481 lines
15 KiB
Python
481 lines
15 KiB
Python
|
"""
|
||
|
|
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.
|
||
|
|
|
||
|
|
"""
|
||
|
|
|
||
|
|
__all__ = ['lsmr']
|
||
|
|
|
||
|
|
from numpy import zeros, infty, atleast_1d, result_type
|
||
|
|
from numpy.linalg import norm
|
||
|
|
from math import sqrt
|
||
|
|
from scipy.sparse.linalg.interface import aslinearoperator
|
||
|
|
|
||
|
|
from .lsqr import _sym_ortho
|
||
|
|
|
||
|
|
|
||
|
|
def lsmr(A, b, damp=0.0, atol=1e-6, btol=1e-6, conlim=1e8,
|
||
|
|
maxiter=None, show=False, x0=None):
|
||
|
|
"""Iterative 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 : {matrix, 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.
|
||
|
|
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 (say),
|
||
|
|
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.
|
||
|
|
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.
|
||
|
|
show : bool, optional
|
||
|
|
Print iterations logs if ``show=True``.
|
||
|
|
x0 : array_like, shape (n,), optional
|
||
|
|
Initial guess of x, if None zeros are used.
|
||
|
|
|
||
|
|
.. 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.
|
||
|
|
https://arxiv.org/abs/1006.0758
|
||
|
|
.. [2] LSMR Software, https://web.stanford.edu/group/SOL/software/lsmr/
|
||
|
|
|
||
|
|
Examples
|
||
|
|
--------
|
||
|
|
>>> 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.
|
||
|
|
"""
|
||
|
|
|
||
|
|
A = aslinearoperator(A)
|
||
|
|
b = atleast_1d(b)
|
||
|
|
if b.ndim > 1:
|
||
|
|
b = b.squeeze()
|
||
|
|
|
||
|
|
msg = ('The exact solution is x = 0, or x = x0, if x0 was given ',
|
||
|
|
'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 ')
|
||
|
|
|
||
|
|
hdg1 = ' itn x(1) norm r norm Ar'
|
||
|
|
hdg2 = ' compatible LS norm A cond A'
|
||
|
|
pfreq = 20 # print frequency (for repeating the heading)
|
||
|
|
pcount = 0 # print counter
|
||
|
|
|
||
|
|
m, n = A.shape
|
||
|
|
|
||
|
|
# stores the num of singular values
|
||
|
|
minDim = min([m, n])
|
||
|
|
|
||
|
|
if maxiter is None:
|
||
|
|
maxiter = minDim
|
||
|
|
|
||
|
|
if x0 is None:
|
||
|
|
dtype = result_type(A, b, float)
|
||
|
|
else:
|
||
|
|
dtype = result_type(A, b, x0, float)
|
||
|
|
|
||
|
|
if show:
|
||
|
|
print(' ')
|
||
|
|
print('LSMR Least-squares solution of Ax = b\n')
|
||
|
|
print(f'The matrix A has {m} rows and {n} columns')
|
||
|
|
print('damp = %20.14e\n' % (damp))
|
||
|
|
print('atol = %8.2e conlim = %8.2e\n' % (atol, conlim))
|
||
|
|
print('btol = %8.2e maxiter = %8g\n' % (btol, maxiter))
|
||
|
|
|
||
|
|
u = b
|
||
|
|
normb = norm(b)
|
||
|
|
if x0 is None:
|
||
|
|
x = zeros(n, dtype)
|
||
|
|
beta = normb.copy()
|
||
|
|
else:
|
||
|
|
x = atleast_1d(x0)
|
||
|
|
u = u - A.matvec(x)
|
||
|
|
beta = norm(u)
|
||
|
|
|
||
|
|
if beta > 0:
|
||
|
|
u = (1 / beta) * u
|
||
|
|
v = A.rmatvec(u)
|
||
|
|
alpha = norm(v)
|
||
|
|
else:
|
||
|
|
v = zeros(n, dtype)
|
||
|
|
alpha = 0
|
||
|
|
|
||
|
|
if alpha > 0:
