What does IFFT return in Python?

Question

I need the inverse Fourier transform of a complex array. ifft should return a real array, but it returns another complex array.

In MATLAB, a=ifft(fft(a)), but in Python it does not work like that.

a = np.arange(6)
m = ifft(fft(a))
m # Google says m should = a, but m is complex

Output :

array([0.+0.00000000e+00j, 1.+3.70074342e-16j, 2.+0.00000000e+00j,
       3.-5.68396583e-17j, 4.+0.00000000e+00j, 5.-3.13234683e-16j])

Answer 1

The imaginary part is result floating precision number calculation error. If it is very small, it rather can be dropped.

Numpy has built-in function real_if_close, to do so:

>>> np.real_if_close(np.fft.ifft(np.fft.fft(a)))
array([0., 1., 2., 3., 4., 5.])

You can read about floating system limitations here: https://docs.python.org/3.8/tutorial/floatingpoint.html

Answer 2

if the imaginary part is close to zero you could discard it:

import numpy as np

arr = np.array(
    [
        0.0 + 0.00000000e00j,
        1.0 + 3.70074342e-16j,
        2.0 + 0.00000000e00j,
        3.0 - 5.68396583e-17j,
        4.0 + 0.00000000e00j,
        5.0 - 3.13234683e-16j,
    ]
)

if all(np.isclose(arr.imag, 0)):
    arr = arr.real
# [ 0.  1.  2.  3.  4.  5.]

(that's what real_if_close does in one line as in R2RT's answer).