The scipy.fftpack.rfft function returns the DFT as a vector of floats, alternating between the real and complex part. This means to multiply to DFTs together (for convolution) I will have to do the complex multiplication "manually" which seems quite tricky. This must be something people do often - I presume/hope there is a simple trick to do this efficiently that I haven't spotted?
Basically I want to fix this code so that both methods give the same answer:
import numpy as np
import scipy.fftpack as sfft
X = np.random.normal(size = 2000)
Y = np.random.normal(size = 2000)
NZ = np.fft.irfft(np.fft.rfft(Y) * np.fft.rfft(X))
SZ = sfft.irfft(sfft.rfft(Y) * sfft.rfft(X)) # This multiplication is wrong
NZ
array([-43.23961083, 53.62608086, 17.92013729, ..., -16.57605207,
8.19605764, 5.23929023])
SZ
array([-19.90115323, 16.98680347, -8.16608202, ..., -47.01643274,
-3.50572376, 58.1961597 ])
N.B. I am aware that fftpack contains a convolve function, but I only need to fft one half of the transform - my filter can be fft'd once in advance and then used over and over again.
You don't have to flip back to np.float64 and hstack. You can create an empty destination array, the same shape as sfft.rfft(Y) and sfft.rfft(X), then create a np.complex128 view of it and fill this view with the result of the multiplication. This will automatically fill the destination array as wanted.
If I retake your example :
import numpy as np
import scipy.fftpack as sfft
X = np.random.normal(size = 2000)
Y = np.random.normal(size = 2000)
Xf = np.fft.rfft(X)
Xf_cpx = Xf[1:-1].view(np.complex128)
Yf = np.fft.rfft(Y)
Yf_cpx = Yf[1:-1].view(np.complex128)
Zf = np.empty(X.shape)
Zf_cpx = Zf[1:-1].view(np.complex128)
Zf[0] = Xf[0]*Yf[0]
# the [...] is important to use the view as a reference to Zf and not overwrite it
Zf_cpx[...] = Xf_cpx * Yf_cpx
Zf[-1] = Xf[-1]*Yf[-1]
Z = sfft.irfft.irfft(Zf)
and that's it! You can use a simple if statement if you want your code to be more general and handle odd lengths as explained in Jaime's answer. Here is a function that does what you want:
def rfft_mult(a,b):
"""Multiplies two outputs of scipy.fftpack.rfft"""
assert a.shape == b.shape
c = np.empty( a.shape )
c[...,0] = a[...,0]*b[...,0]
# To comply with the rfft support of multi dimensional arrays
ar = a.reshape(-1,a.shape[-1])
br = b.reshape(-1,b.shape[-1])
cr = c.reshape(-1,c.shape[-1])
# Note that we cannot use ellipses to achieve that because of
# the way `view` work. If there are many dimensions, one should
# consider to manually perform the complex multiplication with slices.
if c.shape[-1] & 0x1: # if odd
for i in range(len(ar)):
ac = ar[i,1:].view(np.complex128)
bc = br[i,1:].view(np.complex128)
cc = cr[i,1:].view(np.complex128)
cc[...] = ac*bc
else:
for i in range(len(ar)):
ac = ar[i,1:-1].view(np.complex128)
bc = br[i,1:-1].view(np.complex128)
cc = cr[i,1:-1].view(np.complex128)
cc[...] = ac*bc
c[...,-1] = a[...,-1]*b[...,-1]
return c