[Scipy-svn] r6251 - in trunk/scipy/sparse/linalg/isolve: . tests
scipy-svn@scip...
scipy-svn@scip...
Tue Mar 2 13:14:21 CST 2010
Author: stefan
Date: 2010-03-02 13:14:21 -0600 (Tue, 02 Mar 2010)
New Revision: 6251
Added:
trunk/scipy/sparse/linalg/isolve/lsqr.py
trunk/scipy/sparse/linalg/isolve/tests/test_lsqr.py
Modified:
trunk/scipy/sparse/linalg/isolve/__init__.py
Log:
ENH: Add LSQR algorithm for solving sparse least squares problems to scipy.sparse.
Modified: trunk/scipy/sparse/linalg/isolve/__init__.py
===================================================================
--- trunk/scipy/sparse/linalg/isolve/__init__.py 2010-02-24 12:27:43 UTC (rev 6250)
+++ trunk/scipy/sparse/linalg/isolve/__init__.py 2010-03-02 19:14:21 UTC (rev 6251)
@@ -4,6 +4,7 @@
from iterative import *
from minres import minres
from lgmres import lgmres
+from lsqr import lsqr
__all__ = filter(lambda s:not s.startswith('_'),dir())
from numpy.testing import Tester
Added: trunk/scipy/sparse/linalg/isolve/lsqr.py
===================================================================
--- trunk/scipy/sparse/linalg/isolve/lsqr.py (rev 0)
+++ trunk/scipy/sparse/linalg/isolve/lsqr.py 2010-03-02 19:14:21 UTC (rev 6251)
@@ -0,0 +1,495 @@
+"""Sparse Equations and Least Squares.
+
+The original Fortran code was written by C. C. Paige and M. A. Saunders as
+described in
+
+C. C. Paige and M. A. Saunders, LSQR: An algorithm for sparse linear
+equations and sparse least squares, TOMS 8(1), 43--71 (1982).
+
+C. C. Paige and M. A. Saunders, Algorithm 583; LSQR: Sparse linear
+equations and least-squares problems, TOMS 8(2), 195--209 (1982).
+
+It is licensed under the following BSD license:
+
+Copyright (c) 2006, Systems Optimization Laboratory
+All rights reserved.
+
+Redistribution and use in source and binary forms, with or without
+modification, are permitted provided that the following conditions are
+met:
+
+ * Redistributions of source code must retain the above copyright
+ notice, this list of conditions and the following disclaimer.
+
+ * Redistributions in binary form must reproduce the above
+ copyright notice, this list of conditions and the following
+ disclaimer in the documentation and/or other materials provided
+ with the distribution.
+
+ * Neither the name of Stanford University nor the names of its
+ contributors may be used to endorse or promote products derived
+ from this software without specific prior written permission.
+
+THIS SOFTWARE IS PROVIDED BY THE COPYRIGHT HOLDERS AND CONTRIBUTORS
+"AS IS" AND ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT
+LIMITED TO, THE IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR
+A PARTICULAR PURPOSE ARE DISCLAIMED. IN NO EVENT SHALL THE COPYRIGHT
+OWNER OR CONTRIBUTORS BE LIABLE FOR ANY DIRECT, INDIRECT, INCIDENTAL,
+SPECIAL, EXEMPLARY, OR CONSEQUENTIAL DAMAGES (INCLUDING, BUT NOT
+LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS OR SERVICES; LOSS OF USE,
+DATA, OR PROFITS; OR BUSINESS INTERRUPTION) HOWEVER CAUSED AND ON ANY
+THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT LIABILITY, OR TORT
+(INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY OUT OF THE USE
+OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF SUCH DAMAGE.
+
+The Fortran code was translated to Python for use in CVXOPT by Jeffery
+Kline with contributions by Mridul Aanjaneya and Bob Myhill.
+
+Adapted for SciPy by Stefan van der Walt.
