U Gv fA=@s`dZdgZddlmZmZmZmZddlmZddl m Z ddl m Z ddl mZdd dZd S)a Copyright (C) 2010 David Fong and Michael Saunders LSMR uses an iterative method. 07 Jun 2010: Documentation updated 03 Jun 2010: First release version in Python David Chin-lung Fong clfong@stanford.edu Institute for Computational and Mathematical Engineering Stanford University Michael Saunders saunders@stanford.edu Systems Optimization Laboratory Dept of MS&E, Stanford University. lsmr)zerosinfty atleast_1d result_type)norm)sqrt)aslinearoperator) _sym_orthoư>חANFc NCs*t|}t|}|jdkr"|}d} d} d} d} d} |j\}}t||g}|dkrX|}|dkrnt||t}nt|||t}|rtdtd td |d |d td |td||ftd||f|}t |}|dkrt ||}| }n"t| }|| |}t |}|dkrFd||}| |}t |}nt ||}d}|dkrjd||}d}||}|}d}d}d}d}| }t ||} |}!d}"d}#d}$d}%d}&d}'||}(d})d}*t|(}+d},d}-d}.d}/|dkrd|}/|}0||}1|1dkr.|rt| d||.||0|1|+|,|-fS|dkrTd|d<||.||0|1|+|,|-fS|rtdt| | d}2||}3d||df}4d|0|1f}5d|2|3f}6td|4|5|6g||kr|d}|| 9}|| |7}t |}|dkr2|d|9}|| 9}|| |7}t |}|dkr2|d|9}t||\}7}8}9|}:t|9|\};}<}|<|}=|;|}|}>|&}?||}@||}At|||=\}}}||}&| |}| |@||:|> 9} | |7} ||&||| 7}||=| 9}||7}|7|!}B|8 |!}C|;|B}D|< |B}!|%}Et|#|@\}F}G}H|G|}%|F|}#|G |"|F|D}"|?|E|$|H}$|&|%|$|#}I|'|C|C}'t|'|"|Id|!|!}0|(||}(t|(}+|(||}(t|)|>})|dkrt|*|>}*t|)|At|*|A},t|}1t |}-|0|}2|+|0dkr|1|+|0}3nt}3d|,}J|2d|+|-|}K|||+|-|}L||krVd}.d|Jdkrhd}.d|3dkrzd}.d|Kdkrd}.|J|/krd}.|3|krd}.|2|Lkrd}.|r|dks |dks ||dks |ddks |Jd|/ks |3d|ks |2d|Lks |.dkr| | kr@d} tdt| | | d} d||df}4d|0|1f}5d|2|3f}6d|+|,f}Mtd|4|5|6|Mg|.dkrqq|rtdtd t| |.td!|.|0ftd"|+|1ftd#||,ftd$|-t|4|5t|6|M||.||0|1|+|,|-fS)%aIterative solver for least-squares problems. lsmr solves the system of linear equations ``Ax = b``. If the system is inconsistent, it solves the least-squares problem ``min ||b - Ax||_2``. ``A`` is a rectangular matrix of dimension m-by-n, where all cases are allowed: m = n, m > n, or m < n. ``b`` is a vector of length m. The matrix A may be dense or sparse (usually sparse). Parameters ---------- A : {sparse matrix, ndarray, LinearOperator} Matrix A in the linear system. Alternatively, ``A`` can be a linear operator which can produce ``Ax`` and ``A^H x`` using, e.g., ``scipy.sparse.linalg.LinearOperator``. b : array_like, shape (m,) Vector ``b`` in the linear system. damp : float Damping factor for regularized least-squares. `lsmr` solves the regularized least-squares problem:: min ||(b) - ( A )x|| ||(0) (damp*I) ||_2 where damp is a scalar. If damp is None or 0, the system is solved without regularization. Default is 0. atol, btol : float, optional Stopping tolerances. `lsmr` continues iterations until a certain backward error estimate is smaller than some quantity depending on atol and btol. Let ``r = b - Ax`` be the residual vector for the current approximate solution ``x``. If ``Ax = b`` seems to be consistent, `lsmr` terminates when ``norm(r) <= atol * norm(A) * norm(x) + btol * norm(b)``. Otherwise, `lsmr` terminates when ``norm(A^H r) <= atol * norm(A) * norm(r)``. If both tolerances are 1.0e-6 (default), the final ``norm(r)`` should be accurate to about 6 digits. (The final ``x`` will usually have fewer correct digits, depending on ``cond(A)`` and the size of LAMBDA.) If `atol` or `btol` is None, a default value of 1.0e-6 will be used. Ideally, they should be estimates of the relative error in the entries of ``A`` and ``b`` respectively. For example, if the entries of ``A`` have 7 correct digits, set ``atol = 1e-7``. This prevents the algorithm from doing unnecessary work beyond the uncertainty of the input data. conlim : float, optional `lsmr` terminates if an estimate of ``cond(A)`` exceeds `conlim`. For compatible systems ``Ax = b``, conlim could be as large as 1.0e+12 (say). For least-squares problems, `conlim` should be less than 1.0e+8. If `conlim` is None, the default value is 1e+8. Maximum precision can be obtained by setting ``atol = btol = conlim = 0``, but the number of iterations may then be excessive. Default is 1e8. maxiter : int, optional `lsmr` terminates if the number of iterations reaches `maxiter`. The default is ``maxiter = min(m, n)``. For ill-conditioned systems, a larger value of `maxiter` may be needed. Default is False. show : bool, optional Print iterations logs if ``show=True``. Default is False. x0 : array_like, shape (n,), optional Initial guess of ``x``, if None zeros are used. Default is