[SciPy-user] FFT speed ?

Anne Archibald peridot.faceted@gmail....
Wed Feb 7 13:15:23 CST 2007

On 07/02/07, Scott Ransom <sransom@nrao.edu> wrote:
> On Wed, Feb 07, 2007 at 12:13:12PM -0500, Anne Archibald wrote:

> > Can FFTW (or any of the FFT packages numpy/scipy can use) compute an
> > FFT of size 186889 in a reasonable time? I know there are algorithms
> > for large prime factors, and for small prime factors, and that you can
> > combine the two (though perhaps primes of moderate size are a
> > problem).
> I know that FFTW uses O(NlogN) algorithms for any N, even large
> prime factors.  It is quite likely that for large prime N, FFTPACK
> (which is what numpy uses) goes to a standard DFT algorithm
> (O(N^2)).

Indeed, for numbers in the vicinity of 10000, some take 500 times as
long as others (on my machine), so I suspect that it is falling back
to a DFT.

> One important thing to remember is that the constant in front of
> the NlogN is highly dependent on the algorithm.  That is why even
> though FFTW v3 uses NlogN algorithms for all N, some N (like
> powers of 2 and those composed of only small prime factors) are
> _much_ faster than those for other N.  But the bottom line is that
> no matter what the constant, for large N, O(NlogN) is _much_ faster
> than O(N^2).

Of course. But when you say the constant varies, do you mean by a
factor of ten? a hundred? a thousand?

On my machine, scipy.show_config reports that it can find FFTW2 but
not FFTW3; does that mean it is actually *using* FFTW2? How does one

Does scipy provide any access to the special features of FFTW's
interface? (wisdom, for efficiently computing many FFTs on the same
array, for example)

Anne M. Archibald

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