|
||
|
|
v = (1 / alpha) * v
|
||
|
|
|
||
|
|
# Initialize variables for 1st iteration.
|
||
|
|
|
||
|
|
itn = 0
|
||
|
|
zetabar = alpha * beta
|
||
|
|
alphabar = alpha
|
||
|
|
rho = 1
|
||
|
|
rhobar = 1
|
||
|
|
cbar = 1
|
||
|
|
sbar = 0
|
||
|
|
|
||
|
|
h = v.copy()
|
||
|
|
hbar = zeros(n, dtype)
|
||
|
|
|
||
|
|
# Initialize variables for estimation of ||r||.
|
||
|
|
|
||
|
|
betadd = beta
|
||
|
|
betad = 0
|
||
|
|
rhodold = 1
|
||
|
|
tautildeold = 0
|
||
|
|
thetatilde = 0
|
||
|
|
zeta = 0
|
||
|
|
d = 0
|
||
|
|
|
||
|
|
# Initialize variables for estimation of ||A|| and cond(A)
|
||
|
|
|
||
|
|
normA2 = alpha * alpha
|
||
|
|
maxrbar = 0
|
||
|
|
minrbar = 1e+100
|
||
|
|
normA = sqrt(normA2)
|
||
|
|
condA = 1
|
||
|
|
normx = 0
|
||
|
|
|
||
|
|
# Items for use in stopping rules, normb set earlier
|
||
|
|
istop = 0
|
||
|
|
ctol = 0
|
||
|
|
if conlim > 0:
|
||
|
|
ctol = 1 / conlim
|
||
|
|
normr = beta
|
||
|
|
|
||
|
|
# Reverse the order here from the original matlab code because
|
||
|
|
# there was an error on return when arnorm==0
|
||
|
|
normar = alpha * beta
|
||
|
|
if normar == 0:
|
||
|
|
if show:
|
||
|
|
print(msg[0])
|
||
|
|
return x, istop, itn, normr, normar, normA, condA, normx
|
||
|
|
|
||
|
|
if show:
|
||
|
|
print(' ')
|
||
|
|
print(hdg1, hdg2)
|
||
|
|
test1 = 1
|
||
|
|
test2 = alpha / beta
|
||
|
|
str1 = '%6g %12.5e' % (itn, x[0])
|
||
|
|
str2 = ' %10.3e %10.3e' % (normr, normar)
|
||
|
|
str3 = ' %8.1e %8.1e' % (test1, test2)
|
||
|
|
print(''.join([str1, str2, str3]))
|
||
|
|
|
||
|
|
# Main iteration loop.
|
||
|
|
while itn < maxiter:
|
||
|
|
itn = itn + 1
|
||
|
|
|
||
|
|
# Perform the next step of the bidiagonalization to obtain the
|
||
|
|
# next beta, u, alpha, v. These satisfy the relations
|
||
|
|
# beta*u = a*v - alpha*u,
|
||
|
|
# alpha*v = A'*u - beta*v.
|
||
|
|
|
||
|
|
u *= -alpha
|
||
|
|
u += A.matvec(v)
|
||
|
|
beta = norm(u)
|
||
|
|
|
||
|
|
if beta > 0:
|
||
|
|
u *= (1 / beta)
|
||
|
|
v *= -beta
|
||
|
|
v += A.rmatvec(u)
|
||
|
|
alpha = norm(v)
|
||
|
|
if alpha > 0:
|
||
|
|
v *= (1 / alpha)
|
||
|
|
|
||
|
|
# At this point, beta = beta_{k+1}, alpha = alpha_{k+1}.
|
||
|
|
|
||
|
|
# Construct rotation Qhat_{k,2k+1}.
|
||
|
|
|
||
|
|
chat, shat, alphahat = _sym_ortho(alphabar, damp)
|
||
|
|
|
||
|
|
# Use a plane rotation (Q_i) to turn B_i to R_i
|
||
|
|
|
||
|
|
rhoold = rho
|
||
|
|
c, s, rho = _sym_ortho(alphahat, beta)
|
||
|
|
thetanew = s*alpha
|
||
|
|
alphabar = c*alpha
|
||
|
|
|
||
|
|
# Use a plane rotation (Qbar_i) to turn R_i^T to R_i^bar
|
||
|
|
|
||
|
|
rhobarold = rhobar
|
||
|
|
zetaold = zeta
|
||
|
|
thetabar = sbar * rho
|
||
|
|
rhotemp = cbar * rho
|
||
|
|
cbar, sbar, rhobar = _sym_ortho(cbar * rho, thetanew)