None. .. versionadded:: 1.0.0 Returns ------- x : ndarray of float Least-square solution returned. istop : int istop gives the reason for stopping:: istop = 0 means x=0 is a solution. If x0 was given, then x=x0 is a solution. = 1 means x is an approximate solution to A@x = B, according to atol and btol. = 2 means x approximately solves the least-squares problem according to atol. = 3 means COND(A) seems to be greater than CONLIM. = 4 is the same as 1 with atol = btol = eps (machine precision) = 5 is the same as 2 with atol = eps. = 6 is the same as 3 with CONLIM = 1/eps. = 7 means ITN reached maxiter before the other stopping conditions were satisfied. itn : int Number of iterations used. normr : float ``norm(b-Ax)`` normar : float ``norm(A^H (b - Ax))`` norma : float ``norm(A)`` conda : float Condition number of A. normx : float ``norm(x)`` Notes ----- .. versionadded:: 0.11.0 References ---------- .. [1] D. C.-L. Fong and M. A. Saunders, "LSMR: An iterative algorithm for sparse least-squares problems", SIAM J. Sci. Comput., vol. 33, pp. 2950-2971, 2011. :arxiv:`1006.0758` .. [2] LSMR Software, https://web.stanford.edu/group/SOL/software/lsmr/ Examples -------- >>> import numpy as np >>> from scipy.sparse import csc_matrix >>> from scipy.sparse.linalg import lsmr >>> A = csc_matrix([[1., 0.], [1., 1.], [0., 1.]], dtype=float) The first example has the trivial solution ``[0, 0]`` >>> b = np.array([0., 0., 0.], dtype=float) >>> x, istop, itn, normr = lsmr(A, b)[:4] >>> istop 0 >>> x array([0., 0.]) The stopping code `istop=0` returned indicates that a vector of zeros was found as a solution. The returned solution `x` indeed contains ``[0., 0.]``. The next example has a non-trivial solution: >>> b = np.array([1., 0., -1.], dtype=float) >>> x, istop, itn, normr = lsmr(A, b)[:4] >>> istop 1 >>> x array([ 1., -1.]) >>> itn 1 >>> normr 4.440892098500627e-16 As indicated by `istop=1`, `lsmr` found a solution obeying the tolerance limits. The given solution ``[1., -1.]`` obviously solves the equation. The remaining return values include information about the number of iterations (`itn=1`) and the remaining difference of left and right side of the solved equation. The final example demonstrates the behavior in the case where there is no solution for the equation: >>> b = np.array([1., 0.01, -1.], dtype=float) >>> x, istop, itn, normr = lsmr(A, b)[:4] >>> istop 2 >>> x array([ 1.00333333, -0.99666667]) >>> A.dot(x)-b array([ 0.00333333, -0.00333333, 0.00333333]) >>> normr 0.005773502691896255 `istop` indicates that the system is inconsistent and thus `x` is rather an approximate solution to the corresponding least-squares problem. `normr` contains the minimal distance that was found. )z9The exact solution is x = 0, or x = x0, if x0 was given z:Ax - b is small enough, given atol, btol z:The least-squares solution is good enough, given atol z:The estimate of cond(Abar) has exceeded conlim z:Ax - b is small enough for this machine z:The least-squares solution is good enough for this machinez:Cond(Abar) seems to be too large for this machine z:The iteration limit has been reached z( itn x(1) norm r norm Arz% compatible LS norm A cond ArN z2LSMR Least-squares solution of Ax = b zThe matrix A has z rows and z columnszdamp = %20.14e z,atol = %8.2e conlim = %8.2e z'btol = %8.2e maxiter = %8g g}Ô%ITz %6g %12.5ez %10.3e %10.3ez %8.1e %8.1e( g?z %8.1e %8.1ez LSMR finishedzistop =%8g normr =%8.1ez! normA =%8.1e normAr =%8.1ezitn =%8g condA =%8.1ez normx =%8.1e)r rndimZsqueezeshapeminrfloatprintrrcopyZmatvecZrmatvecrjoinr maxabsr)NAbZdampZatolZbtolZconlimmaxitershowZx0msgZhdg1Zhdg2ZpfreqZpcountmnZminDimZdtypeuZnormbxbetavalphaitnZzetabarZalphabarrhoZrhobarZcbarZsbarhZhbarZbetaddZbetadZrhodoldZ tautildeoldZ thetatildezetadZnormA2ZmaxrbarZminrbarZnormAZcondAZnormxistopZctolZnormrZnormarZtest1Ztest2Zstr1Zstr2Zstr3ZchatZshatZalphahatZrhooldcsZthetanewZ rhobaroldZzetaoldZthetabarZrhotempZ betaacuteZ betacheckZbetahatZ thetatildeoldZ ctildeoldZ stildeoldZ rhotildeoldZtaudZtest3t1ZrtolZstr4rrD/tmp/pip-unpacked-wheel-96ln3f52/scipy/sparse/linalg/_isolve/lsmr.pyrs~)                                           "              )r r r r NFN)__doc____all__ZnumpyrrrrZ numpy.linalgrmathrZscipy.sparse.linalg._interfacer Z scipy.sparse.linalg._isolve.lsqrr rrrrr9s