+
+"""
+
+__all__ = ['lsqr']
+
+import numpy as np
+from math import sqrt
+from scipy.sparse.linalg.interface import aslinearoperator
+
+def _sym_ortho(a,b):
+ """
+ Jeffery Kline noted: I added the routine 'SymOrtho' for numerical
+ stability. This is recommended by S.-C. Choi in [1]_. It removes
+ the unpleasant potential of ``1/eps`` in some important places
+ (see, for example text following "Compute the next
+ plane rotation Qk" in minres_py).
+
+ References
+ ----------
+ .. [1] S.-C. Choi, "Iterative Methods for Singular Linear Equations
+ and Least-Squares Problems", Dissertation,
+ http://www.stanford.edu/group/SOL/dissertations/sou-cheng-choi-thesis.pdf
+
+ """
+ aa = abs(a)
+ ab = abs(b)
+ if b == 0.:
+ s = 0.
+ r = aa
+ if aa == 0.:
+ c = 1.
+ else:
+ c = a/aa
+ elif a == 0.:
+ c = 0.
+ s = b / ab
+ r = ab
+ elif ab >= aa:
+ sb = 1
+ if b < 0: sb=-1
+ tau = a/b
+ s = sb * (1 + tau**2)**-0.5
+ c = s * tau
+ r = b / s
+ elif aa > ab:
+ sa = 1
+ if a < 0: sa = -1
+ tau = b / a
+ c = sa * (1 + tau**2)**-0.5
+ s = c * tau
+ r = a / c
+
+ return c, s, r
+
+
+def lsqr(A, b, damp=0.0, atol=1e-8, btol=1e-8, conlim=1e8,
+ iter_lim=None, show=False, calc_var=False):
+ """Find the least-squares solution to a large, sparse, linear system
+ of equations.
+
+ The function solves ``Ax = b`` or ``min ||b - Ax||^2`` or
+ ``min ||Ax - b||^2 + d^2 ||x||^2.
+
+ The matrix A may be square or rectangular (over-determined or
+ under-determined), and may have any rank.
+
+ ::
+
+ 1. Unsymmetric equations -- solve A*x = b
+
+ 2. Linear least squares -- solve A*x = b
+ in the least-squares sense
+
+ 3. Damped least squares -- solve ( A )*x = ( b )
+ ( damp*I ) ( 0 )
+ in the least-squares sense
+
+ Parameters
+ ----------
+ A : {sparse matrix, ndarray, LinearOperatorLinear}
+ Representation of an mxn matrix. It is required that
+ the linear operator can produce ``Ax`` and ``A^T x``.
+ b : (m,) ndarray
+ Right-hand side vector ``b``.
+ damp : float
+ Damping coefficient.
+ atol, btol : float
+ Stopping tolerances. If both are 1.0e-9 (say), the final
+ residual norm should be accurate to about 9 digits. (The
+ final x will usually have fewer correct digits, depending on
+ cond(A) and the size of damp.)
+ conlim : float
+ Another stopping tolerance. lsqr 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.
+ Maximum precision can be obtained by setting ``atol = btol =
+ conlim = zero``, but the number of iterations may then be
+ excessive.
+ iter_lim : int
+ Explicit limitation on number of iterations (for safety).
+ show : bool
+ Display an iteration log.
+ calc_var : bool
+ Whether to estimate diagonals of ``(A'A + damp^2*I)^{-1}``.
+
+ Returns
+ -------
+ x : ndarray of float
+ The final solution.
+ istop : int
+ Gives the reason for termination.
+ 1 means x is an approximate solution to Ax = b.
+ 2 means x approximately solves the least-squares problem.
+ itn : int
+ Iteration number upon termination.
+ r1norm : float
+ ``norm(r)``, where ``r = b - Ax``.
+ r2norm : float
+ ``sqrt( norm(r)^2 + damp^2 * norm(x)^2 )``. Equal to `r1norm` if
+ ``damp == 0``.
+ anorm : float
+ Estimate of Frobenius norm of ``Abar = [[A]; [damp*I]]``.
