Does Mathematica reuse previous computations?
$begingroup$
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 ListConvolve
s instead of hundreds of GaussianFilter
s?
numerics convolution caching
$endgroup$
|
show 2 more comments
$begingroup$
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 ListConvolve
s instead of hundreds of GaussianFilter
s?
numerics convolution caching
$endgroup$
2
$begingroup$
My first guess is thatGaussianFilter
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 theRepeatedTimings
.
$endgroup$
– 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
|
show 2 more comments
$begingroup$
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 ListConvolve
s instead of hundreds of GaussianFilter
s?
numerics convolution caching
$endgroup$
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 ListConvolve
s instead of hundreds of GaussianFilter
s?
numerics convolution caching
numerics convolution caching
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 thatGaussianFilter
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 theRepeatedTimings
.
$endgroup$
– 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
|
show 2 more comments
2
$begingroup$
My first guess is thatGaussianFilter
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 theRepeatedTimings
.
$endgroup$
– 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
|
show 2 more comments
2 Answers
2
active
oldest
votes
$begingroup$
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}]
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.
$endgroup$
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 addMethod -> "Gaussian"
to theGaussianFilter
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
|
show 1 more comment
$begingroup$
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.
$endgroup$
add a comment |
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2 Answers
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2 Answers
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active
oldest
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oldest
votes
$begingroup$
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}]
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.
$endgroup$
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 addMethod -> "Gaussian"
to theGaussianFilter
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
|
show 1 more comment
$begingroup$
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}]
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.
$endgroup$
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 addMethod -> "Gaussian"
to theGaussianFilter
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
|
show 1 more comment
$begingroup$
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}]
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.
$endgroup$
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}]
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.
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 addMethod -> "Gaussian"
to theGaussianFilter
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
|
show 1 more comment
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 addMethod -> "Gaussian"
to theGaussianFilter
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
|
show 1 more comment
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
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.
answered Mar 15 at 15:31
bill sbill s
54.7k377157
54.7k377157
add a comment |
add a comment |
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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