[SciPy-dev] fftfreq very slow; rfftfreq incorrect?
a.h.jaffe at gmail.com
Wed Aug 30 14:00:19 CDT 2006
Stefan van der Walt wrote:
> On Wed, Aug 30, 2006 at 12:04:22PM +0100, Andrew Jaffe wrote:
>> the current implementation of fftfreq (which is meant to return the
>> appropriate frequencies for an FFT) does the following:
>> k = range(0,(n-1)/2+1)+range(-(n/2),0)
>> return array(k,'d')/(n*d)
>> I have tried this with very long (2**24) arrays, and it is ridiculously
>> slow. Should this instead use arange (or linspace?) and concatenate
>> rather than converting the above list? This seems to result in
>> acceptable performance, but we could also perhaps even pre-allocate the
> Please try the attached benchmark.
Results attached. Bottom line is that both new versions are several
times faster than the old, and the concat (hstack) version is somewhat
faster than the other, but it depends on n.
I'm on OSX with the 2.4.3 universal build on PPC and the latest SVN numpy.
>> The numpy.fft.rfftfreq seems just plain incorrect to me. It seems to
>> produce lots of duplicated frequencies, contrary to the actual output of
<< removed >>
> Please produce a code snippet to demonstrate the problem. We can then
> fix the bug and use your code as a unit test.
Aha, the problem is that scipy and numpy define rfft differently!
numpy returns n/2+1 complex numbers (so the first and last numbers are
actually real) with the frequencies equivalent to the positive part of
the fftfreq, whereas scipy returns n real numbers with the frequencies
as in rfftfreq.
I think the numpy behavior makes more sense, as it doesn't require any
unpacking after the fact, at the expense of a tiny amount of wasted
space. But would this in fact require scipy doing extra work from
whatever the 'native' real_fft (fftw, I assume) produces?
Anyone else have an opinion?
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