Question

I would like to use scipy’s DCT-II since it is already coded and fast. Looking at the doc, it seems it is the 1D implementation. Is it possible to use it in such a way to use it as a 3D implementation? I’m not sure about the math. Are the 2D and 3D implementations the equivalent of multiplying 2 or 3 times the 1D using different dimensions in the calculation?

Answer 1

Basically, the following does the trick:

import numpy as np
from scipy.fftpack import dct, idct

# Lets create a 3D array and fill it with some values
a = np.random.rand(3,3,3)

b = dct(dct(dct(a).transpose(0,2,1)).transpose(1,2,0)).transpose(1,2,0).transpose(0,2,1)