pyfftw rfft/irfft not symmetric for odd number of samples?

Question

I'm trying to call forward and inverse rfft using the pyfftw module.
For even number of samples, irfft(rfft) = identity as I was expecting,
But this is not true for an odd number of samples, why?

example below :

import numpy as np
from pyfftw.builders import rfft as rfft_builder, irfft as irfft_builder
from pyfftw import empty_aligned

n = 5  # number of samples

# >>>>> forward
a = empty_aligned((n, ), dtype="float32")
rfft = rfft_builder(a)

a[:] = np.arange(n)
b = rfft()
print(a, "=>", b)

# <<<<< backward
B = empty_aligned((len(b), ), dtype=b.dtype)
irfft = irfft_builder(B)

B[:] = b[:]
A = irfft()
print(B, "=>", A)

For n=4 => ok

[0. 1. 2. 3.] => [ 6.+0.j -2.+2.j -2.+0.j]
[ 6.+0.j -2.+2.j -2.+0.j] => [0. 1. 2. 3.]

For n=5 => ???

[0. 1. 2. 3. 4.] => [10. +0.j        -2.5+3.440955j  -2.5+0.8122992j]
[10. +0.j        -2.5+3.440955j  -2.5+0.8122992j] => [0.625     1.4045225 3.125     4.8454776]

Answer 1

This matches the behavior of np.fft.irfft() , which should be called as np.fft.irfft(b, len(a)) to get the original a back. The numpy docs go into more detail in the Notes section.

For pyFFTW, building the inverse transform with irfft = irfft_builder(B, n) should fix it.