Using scipy, is there an easy way to emulate the behaviour of MATLAB's dctmtx function which returns a NxN DCT matrix for some given N? There's scipy.fftpack.dctn but that only applies the DCT.
Do I have to implement this from scratch if I don't want use another dependency besides scipy?
The DCT is a linear transformation, so one way to get the matrix for the transformation is to apply it to the identity matrix. Here's an example where I find the matrix for sequences of length 8 (change 8 to N for the general case):
In [124]: import numpy as np
In [125]: from scipy.fft import dct
In [126]: D = dct(np.eye(8), axis=0)
D is the matrix:
In [127]: D
Out[127]:
array([[ 2. , 2. , 2. , 2. , 2. , 2. , 2. , 2. ],
[ 1.96157056, 1.66293922, 1.11114047, 0.39018064, -0.39018064, -1.11114047, -1.66293922, -1.96157056],
[ 1.84775907, 0.76536686, -0.76536686, -1.84775907, -1.84775907, -0.76536686, 0.76536686, 1.84775907],
[ 1.66293922, -0.39018064, -1.96157056, -1.11114047, 1.11114047, 1.96157056, 0.39018064, -1.66293922],
[ 1.41421356, -1.41421356, -1.41421356, 1.41421356, 1.41421356, -1.41421356, -1.41421356, 1.41421356],
[ 1.11114047, -1.96157056, 0.39018064, 1.66293922, -1.66293922, -0.39018064, 1.96157056, -1.11114047],
[ 0.76536686, -1.84775907, 1.84775907, -0.76536686, -0.76536686, 1.84775907, -1.84775907, 0.76536686],
[ 0.39018064, -1.11114047, 1.66293922, -1.96157056, 1.96157056, -1.66293922, 1.11114047, -0.39018064]])
Verify that D @ x is equivalent to dct(x):
In [128]: x = np.array([1, 2, 0, -1, 3, 0, 1, -1])
In [129]: dct(x)
Out[129]: array([10. , 4.02535777, -1.39941754, 7.38025967, -1.41421356, -6.39104653, -7.07401092, 7.51550307])
In [130]: D @ x
Out[130]: array([10. , 4.02535777, -1.39941754, 7.38025967, -1.41421356, -6.39104653, -7.07401092, 7.51550307])
Note that D @ x will generally be much slower than dct(x).
To get exact agreement with Matlab's dctmtx, add the argument norm='ortho'. For example, here's dctmtx in Octave (which returns the same array as in Matlab):
octave:1> pkg load image
octave:2> dctmtx(4)
ans =
0.50000 0.50000 0.50000 0.50000
0.65328 0.27060 -0.27060 -0.65328
0.50000 -0.50000 -0.50000 0.50000
0.27060 -0.65328 0.65328 -0.27060
Here's the calculation using scipy.fft.dct:
In [56]: from scipy.fft import dct
In [57]: dct(np.eye(4), axis=0, norm='ortho')
Out[57]:
array([[ 0.5 , 0.5 , 0.5 , 0.5 ],
[ 0.65328148, 0.27059805, -0.27059805, -0.65328148],
[ 0.5 , -0.5 , -0.5 , 0.5 ],
[ 0.27059805, -0.65328148, 0.65328148, -0.27059805]])