[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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