sturlamolden wrote:
>
> Actually, there was a typo in the original code. I used d1[l-1] where I
> should have used d1[l+1]. Arrgh. Here is the corrected version, the
> Matlab code must be changed similarly. It has no relevance for the
> performance timings though.
>
>
> def D4_Transform(x, s1=None, d1=None, d2=None):
> """
> D4 Wavelet transform in NumPy
> (C) Sturla Molden
> """
> C1 = 1.7320508075688772
> C2 = 0.4330127018922193
> C3 = -0.066987298107780702
> C4 = 0.51763809020504137
> C5 = 1.9318516525781364
> if d1 == None:
> d1 = numpy.zeros(x.size/2)
> s1 = numpy.zeros(x.size/2)
> d2 = numpy.zeros(x.size/2)
> odd = x[1::2]
> even = x[:-1:2]
> d1[:] = odd[:] - C1*even[:]
> s1[:-1] = even[:-1] + C2*d1[:-1] + C3*d1[1:]
> s1[-1] = even[-1] + C2*d1[-1] + C3*d1[0]
> d2[0] = d1[0] + s1[-1]
> d2[1:] = d1[1:] + s1[:-1]
> even[:] = C4 * s1[:]
> odd[:] = C5 * d2[:]
If you swap C4 with C5 then our solutions give identical results and
both match the classic dwt approach:
even[:] = C5 * s1[:]
odd[:] = C4 * d2[:]
> if x.size > 2:
> D4_Transform(even,s1[0:even.size/2],
> d1[0:even.size/2],d2[0:even.size/2])
cheers,
fw
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