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108 lines
2.6 KiB
Python
108 lines
2.6 KiB
Python
15 years ago
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import numpy as np
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__all__ = ['dct', 'idct']
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def dct(x, n=None):
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"""
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Discrete Cosine Transform
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N-1
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y[k] = 2* sum x[n]*cos(pi*k*(2n+1)/(2*N)), 0 <= k < N.
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n=0
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Examples
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--------
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>>> import numpy as np
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>>> x = np.arange(5)
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>>> np.abs(x-idct(dct(x)))<1e-14
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array([ True, True, True, True, True], dtype=bool)
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>>> np.abs(x-dct(idct(x)))<1e-14
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array([ True, True, True, True, True], dtype=bool)
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Reference
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---------
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http://en.wikipedia.org/wiki/Discrete_cosine_transform
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http://users.ece.utexas.edu/~bevans/courses/ee381k/lectures/
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"""
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fft = np.fft.fft
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x = np.atleast_1d(x)
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if n is None:
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n = x.shape[-1]
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if x.shape[-1] < n:
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n_shape = x.shape[:-1] + (n - x.shape[-1],)
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xx = np.hstack((x, np.zeros(n_shape)))
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else:
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xx = x[..., :n]
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real_x = np.all(np.isreal(xx))
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if (real_x and (np.remainder(n, 2) == 0)):
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xp = 2 * fft(np.hstack((xx[..., ::2], xx[..., ::-2])))
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else:
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xp = fft(np.hstack((xx, xx[..., ::-1])))
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xp = xp[..., :n]
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w = np.exp(-1j * np.arange(n) * np.pi / (2 * n))
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y = xp * w
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if real_x:
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return y.real
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else:
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return y
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def idct(x, n=None):
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"""
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Inverse Discrete Cosine Transform
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N-1
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x[k] = 1/N sum w[n]*y[n]*cos(pi*k*(2n+1)/(2*N)), 0 <= k < N.
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n=0
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w(0) = 1/2
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w(n) = 1 for n>0
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Examples
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--------
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>>> import numpy as np
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>>> x = np.arange(5)
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>>> np.abs(x-idct(dct(x)))<1e-14
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array([ True, True, True, True, True], dtype=bool)
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>>> np.abs(x-dct(idct(x)))<1e-14
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array([ True, True, True, True, True], dtype=bool)
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Reference
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---------
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http://en.wikipedia.org/wiki/Discrete_cosine_transform
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http://users.ece.utexas.edu/~bevans/courses/ee381k/lectures/
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"""
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ifft = np.fft.ifft
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x = np.atleast_1d(x)
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if n is None:
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n = x.shape[-1]
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w = np.exp(1j * np.arange(n) * np.pi / (2 * n))
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if x.shape[-1] < n:
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n_shape = x.shape[:-1] + (n - x.shape[-1],)
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xx = np.hstack((x, np.zeros(n_shape))) * w
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else:
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xx = x[..., :n] * w
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real_x = np.all(np.isreal(x))
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if (real_x and (np.remainder(n, 2) == 0)):
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xx[..., 0] = xx[..., 0] * 0.5
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yp = ifft(xx)
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y = np.zeros(xx.shape, dtype=complex)
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y[..., ::2] = yp[..., :n / 2]
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y[..., ::-2] = yp[..., n / 2::]
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else:
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yp = ifft(np.hstack((xx, np.zeros_like(xx[..., 0]), np.conj(xx[..., :0:-1]))))
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y = yp[..., :n]
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if real_x:
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return y.real
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else:
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return y
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