+ acond : float
+ Estimate of ``cond(Abar)``.
+ arnorm : float
+ Estimate of ``norm(A'*r - damp^2*x)``.
+ xnorm : float
+ ``norm(x)``
+ var : ndarray of float
+ If ``calc_var`` is True, estimates all diagonals of
+ ``(A'A)^{-1}`` (if ``damp == 0``) or more generally ``(A'A +
+ damp^2*I)^{-1}``. This is well defined if A has full column
+ rank or ``damp > 0``. (Not sure what var means if ``rank(A)
+ < n`` and ``damp = 0.``)
+
+ Notes
+ -----
+ LSQR uses an iterative method to approximate the solution. The
+ number of iterations required to reach a certain accuracy depends
+ strongly on the scaling of the problem. Poor scaling of the rows
+ or columns of A should therefore be avoided where possible.
+
+ For example, in problem 1 the solution is unaltered by
+ row-scaling. If a row of A is very small or large compared to
+ the other rows of A, the corresponding row of ( A b ) should be
+ scaled up or down.
+
+ In problems 1 and 2, the solution x is easily recovered
+ following column-scaling. Unless better information is known,
+ the nonzero columns of A should be scaled so that they all have
+ the same Euclidean norm (e.g., 1.0).
+
+ In problem 3, there is no freedom to re-scale if damp is
+ nonzero. However, the value of damp should be assigned only
+ after attention has been paid to the scaling of A.
+
+ The parameter damp is intended to help regularize
+ ill-conditioned systems, by preventing the true solution from
+ being very large. Another aid to regularization is provided by
+ the parameter acond, which may be used to terminate iterations
+ before the computed solution becomes very large.
+
+ If some initial estimate ``x0`` is known and if ``damp == 0``,
+ one could proceed as follows:
+
+ 1. Compute a residual vector ``r0 = b - A*x0``.
+ 2. Use LSQR to solve the system ``A*dx = r0``.
+ 3. Add the correction dx to obtain a final solution ``x = x0 + dx``.
+
+ This requires that ``x0`` be available before and after the call
+ to LSQR. To judge the benefits, suppose LSQR takes k1 iterations
+ to solve A*x = b and k2 iterations to solve A*dx = r0.
+ If x0 is "good", norm(r0) will be smaller than norm(b).
+ If the same stopping tolerances atol and btol are used for each
+ system, k1 and k2 will be similar, but the final solution x0 + dx
+ should be more accurate. The only way to reduce the total work
+ is to use a larger stopping tolerance for the second system.
+ If some value btol is suitable for A*x = b, the larger value
+ btol*norm(b)/norm(r0) should be suitable for A*dx = r0.
+
+ Preconditioning is another way to reduce the number of iterations.
+ If it is possible to solve a related system ``M*x = b``
+ efficiently, where M approximates A in some helpful way (e.g. M -
+ A has low rank or its elements are small relative to those of A),
+ LSQR may converge more rapidly on the system ``A*M(inverse)*z =
+ b``, after which x can be recovered by solving M*x = z.
+
+ If A is symmetric, LSQR should not be used!
+
+ Alternatives are the symmetric conjugate-gradient method (cg)
+ and/or SYMMLQ. SYMMLQ is an implementation of symmetric cg that
+ applies to any symmetric A and will converge more rapidly than
+ LSQR. If A is positive definite, there are other implementations
+ of symmetric cg that require slightly less work per iteration than
+ SYMMLQ (but will take the same number of iterations).
+
+ References
+ ----------
+ .. [1] C. C. Paige and M. A. Saunders (1982a).
+ "LSQR: An algorithm for sparse linear equations and
+ sparse least squares", ACM TOMS 8(1), 43-71.
+ .. [2] C. C. Paige and M. A. Saunders (1982b).
+ "Algorithm 583. LSQR: Sparse linear equations and least
+ squares problems", ACM TOMS 8(2), 195-209.
