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## $Id$
## -*- coding: utf-8 -*-
# jpeg.dct
# ========
from numpy import dot, linalg
import numpy
def auxcos(x, u):
return numpy.cos((numpy.pi / 8) * (x + 0.5) * u)
def cosmat(M=8, N=8):
C = numpy.array([[auxcos(x, u) for u in range(N)]
for x in range(M)]) / 2
C[:, 0] = C[:, 0] / numpy.sqrt(2)
# C[0,:] = C[0,:] / numpy.sqrt(2)
return C
auxM = cosmat(8, 8)
invM = linalg.inv(auxM)
auxT = numpy.transpose(auxM)
invT = numpy.transpose(invM)
def dct2(g):
"""
Perform a 2D DCT transform on g, assuming that g is 8x8.
"""
assert (8, 8) == numpy.shape(g)
return dot(auxT, dot(g, auxM))
def idct2(g):
"""
Perform a 2D inverse DCT transform on g, assuming that g is 8x8.
"""
assert (8, 8) == numpy.shape(g)
# return dot( invM, dot( g, invT ) )
return dot(invT, dot(g, invM))
def bdct(C, f=dct2):
"""
Make a blockwise (8x8 blocks) 2D DCT transform on the matrix C.
The optional second parameter f specifies the DCT transform function.
The height and width of C have to be divisible by 8.
"""
(M, N) = numpy.shape(C)
assert M % 8 == 0
assert N % 8 == 0
S = numpy.ndarray((M, N))
for i in range(0, M, 8):
for j in range(0, N, 8):
S[i:(i + 8), j:(j + 8)] = f(C[i:(i + 8), j:(j + 8)])
return S
def ibdct(C): return bdct(C, f=idct2)
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