#
# SIMPLE_FFT.py
# For Python Version(s) 2.0.x to 3.7.0...
# 20 August 2018, CC0, Public Domain Licence.
#
# TESTED ON:
#
# Python 2.0 (#1, Oct 29 2000, 23:53:20)  (SAS/C 6.x] on amiga
#
# Python 2.7.10 (default, Oct  6 2017, 22:29:07)
# [GCC 4.2.1 Compatible Apple LLVM 9.0.0 (clang-900.0.31)] on darwin
#
# Python 3.5.2 (v3.5.2:4def2a2901a5, Jun 26 2016, 10:47:25)
# [GCC 4.2.1 (Apple Inc. build 5666) (dot 3)] on darwin
#
# Python 3.7.0 (v7.7.0:1bf9cc5093, Jun 27 2018,04:59:51) [MSC v.1914 64 bit
# (AMD64)] on win32
#
# Python 2.7.15rc1 (default, Apr 15 2018, 21:51:34)
# [GCC 7.3.0] on linux2

# This is a builtin, no external libraries required.
import cmath

def fft(DATA):
	N=len(DATA)
	if N<=1: return DATA
	EVEN=fft([DATA[K] for K in range(0,N,2)])
	ODD=fft([DATA[K] for K in range(1,N,2)])
	L=[EVEN[K]+cmath.exp(-2j*cmath.pi*K/N)*ODD[K] for K in range(int(N/2))]
	R=[EVEN[K]-cmath.exp(-2j*cmath.pi*K/N)*ODD[K] for K in range(int(N/2))]
	return L+R

# NOTE: Although not necessary, for best results, DATA sizes be in powers of 2.
# http://www.bitweenie.com/listings/fft-zero-padding/
# Single cycle square wave, 8 samples.
FFT_LIST=[1.0, 1.0, 1.0, 1.0, 0.0, 0.0, 0.0, 0.0]

FFT=fft(FFT_LIST)
print(FFT)

# Correct results, within limits of floating point accuracy, should read:
# -----------------------------------------------------------------------
#
# [(4+0j), (1-2.414213562373095j), 0j, (1-0.4142135623730949j), 0j,
# (0.9999999999999999+0.4142135623730949j), 0j,
# (0.9999999999999997+2.414213562373095j)]