90 lines
2.7 KiB
Python
90 lines
2.7 KiB
Python
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"""
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Solve the orthogonal Procrustes problem.
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"""
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import numpy as np
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from .decomp_svd import svd
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__all__ = ['orthogonal_procrustes']
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def orthogonal_procrustes(A, B, check_finite=True):
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"""
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Compute the matrix solution of the orthogonal Procrustes problem.
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Given matrices A and B of equal shape, find an orthogonal matrix R
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that most closely maps A to B using the algorithm given in [1]_.
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Parameters
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----------
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A : (M, N) array_like
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Matrix to be mapped.
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B : (M, N) array_like
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Target matrix.
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check_finite : bool, optional
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Whether to check that the input matrices contain only finite numbers.
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Disabling may give a performance gain, but may result in problems
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(crashes, non-termination) if the inputs do contain infinities or NaNs.
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Returns
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-------
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R : (N, N) ndarray
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The matrix solution of the orthogonal Procrustes problem.
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Minimizes the Frobenius norm of ``(A @ R) - B``, subject to
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``R.T @ R = I``.
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scale : float
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Sum of the singular values of ``A.T @ B``.
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Raises
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------
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ValueError
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If the input array shapes don't match or if check_finite is True and
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the arrays contain Inf or NaN.
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Notes
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-----
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Note that unlike higher level Procrustes analyses of spatial data, this
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function only uses orthogonal transformations like rotations and
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reflections, and it does not use scaling or translation.
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.. versionadded:: 0.15.0
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References
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----------
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.. [1] Peter H. Schonemann, "A generalized solution of the orthogonal
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Procrustes problem", Psychometrica -- Vol. 31, No. 1, March, 1996.
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Examples
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--------
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>>> from scipy.linalg import orthogonal_procrustes
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>>> A = np.array([[ 2, 0, 1], [-2, 0, 0]])
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Flip the order of columns and check for the anti-diagonal mapping
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>>> R, sca = orthogonal_procrustes(A, np.fliplr(A))
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>>> R
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array([[-5.34384992e-17, 0.00000000e+00, 1.00000000e+00],
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[ 0.00000000e+00, 1.00000000e+00, 0.00000000e+00],
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[ 1.00000000e+00, 0.00000000e+00, -7.85941422e-17]])
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>>> sca
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9.0
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"""
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if check_finite:
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A = np.asarray_chkfinite(A)
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B = np.asarray_chkfinite(B)
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else:
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A = np.asanyarray(A)
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B = np.asanyarray(B)
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if A.ndim != 2:
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raise ValueError('expected ndim to be 2, but observed %s' % A.ndim)
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if A.shape != B.shape:
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raise ValueError('the shapes of A and B differ (%s vs %s)' % (
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A.shape, B.shape))
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# Be clever with transposes, with the intention to save memory.
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u, w, vt = svd(B.T.dot(A).T)
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R = u.dot(vt)
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scale = w.sum()
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return R, scale
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