+ .. [3] M. A. Saunders (1995). "Solution of sparse rectangular
+ systems using LSQR and CRAIG", BIT 35, 588-604.
+
+ """
+ A = aslinearoperator(A)
+ b = b.squeeze()
+
+ m, n = A.shape
+ if iter_lim is None: iter_lim = 2 * n
+ var = np.zeros(n)
+
+ msg=('The exact solution is x = 0 ',
+ 'Ax - b is small enough, given atol, btol ',
+ 'The least-squares solution is good enough, given atol ',
+ 'The estimate of cond(Abar) has exceeded conlim ',
+ 'Ax - b is small enough for this machine ',
+ 'The least-squares solution is good enough for this machine',
+ 'Cond(Abar) seems to be too large for this machine ',
+ 'The iteration limit has been reached ');
+
+ if show:
+ print ' '
+ print 'LSQR Least-squares solution of Ax = b'
+ str1 = 'The matrix A has %8g rows and %8g cols' % (m, n)
+ str2 = 'damp = %20.14e calc_var = %8g' % (damp, calc_var)
+ str3 = 'atol = %8.2e conlim = %8.2e'%( atol, conlim)
+ str4 = 'btol = %8.2e iter_lim = %8g' %( btol, iter_lim)
+ print str1
+ print str2
+ print str3
+ print str4
+
+ itn = 0
+ istop = 0
+ nstop = 0
+ ctol = 0
+ if conlim > 0: ctol = 1/conlim
+ anorm = 0
+ acond = 0
+ dampsq = damp**2
+ ddnorm = 0
+ res2 = 0
+ xnorm = 0
+ xxnorm = 0
+ z = 0
+ cs2 = -1
+ sn2 = 0
+
+ """
+ Set up the first vectors u and v for the bidiagonalization.
+ These satisfy beta*u = b, alfa*v = A'u.
+ """
+ __xm = np.zeros(m) # a matrix for temporary holding
+ __xn = np.zeros(n) # a matrix for temporary holding
+ v = np.zeros(n)
+ u = b
+ x = np.zeros(n)
+ alfa = 0
+ beta = np.linalg.norm(u)
+ w = np.zeros(n)
+
+ if beta > 0:
+ u = (1/beta) * u
+ v = A.rmatvec(u)
+ alfa = np.linalg.norm(v)
+
+ if alfa > 0:
+ v = (1/alfa) * v
+ w = v.copy()
+
+ rhobar = alfa
+ phibar = beta
+ bnorm = beta
+ rnorm = beta
+ r1norm = rnorm
+ r2norm = rnorm
+
+ # Reverse the order here from the original matlab code because
+ # there was an error on return when arnorm==0
+ arnorm = alfa * beta
+ if arnorm == 0:
+ print msg[0];
+ return x, istop, itn, r1norm, r2norm, anorm, acond, arnorm, xnorm, var
+
+ head1 = ' Itn x[0] r1norm r2norm ';
+ head2 = ' Compatible LS Norm A Cond A';
+
+ if show:
+ print ' '
+ print head1, head2
+ test1 = 1; test2 = alfa / beta;
+ str1 = '%6g %12.5e' %( itn, x[0] );
+ str2 = ' %10.3e %10.3e'%( r1norm, r2norm );
+ str3 = ' %8.1e %8.1e' %( test1, test2 );
+ print str1, str2, str3
+
+ # Main iteration loop.
+ while itn < iter_lim:
+ itn = itn + 1
+ """
+ % Perform the next step of the bidiagonalization to obtain the
+ % next beta, u, alfa, v. These satisfy the relations
+ % beta*u = a*v - alfa*u,
+ % alfa*v = A'*u - beta*v.
+ """
+ u = A.matvec(v) - alfa * u
+ beta = np.linalg.norm(u)
+
+ if beta > 0:
+ u = (1/beta) * u
+ anorm = sqrt(anorm**2 + alfa**2 + beta**2 + damp**2)
+ v = A.rmatvec(u) - beta * v
+ alfa = np.linalg.norm(v)
+ if alfa > 0:
+ v = (1 / alfa) * v
+
+ # Use a plane rotation to eliminate the damping parameter.
