Hi, I cannot see how the following would work when it is np.fft.fft() that takes a long time based on the length of data. In my case my data is non-periodic.> from numpy.fft import fft> from numpy.random import rand> from math import log, ceil> seq_A = rand(2649674)> seq_B = rand(2646070)> fft_A = fft(seq_A) #Long> fft_B = fft(seq_B)>zeropadded_fft_A = fft(seq_A, n=2**(ceil(log(len(seq_A),2))+1))>zeropadded_fft_B = fft(seq_B, n=2**(ceil(log(len(seq_B),2))+1))Ideally I need to calculate a favourable length to pad, prior to calling np.fft.fft() as Stéfan pointed out by calculating the factors. In : from sympy import factorint In : max(factorint(2646070)) Out: 367 In : max(factorint(2649674)) Out: 1324837 I will try adding some code to calculate the next power of two above my array length as you suggest;> And while you zero-pad, you can zero-pad to a sequence that is a power of two, thus preventing awkward factorizations.Does numpy have an easy way to do this, i.e. for a given number, find the next highest number (within a range) that could be factored into small, prime numbers as Phil explained? It would help if it gave a list, prioritised by number of factors.Thanks,Joseph Date: Fri, 28 Aug 2015 17:26:32 -0700 From: email@example.com To: firstname.lastname@example.org Subject: Re: [Numpy-discussion] Numpy FFT.FFT slow with certain samples
On Aug 28, 2015 5:17 PM, "Pierre-Andre Noel" email@example.com wrote:
I had in mind the use of FFT to do convolutions (
https://en.wikipedia.org/wiki/Convolution_theorem ). If you do not
zero-pad properly, then the end of the signal may "bleed" on the
beginning, and vice versa.
Ah, gotcha! All these things should also be handled nicely in scipy.signal.fftconvolve. Stéfan
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