Replacing numpy.fft routines with pyfftw, not working as expected

Question

I have some working python code making use of the numpy.fft package, here is a snippet:

for i in range(steps):
    print i
    psixvec = Ux * psixvec
    psikvec = Uk * np.fft.fftn(psixvec)
    psixvec = np.fft.ifftn(psikvec)

return psixvec

I tried to rewrite this code to make use of the pyfftw package. What I came up with is the following code, which should work:

fft = fftw.builders.fftn(psix_align, auto_align_input = True, auto_contiguous = True,
                      overwrite_input = False, threads = 1, avoid_copy = False)

ifft = fftw.builders.ifftn(psik_align, auto_align_input = True, auto_contiguous = True,
                      overwrite_input = False, threads = 1, avoid_copy = False) 


for i in range(steps):
    psix_align[:] = Ux * psix_align
    psik_align[:] = Uk * fft()
    psix_align[:] = ifft()

return psix_align

The problem is, this code does not produce the same result as using the numpy.fft package. See the attached images.

numpy fft package pyfftw package

Which image is the result of which computation? – Bakuriu Sep 04 '13 at 14:34 — Bakuriu, Sep 04 '13 at 14:34
First image is numpy, second is pyfftw. – Micha Sep 04 '13 at 14:37 — Micha, Sep 04 '13 at 14:37

score 3 · Accepted Answer · answered Sep 04 '13 at 18:16

Solved. For the initialization I was using

psix_align = fftw.n_byte_align(psi0, fftw.simd_alignment, dtype='complex64')
psik_align = fftw.n_byte_align(np.zeros_like(psi0), fftw.simd_alignment, dtype='complex64')

I needed to replace complex64 with complex128. Now I get the same result. That's probably due to the fact that the numbers involved are very small (see the 1e-11 on the z-axis).

Edit: Maybe someone can add pyfftw to the tags of the question?

Replacing numpy.fft routines with pyfftw, not working as expected

1 Answers1