+ # This alters the diagonal (rhobar) of the lower-bidiagonal matrix.
+ rhobar1 = sqrt(rhobar**2 + damp**2)
+ cs1 = rhobar / rhobar1
+ sn1 = damp / rhobar1
+ psi = sn1 * phibar
+ phibar = cs1 * phibar
+
+ # Use a plane rotation to eliminate the subdiagonal element (beta)
+ # of the lower-bidiagonal matrix, giving an upper-bidiagonal matrix.
+ cs, sn, rho = _sym_ortho(rhobar1, beta)
+
+ theta = sn * alfa
+ rhobar = -cs * alfa
+ phi = cs * phibar
+ phibar = sn * phibar
+ tau = sn * phi
+
+ # Update x and w.
+ t1 = phi / rho
+ t2 = -theta / rho
+ dk = (1 / rho) * w
+
+ x = x + t1 * w
+ w = v + t2 * w
+ ddnorm = ddnorm + np.linalg.norm(dk)**2
+
+ if calc_var:
+ var = var + dk**2
+
+ # Use a plane rotation on the right to eliminate the
+ # super-diagonal element (theta) of the upper-bidiagonal matrix.
+ # Then use the result to estimate norm(x).
+ delta = sn2 * rho
+ gambar = -cs2 * rho
+ rhs = phi - delta * z
+ zbar = rhs / gambar
+ xnorm = sqrt(xxnorm + zbar**2)
+ gamma = sqrt(gambar**2 +theta**2)
+ cs2 = gambar / gamma
+ sn2 = theta / gamma
+ z = rhs / gamma
+ xxnorm = xxnorm + z**2
+
+ # Test for convergence.
+ # First, estimate the condition of the matrix Abar,
+ # and the norms of rbar and Abar'rbar.
+ acond = anorm * sqrt(ddnorm)
+ res1 = phibar**2
+ res2 = res2 + psi**2
+ rnorm = sqrt(res1 + res2)
+ arnorm = alfa * abs(tau)
+
+ # Distinguish between
+ # r1norm = ||b - Ax|| and
+ # r2norm = rnorm in current code
+ # = sqrt(r1norm^2 + damp^2*||x||^2).
+ # Estimate r1norm from
+ # r1norm = sqrt(r2norm^2 - damp^2*||x||^2).
+ # Although there is cancellation, it might be accurate enough.
+ r1sq = rnorm**2 - dampsq * xxnorm
+ r1norm = sqrt(abs(r1sq))
+ if r1sq < 0:
+ r1norm = -r1norm
+ r2norm = rnorm
+
+ # Now use these norms to estimate certain other quantities,
+ # some of which will be small near a solution.
+ test1 = rnorm / bnorm
+ test2 = arnorm / (anorm * rnorm)
+ test3 = 1 / acond
+ t1 = test1 / (1 + anorm * xnorm / bnorm)
+ rtol = btol + atol * anorm * xnorm / bnorm
+
+ # The following tests guard against extremely small values of
+ # atol, btol or ctol. (The user may have set any or all of
+ # the parameters atol, btol, conlim to 0.)
+ # The effect is equivalent to the normal tests using
+ # atol = eps, btol = eps, conlim = 1/eps.
+ if itn >= iter_lim: istop = 7
+ if 1 + test3 <= 1: istop = 6
+ if 1 + test2 <= 1: istop = 5
+ if 1 + t1 <= 1: istop = 4
+
+ # Allow for tolerances set by the user.
+ if test3 <= ctol: istop = 3
+ if test2 <= atol: istop = 2
+ if test1 <= rtol: istop = 1
+
+ # See if it is time to print something.
