor 10 is somewhat an indicator for that) or does not use the fastest FFT implementation possible. A bit weird.



    So, to my own surprise, your idea of computing the kernel once does help but for quite unexpected reasons.






    share|improve this answer











    $endgroup$



    Here I implemented three different versions of Gaussian filtering (for periodic data). It took me a while to adjust the constants and still some of them might be wrong.



    Prepare the Gaussian kernel



    n = 200000;
    σ = .1;
    t = Subdivide[-1. Pi, 1. Pi, n - 1];
    ker = 1/Sqrt[2 Pi]/ σ Exp[-(t/σ)^2/2];
    ker = Join[ker[[Quotient[n,2] + 1 ;;]], ker[[;; Quotient[n,2]]]];


    Generate noisy function



    u = Sin[t] + Cos[2 t] + 1.5 Cos[3 t] + .5 RandomReal[{-1, 1}, Length[t]];


    The three methods with their timings. As Niki Estner pointed out, GaussianFilter with the option Method -> "Gaussian" performs much batter than GaussianFilter with the default emthod.



    kerhat = 2 Pi/Sqrt[N@n] Fourier[ker];
    vConvolve = (2. Pi/n) ListConvolve[ker, u, {-1, -1}]; // RepeatedTiming // First
    vFFT = Re[Fourier[InverseFourier[u] kerhat]]; // RepeatedTiming // First
    vFilter = GaussianFilter[u, 1./(Pi) σ n, Padding -> "Periodic"]; // RepeatedTiming // First
    vGaussian = GaussianFilter[u, 1./(Pi) σ n, Padding -> "Periodic", Method -> "Gaussian"]; // RepeatedTiming // First



    0.0038



    0.0058



    0.055



    0.0072




    ListLinePlot[{u, vFFT, vFilter, vConvolve}]


    enter image description here



    From further experiments with different values for n, GaussianFilter seems to be slower by a factor 10-20 over a wide range of n (from n = 1000 to n = 1000000). So it seems that it does use some FFT-based method (because it has the same speed asymptotics) but maybe some crucial part of the algorithm is not compiled (the factor 10 is somewhat an indicator for that) or does not use the fastest FFT implementation possible. A bit weird.



    So, to my own surprise, your idea of computing the kernel once does help but for quite unexpected reasons.







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Mar 17 at 13:15

























    answered Mar 15 at 16:24









    Henrik SchumacherHenrik Schumacher

    57.6k578158




    57.6k578158








    • 1




      $begingroup$
      IIRC, it's using a discrete Gaussian kernel, which involves non-compilable modified Bessel functions, so that might be the reason for the added computational effort you observe.
      $endgroup$
      – J. M. is slightly pensive
      Mar 15 at 16:35










    • $begingroup$
      Yes, that is one reason I proposed using GaussianFilter itself to generate the kernel.
      $endgroup$
      – Leon Avery
      Mar 15 at 16:36






    • 1




      $begingroup$
      If I add Method -> "Gaussian" to the GaussianFilter call, it's about as fast as the other two
      $endgroup$
      – Niki Estner
      Mar 16 at 12:32






    • 2




      $begingroup$
      @NikiEstner Ah, that's good to know! Somewhat unexpected that we have to call out twice for a "Gaussian" in order to get one, isn't it?
      $endgroup$
      – Henrik Schumacher
      Mar 16 at 13:01










    • $begingroup$
      @Henrik Schumacher: Your results are about 6 times faster than mine. My platform is Xeon E3-1270, V3, 3.695 MHz; 4 cores, 8 threads; 32 GB ECC memory; Windows 8.1 Prof, 64-bit. I'm not used to such shenanigans. Can you suggest a reason? In any case, thanks for the good work.
      $endgroup$
      – CElliott
      Mar 20 at 16:39














    • 1




      $begingroup$
      IIRC, it's using a discrete Gaussian kernel, which involves non-compilable modified Bessel functions, so that might be the reason for the added computational effort you observe.
      $endgroup$
      – J. M. is slightly pensive
      Mar 15 at 16:35










    • $begingroup$
      Yes, that is one reason I proposed using GaussianFilter itself to generate the kernel.
      $endgroup$
      – Leon Avery
      Mar 15 at 16:36






    • 1




      $begingroup$
      If I add Method -> "Gaussian" to the GaussianFilter call, it's about as fast as the other two
      $endgroup$
      – Niki Estner
      Mar 16 at 12:32






    • 2




      $begingroup$
      @NikiEstner Ah, that's good to know! Somewhat unexpected that we have to call out twice for a "Gaussian" in order to get one, isn't it?
      $endgroup$
      – Henrik Schumacher
      Mar 16 at 13:01










    • $begingroup$
      @Henrik Schumacher: Your results are about 6 times faster than mine. My platform is Xeon E3-1270, V3, 3.695 MHz; 4 cores, 8 threads; 32 GB ECC memory; Windows 8.1 Prof, 64-bit. I'm not used to such shenanigans. Can you suggest a reason? In any case, thanks for the good work.
      $endgroup$
      – CElliott
      Mar 20 at 16:39








    1




    1




    $begingroup$
    IIRC, it's using a discrete Gaussian kernel, which involves non-compilable modified Bessel functions, so that might be the reason for the added computational effort you observe.
    $endgroup$
    – J. M. is slightly pensive
    Mar 15 at 16:35




    $begingroup$
    IIRC, it's using a discrete Gaussian kernel, which involves non-compilable modified Bessel functions, so that might be the reason for the added computational effort you observe.
    $endgroup$
    – J. M. is slightly pensive
    Mar 15 at 16:35












    $begingroup$
    Yes, that is one reason I proposed using GaussianFilter itself to generate the kernel.
    $endgroup$
    – Leon Avery
    Mar 15 at 16:36




