Does Mathematica reuse previous computations?












6












$begingroup$


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
    yesterday








  • 2




    $begingroup$
    Best way would be to just try it and compare the RepeatedTimings.
    $endgroup$
    – Henrik Schumacher
    yesterday






  • 3




    $begingroup$
    @HenrikSchumacher I'm pretty sure ListConvolve also uses FFT (based on its performance).
    $endgroup$
    – Szabolcs
    yesterday






  • 1




    $begingroup$
    ListConvolve does indeed use FFT.
    $endgroup$
    – Leon Avery
    yesterday








  • 4




    $begingroup$
    Indeed it does.
    $endgroup$
    – J. M. is slightly pensive
    yesterday
















6












$begingroup$


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
    yesterday








  • 2




    $begingroup$
    Best way would be to just try it and compare the RepeatedTimings.
    $endgroup$
    – Henrik Schumacher
    yesterday






  • 3




    $begingroup$
    @HenrikSchumacher I'm pretty sure ListConvolve also uses FFT (based on its performance).
    $endgroup$
    – Szabolcs
    yesterday






  • 1




    $begingroup$
    ListConvolve does indeed use FFT.
    $endgroup$
    – Leon Avery
    yesterday








  • 4




    $begingroup$
    Indeed it does.
    $endgroup$
    – J. M. is slightly pensive
    yesterday














6












6








6


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






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked yesterday









Leon AveryLeon Avery

695318




695318








  • 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
    yesterday








  • 2




    $begingroup$
    Best way would be to just try it and compare the RepeatedTimings.
    $endgroup$
    – Henrik Schumacher
    yesterday






  • 3




    $begingroup$
    @HenrikSchumacher I'm pretty sure ListConvolve also uses FFT (based on its performance).
    $endgroup$
    – Szabolcs
    yesterday






  • 1




    $begingroup$
    ListConvolve does indeed use FFT.
    $endgroup$
    – Leon Avery
    yesterday








  • 4




    $begingroup$
    Indeed it does.
    $endgroup$
    – J. M. is slightly pensive
    yesterday














  • 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
    yesterday








  • 2




    $begingroup$
    Best way would be to just try it and compare the RepeatedTimings.
    $endgroup$
    – Henrik Schumacher
    yesterday






  • 3




    $begingroup$
    @HenrikSchumacher I'm pretty sure ListConvolve also uses FFT (based on its performance).
    $endgroup$
    – Szabolcs
    yesterday






  • 1




    $begingroup$
    ListConvolve does indeed use FFT.
    $endgroup$
    – Leon Avery
    yesterday








  • 4




    $begingroup$
    Indeed it does.
    $endgroup$
    – J. M. is slightly pensive
    yesterday








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
yesterday






$begingroup$
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
yesterday






2




2




$begingroup$
Best way would be to just try it and compare the RepeatedTimings.
$endgroup$
– Henrik Schumacher
yesterday




$begingroup$
Best way would be to just try it and compare the RepeatedTimings.
$endgroup$
– Henrik Schumacher
yesterday




3




3




$begingroup$
@HenrikSchumacher I'm pretty sure ListConvolve also uses FFT (based on its performance).
$endgroup$
– Szabolcs
yesterday




$begingroup$
@HenrikSchumacher I'm pretty sure ListConvolve also uses FFT (based on its performance).
$endgroup$
– Szabolcs
yesterday




1




1




$begingroup$
ListConvolve does indeed use FFT.
$endgroup$
– Leon Avery
yesterday






$begingroup$
ListConvolve does indeed use FFT.
$endgroup$
– Leon Avery
yesterday






4




4




$begingroup$
Indeed it does.
$endgroup$
– J. M. is slightly pensive
yesterday




$begingroup$
Indeed it does.
$endgroup$
– J. M. is slightly pensive
yesterday










2 Answers
2






active

oldest

votes


















12












$begingroup$

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:



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

0.0043

0.0057

0.054

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
    yesterday










  • $begingroup$
    Yes, that is one reason I proposed using GaussianFilter itself to generate the kernel.
    $endgroup$
    – Leon Avery
    yesterday










  • $begingroup$
    If I add Method -> "Gaussian" to the GaussianFilter call, it's about as fast as the other two
    $endgroup$
    – Niki Estner
    12 hours ago






  • 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
    11 hours ago



















5












$begingroup$

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









$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "387"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: false,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: null,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f193328%2fdoes-mathematica-reuse-previous-computations%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    12












    $begingroup$

    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:



    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

    0.0043

    0.0057

    0.054

    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
      yesterday










    • $begingroup$
      Yes, that is one reason I proposed using GaussianFilter itself to generate the kernel.
      $endgroup$
      – Leon Avery
      yesterday










    • $begingroup$
      If I add Method -> "Gaussian" to the GaussianFilter call, it's about as fast as the other two
      $endgroup$
      – Niki Estner
      12 hours ago






    • 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
      11 hours ago
















    12












    $begingroup$

    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:



    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

    0.0043

    0.0057

    0.054

    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
      yesterday










    • $begingroup$
      Yes, that is one reason I proposed using GaussianFilter itself to generate the kernel.
      $endgroup$
      – Leon Avery
      yesterday










    • $begingroup$
      If I add Method -> "Gaussian" to the GaussianFilter call, it's about as fast as the other two
      $endgroup$
      – Niki Estner
      12 hours ago






    • 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
      11 hours ago














    12












    12








    12





    $begingroup$

    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:



    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

    0.0043

    0.0057

    0.054

    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:



    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

    0.0043

    0.0057

    0.054

    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 yesterday

























    answered yesterday









    Henrik SchumacherHenrik Schumacher

    56.7k577157




    56.7k577157








    • 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
      yesterday










    • $begingroup$
      Yes, that is one reason I proposed using GaussianFilter itself to generate the kernel.
      $endgroup$
      – Leon Avery
      yesterday










    • $begingroup$
      If I add Method -> "Gaussian" to the GaussianFilter call, it's about as fast as the other two
      $endgroup$
      – Niki Estner
      12 hours ago






    • 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
      11 hours ago














    • 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
      yesterday










    • $begingroup$
      Yes, that is one reason I proposed using GaussianFilter itself to generate the kernel.
      $endgroup$
      – Leon Avery
      yesterday










    • $begingroup$
      If I add Method -> "Gaussian" to the GaussianFilter call, it's about as fast as the other two
      $endgroup$
      – Niki Estner
      12 hours ago






    • 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
      11 hours ago








    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
    yesterday




    $begingroup$
    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
    yesterday












    $begingroup$
    Yes, that is one reason I proposed using GaussianFilter itself to generate the kernel.
    $endgroup$
    – Leon Avery
    yesterday




    $begingroup$
    Yes, that is one reason I proposed using GaussianFilter itself to generate the kernel.
    $endgroup$
    – Leon Avery
    yesterday












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




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




    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
    11 hours ago




    $begingroup$
    @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
    11 hours ago











    5












    $begingroup$

    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









    $endgroup$


















      5












      $begingroup$

      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









      $endgroup$
















        5












        5








        5





        $begingroup$

        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 yesterday









        bill sbill s

        54.3k377156




        54.3k377156






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematica Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f193328%2fdoes-mathematica-reuse-previous-computations%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            "Incorrect syntax near the keyword 'ON'. (on update cascade, on delete cascade,)

            Alcedinidae

            RAC Tourist Trophy