[Numpy-discussion] fftfreq very slow; rfftfreq incorrect?
Andrew Jaffe
a.h.jaffe at gmail.com
Wed Aug 30 06:17:51 CDT 2006
[copied to the scipy list since rfftfreq is only in scipy]
Andrew Jaffe wrote:
> Hi all,
>
> 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
> space.
>
> The numpy.fft.rfftfreq seems just plain incorrect to me. It seems to
> produce lots of duplicated frequencies, contrary to the actual output of
> rfft:
>
> def rfftfreq(n,d=1.0):
> """ rfftfreq(n, d=1.0) -> f
>
> DFT sample frequencies (for usage with rfft,irfft).
>
> The returned float array contains the frequency bins in
> cycles/unit (with zero at the start) given a window length n and a
> sample spacing d:
>
> f = [0,1,1,2,2,...,n/2-1,n/2-1,n/2]/(d*n) if n is even
> f = [0,1,1,2,2,...,n/2-1,n/2-1,n/2,n/2]/(d*n) if n is odd
>
> **** None of these should be doubled, right?
>
> """
> assert isinstance(n,int)
> return array(range(1,n+1),dtype=int)/2/float(n*d)
>
> Thanks,
>
> Andrew
>
>
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