Scipy's Fftpack Dct And Idct

Let say you use the dct function, then do no manipulation of the data and use the invert transform; wouldn't the inverted data be the same as the pre-transformed data? Why the floa

Solution 1:

It looks like dct and idct do not normalize by default. define dct to call fftpack.dct in the following manner. Do the same for idct.

In [13]: dct = lambda x: fftpack.dct(x, norm='ortho')

In [14]: idct = lambda x: fftpack.idct(x, norm='ortho')

Once done, you will get back the original answers after performing the transforms.

In [19]: import numpy

In [20]: a = numpy.random.rand(3,3,3)

In [21]: d = dct(dct(dct(a).transpose(0,2,1)).transpose(2,1,0)).transpose(2,1,0).transpose(0,2,1)

In [22]: e = idct(idct(idct(d).transpose(0,2,1)).transpose(2,1,0)).transpose(2,1,0).transpose(0,2,1)

In [23]: a
Out[23]: 
array([[[ 0.51699637,  0.42946223,  0.89843545],
        [ 0.27853391,  0.8931508 ,  0.34319118],
        [ 0.51984431,  0.09217771,  0.78764716]],

       [[ 0.25019845,  0.92622331,  0.06111409],
        [ 0.81363641,  0.06093368,  0.13123373],
        [ 0.47268657,  0.39635091,  0.77978269]],

       [[ 0.86098829,  0.07901332,  0.82169182],
        [ 0.12560088,  0.78210188,  0.69805434],
        [ 0.33544628,  0.81540172,  0.9393219 ]]])

In [24]: e
Out[24]: 
array([[[ 0.51699637,  0.42946223,  0.89843545],
        [ 0.27853391,  0.8931508 ,  0.34319118],
        [ 0.51984431,  0.09217771,  0.78764716]],

       [[ 0.25019845,  0.92622331,  0.06111409],
        [ 0.81363641,  0.06093368,  0.13123373],
        [ 0.47268657,  0.39635091,  0.77978269]],

       [[ 0.86098829,  0.07901332,  0.82169182],
        [ 0.12560088,  0.78210188,  0.69805434],
        [ 0.33544628,  0.81540172,  0.9393219 ]]])

I am not sure why no normalization was chosen by default. But when using ortho, dct and idct each seem to normalize by a factor of 1/sqrt(2 * N) or 1/sqrt(4 * N). There may be applications where the normalization is needed for dct and not idct and vice versa.