+ prnt = False;
+ if n <= 40: prnt = True
+ if itn <= 10: prnt = True
+ if itn >= iter_lim-10: prnt = True
+ # if itn%10 == 0: prnt = True
+ if test3 <= 2*ctol: prnt = True
+ if test2 <= 10*atol: prnt = True
+ if test1 <= 10*rtol: prnt = True
+ if istop != 0: prnt = True
+
+ if prnt:
+ if show:
+ str1 = '%6g %12.5e' % (itn, x[0])
+ str2 = ' %10.3e %10.3e' % (r1norm, r2norm)
+ str3 = ' %8.1e %8.1e' % (test1, test2)
+ str4 = ' %8.1e %8.1e' % (anorm, acond)
+ print str1, str2, str3, str4
+
+ if istop != 0: break
+
+ # End of iteration loop.
+ # Print the stopping condition.
+ if show:
+ print ' '
+ print 'LSQR finished'
+ print msg[istop]
+ print ' '
+ str1 = 'istop =%8g r1norm =%8.1e' % (istop, r1norm)
+ str2 = 'anorm =%8.1e arnorm =%8.1e' % (anorm, arnorm)
+ str3 = 'itn =%8g r2norm =%8.1e' % (itn, r2norm)
+ str4 = 'acond =%8.1e xnorm =%8.1e' % (acond, xnorm)
+ print str1+ ' ' + str2
+ print str3+ ' ' + str4
+ print ' '
+
+ return x, istop, itn, r1norm, r2norm, anorm, acond, arnorm, xnorm, var
Added: trunk/scipy/sparse/linalg/isolve/tests/test_lsqr.py
===================================================================
--- trunk/scipy/sparse/linalg/isolve/tests/test_lsqr.py (rev 0)
+++ trunk/scipy/sparse/linalg/isolve/tests/test_lsqr.py 2010-03-02 19:14:21 UTC (rev 6251)
@@ -0,0 +1,59 @@
+import numpy as np
+
+from scipy.sparse.linalg import lsqr
+from time import time
+
+# Set up a test problem
+n = 35
+G = np.eye(n)
+normal = np.random.normal
+norm = np.linalg.norm
+
+for jj in range(5):
+ gg = normal(size=n)
+ hh = gg * gg.T
+ G += (hh + hh.T) * 0.5
+ G += normal(size=n) * normal(size=n)
+
+b = normal(size=n)
+
+tol = 1e-10
+show = False
+maxit = None
+
+def test_basic():
+ svx = np.linalg.solve(G, b)
+ X = lsqr(G, b, show=show, atol=tol, btol=tol, iter_lim=maxit)
+ xo = X[0]
+ assert norm(svx - xo) < 1e-5
+
+if __name__ == "__main__":
+ svx = np.linalg.solve(G, b)
+
+ tic = time()
+ X = lsqr(G, b, show=show, atol=tol, btol=tol, iter_lim=maxit)
+ xo = X[0]
+ phio = X[3]
+ psio = X[7]
+ k = X[2]
+ chio = X[8]
+ mg = np.amax(G - G.T)
+ if mg > 1e-14:
+ sym='No'
+ else:
+ sym='Yes'
+
+ print 'LSQR'
+ print "Is linear operator symmetric? " + sym
+ print "n: %3g iterations: %3g" % (n, k)
+ print "Norms computed in %.2fs by LSQR" % (time() - tic)
+ print " ||x|| %9.4e ||r|| %9.4e ||Ar|| %9.4e " %( chio, phio, psio)
+ print "Residual norms computed directly:"
+ print " ||x|| %9.4e ||r|| %9.4e ||Ar|| %9.4e" % (norm(xo),
+ norm(G*xo - b),
+ norm(G.T*(G*xo-b)))
+ print "Direct solution norms:"
+ print " ||x|| %9.4e ||r|| %9.4e " % (norm(svx), norm(G*svx -b))
+ print ""
+ print " || x_{direct} - x_{LSQR}|| %9.4e " % norm(svx-xo)
+ print ""
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