    $begingroup$
    Yes, that is one reason I proposed using GaussianFilter itself to generate the kernel.
    $endgroup$
    – Leon Avery
    Mar 15 at 16:36




    1




    1




    $begingroup$
    If I add Method -> "Gaussian" to the GaussianFilter call, it's about as fast as the other two
    $endgroup$
    – Niki Estner
    Mar 16 at 12:32




    $begingroup$
    If I add Method -> "Gaussian" to the GaussianFilter call, it's about as fast as the other two
    $endgroup$
    – Niki Estner
    Mar 16 at 12:32




    2




    2




    $begingroup$
    @NikiEstner Ah, that's good to know! Somewhat unexpected that we have to call out twice for a "Gaussian" in order to get one, isn't it?
    $endgroup$
    – Henrik Schumacher
    Mar 16 at 13:01




    $begingroup$
    @NikiEstner Ah, that's good to know! Somewhat unexpected that we have to call out twice for a "Gaussian" in order to get one, isn't it?
    $endgroup$
    – Henrik Schumacher
    Mar 16 at 13:01












    $begingroup$
    @Henrik Schumacher: Your results are about 6 times faster than mine. My platform is Xeon E3-1270, V3, 3.695 MHz; 4 cores, 8 threads; 32 GB ECC memory; Windows 8.1 Prof, 64-bit. I'm not used to such shenanigans. Can you suggest a reason? In any case, thanks for the good work.
    $endgroup$
    – CElliott
    Mar 20 at 16:39




    $begingroup$
    @Henrik Schumacher: Your results are about 6 times faster than mine. My platform is Xeon E3-1270, V3, 3.695 MHz; 4 cores, 8 threads; 32 GB ECC memory; Windows 8.1 Prof, 64-bit. I'm not used to such shenanigans. Can you suggest a reason? In any case, thanks for the good work.
    $endgroup$
    – CElliott
    Mar 20 at 16:39











    7












    $begingroup$

    It's hard to know for sure, but one way to test for caching is to apply a single command to lots of data sets, or to apply the command separately to each set. For instance:



    n = 5000;
    data = RandomReal[{-1, 1}, {n, 10000}];
    GaussianFilter[#, 100] & /@ data; // AbsoluteTiming
    Do[GaussianFilter[data[[i]], 100], {i, n}] // AbsoluteTiming
    Do[GaussianFilter[data[[i]], 100 + RandomInteger[{-15, 15}]], {i, n}] // AbsoluteTiming


    The second line generates 5000 different sets of data, each 10000 length. The third applies one Gaussian filter to all the data sets. The third line applies a separate GaussianFilter to each set. The final line forces the GaussianFilter to recompute its kernel. The timings are pretty much the same. This suggests that whatever is happening, the time needed to calculate the Gaussian filter parameters is pretty negligeable.






    share|improve this answer









    $endgroup$


















      7












      $begingroup$

      It's hard to know for sure, but one way to test for caching is to apply a single command to lots of data sets, or to apply the command separately to each set. For instance:



      n = 5000;
      data = RandomReal[{-1, 1}, {n, 10000}];
      GaussianFilter[#, 100] & /@ data; // AbsoluteTiming
      Do[GaussianFilter[data[[i]], 100], {i, n}] // AbsoluteTiming
      Do[GaussianFilter[data[[i]], 100 + RandomInteger[{-15, 15}]], {i, n}] // AbsoluteTiming


      The second line generates 5000 different sets of data, each 10000 length. The third applies one Gaussian filter to all the data sets. The third line applies a separate GaussianFilter to each set. The final line forces the GaussianFilter to recompute its kernel. The timings are pretty much the same. This suggests that whatever is happening, the time needed to calculate the Gaussian filter parameters is pretty negligeable.






      share|improve this answer









      $endgroup$
















        7












        7








        7





        $begingroup$

        It's hard to know for sure, but one way to test for caching is to apply a single command to lots of data sets, or to apply the command separately to each set. For instance:



        n = 5000;
        data = RandomReal[{-1, 1}, {n, 10000}];
        GaussianFilter[#, 100] & /@ data; // AbsoluteTiming
        Do[GaussianFilter[data[[i]], 100], {i, n}] // AbsoluteTiming
        Do[GaussianFilter[data[[i]], 100 + RandomInteger[{-15, 15}]], {i, n}] // AbsoluteTiming


        The second line generates 5000 different sets of data, each 10000 length. The third applies one Gaussian filter to all the data sets. The third line applies a separate GaussianFilter to each set. The final line forces the GaussianFilter to recompute its kernel. The timings are pretty much the same. This suggests that whatever is happening, the time needed to calculate the Gaussian filter parameters is pretty negligeable.






        share|improve this answer









        $endgroup$



        It's hard to know for sure, but one way to test for caching is to apply a single command to lots of data sets, or to apply the command separately to each set. For instance:



        n = 5000;
        data = RandomReal[{-1, 1}, {n, 10000}];
        GaussianFilter[#, 100] & /@ data; // AbsoluteTiming
        Do[GaussianFilter[data[[i]], 100], {i, n}] // AbsoluteTiming
        Do[GaussianFilter[data[[i]], 100 + RandomInteger[{-15, 15}]], {i, n}] // AbsoluteTiming


        The second line generates 5000 different sets of data, each 10000 length. The third applies one Gaussian filter to all the data sets. The third line applies a separate GaussianFilter to each set. The final line forces the GaussianFilter to recompute its kernel. The timings are pretty much the same. This suggests that whatever is happening, the time needed to calculate the Gaussian filter parameters is pretty negligeable.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Mar 15 at 15:31









        bill sbill s

        54.7k377157




        54.7k377157






























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