Scipy的ftpack dct和idct [英] Scipy's fftpack dct and idct
问题描述
假设您使用dct函数，然后不对数据进行任何操作并使用反变换；反转后的数据与预先转换后的数据不一样吗？为什么浮点问题？是报告的问题还是正常行为？
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 floating point issue? Is it a reported issue or is it a normal behavior?
In [21]: a = [1.2, 3.4, 5.1, 2.3, 4.5]
In [22]: b = dct(a)
In [23]: b
Out[23]: array([ 33. , -4.98384545, -4.5 , -5.971707 , 4.5 ])
In [24]: c = idct(b)
In [25]: c
Out[25]: array([ 12., 34., 51., 23., 45.])
有人对此有一个解释吗？当然，简单的 c * 10 **-1 可以解决问题，但是如果您重复调用该函数以在多个维度上使用它，则误差会更大：
Anyone has an explanation as why? Of course, a simple c*10**-1 would do the trick, but if you repeat the call of the function to use it on several dimensions, the error gets bigger:
In [37]: a = np.random.rand(3,3,3)
In [38]: d = dct(dct(dct(a).transpose(0,2,1)).transpose(2,1,0)).transpose(2,1,0).transpose(0,2,1)
In [39]: e = idct(idct(idct(d).transpose(0,2,1)).transpose(2,1,0)).transpose(2,1,0).transpose(0,2,1)
In [40]: a
Out[40]:
array([[[ 0.48709809, 0.50624831, 0.91190972],
[ 0.56545798, 0.85695062, 0.62484782],
[ 0.96092354, 0.17453537, 0.17884233]],
[[ 0.29433402, 0.08540074, 0.18574437],
[ 0.09942075, 0.78902363, 0.62663572],
[ 0.20372951, 0.67039551, 0.52292875]],
[[ 0.79952289, 0.48221372, 0.43838685],
[ 0.25559683, 0.39549153, 0.84129493],
[ 0.69093533, 0.71522961, 0.16522915]]])
In [41]: e
Out[41]:
array([[[ 105.21318703, 109.34963575, 196.97249887],
[ 122.13892469, 185.10133376, 134.96712825],
[ 207.55948396, 37.69964085, 38.62994399]],
[[ 63.57614855, 18.44656009, 40.12078466],
[ 21.47488098, 170.42910452, 135.35331646],
[ 44.00557341, 144.80543099, 112.95260949]],
[[ 172.69694529, 104.15816275, 94.69156014],
[ 55.20891593, 85.42617016, 181.71970442],
[ 149.2420308 , 154.48959477, 35.68949734]]])
此处是文档。
推荐答案
看起来dct和idct不能通过默认。定义 dct 以以下方式调用 fftpack.dct 。对 idct 做同样的事情。
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')
完成后，您将在执行转换后获得原始答案。
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 ]]])
我不确定为什么默认情况下没有选择标准化。但是当使用 ortho 时， dct 和 idct 似乎都通过 1 / sqrt（2 * N）或 1 / sqrt（4 * N）进行归一化。在某些应用程序中， dct 而不是 idct 需要规范化，反之亦然。
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.
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