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Sep 10, 2017 at 14:23 comment added Jonathan Frech You can replace for s in(1,-1)for with for s in 1,-1for or even, if order does not matter, for s in-1,1for.
Aug 12, 2017 at 14:26 comment added miles 101 bytes with f=lambda x:x*(len(x)<2)or[u+v/1j**(4*i/len(x))for i,(u,v)in enumerate(zip(f(x[::2])*2,f(x[1::2])*2))]
Apr 9, 2014 at 3:52 comment added boothby It appears that I cannot downvote my own post. You are correct, I exchanged the loop order and killed the performance. I will leave this up for now, in case I find a way to fix it.
Apr 2, 2014 at 20:38 comment added jakevdp I think that this is no longer a fast fourier transform, though... by "killing t" you added in some unnecessary calculations which move it from O[N log(N)] to O[N^2]
S Sep 8, 2013 at 18:32 history answered boothby CC BY-SA 3.0
S Sep 8, 2013 at 18:32 history made wiki Post Made Community Wiki by boothby