|
||
|
|
zeta = cbar * zetabar
|
||
|
|
zetabar = - sbar * zetabar
|
||
|
|
|
||
|
|
# Update h, h_hat, x.
|
||
|
|
|
||
|
|
hbar *= - (thetabar * rho / (rhoold * rhobarold))
|
||
|
|
hbar += h
|
||
|
|
x += (zeta / (rho * rhobar)) * hbar
|
||
|
|
h *= - (thetanew / rho)
|
||
|
|
h += v
|
||
|
|
|
||
|
|
# Estimate of ||r||.
|
||
|
|
|
||
|
|
# Apply rotation Qhat_{k,2k+1}.
|
||
|
|
betaacute = chat * betadd
|
||
|
|
betacheck = -shat * betadd
|
||
|
|
|
||
|
|
# Apply rotation Q_{k,k+1}.
|
||
|
|
betahat = c * betaacute
|
||
|
|
betadd = -s * betaacute
|
||
|
|
|
||
|
|
# Apply rotation Qtilde_{k-1}.
|
||
|
|
# betad = betad_{k-1} here.
|
||
|
|
|
||
|
|
thetatildeold = thetatilde
|
||
|
|
ctildeold, stildeold, rhotildeold = _sym_ortho(rhodold, thetabar)
|
||
|
|
thetatilde = stildeold * rhobar
|
||
|
|
rhodold = ctildeold * rhobar
|
||
|
|
betad = - stildeold * betad + ctildeold * betahat
|
||
|
|
|
||
|
|
# betad = betad_k here.
|
||
|
|
# rhodold = rhod_k here.
|
||
|
|
|
||
|
|
tautildeold = (zetaold - thetatildeold * tautildeold) / rhotildeold
|
||
|
|
taud = (zeta - thetatilde * tautildeold) / rhodold
|
||
|
|
d = d + betacheck * betacheck
|
||
|
|
normr = sqrt(d + (betad - taud)**2 + betadd * betadd)
|
||
|
|
|
||
|
|
# Estimate ||A||.
|
||
|
|
normA2 = normA2 + beta * beta
|
||
|
|
normA = sqrt(normA2)
|
||
|
|
normA2 = normA2 + alpha * alpha
|
||
|
|
|
||
|
|
# Estimate cond(A).
|
||
|
|
maxrbar = max(maxrbar, rhobarold)
|
||
|
|
if itn > 1:
|
||
|
|
minrbar = min(minrbar, rhobarold)
|
||
|
|
condA = max(maxrbar, rhotemp) / min(minrbar, rhotemp)
|
||
|
|
|
||
|
|
# Test for convergence.
|
||
|
|
|
||
|
|
# Compute norms for convergence testing.
|
||
|
|
normar = abs(zetabar)
|
||
|
|
normx = norm(x)
|
||
|
|
|
||
|
|
# Now use these norms to estimate certain other quantities,
|
||
|
|
# some of which will be small near a solution.
|
||
|
|
|
||
|
|
test1 = normr / normb
|
||
|
|
if (normA * normr) != 0:
|
||
|
|
test2 = normar / (normA * normr)
|
||
|
|
else:
|
||
|
|
test2 = infty
|
||
|
|
test3 = 1 / condA
|
||
|
|
t1 = test1 / (1 + normA * normx / normb)
|
||
|
|
rtol = btol + atol * normA * normx / normb
|
||
|
|
|
||
|
|
# 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 >= maxiter:
|
||
|
|
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.
|
||
|
|
|
||
|
|
if show:
|
||
|
|
if (n <= 40) or (itn <= 10) or (itn >= maxiter - 10) or \
|
||
|
|
(itn % 10 == 0) or (test3 <= 1.1 * ctol) or \
|
||
|
|
(test2 <= 1.1 * atol) or (test1 <= 1.1 * rtol) or \
|
||
|
|
(istop != 0):
|
||
|
|
|
||
|
|
if pcount >= pfreq:
|
||
|
|
pcount = 0
|
||
|
|
print(' ')
|
||
|
|
print(hdg1, hdg2)
|
||
|
|
pcount = pcount + 1
|
||
|
|
str1 = '%6g %12.5e' % (itn, x[0])
|
||
|
|
str2 = ' %10.3e %10.3e' % (normr, normar)
|
||
|
|
str3 = ' %8.1e %8.1e' % (test1, test2)
|
||
|
|
str4 = ' %8.1e %8.1e' % (normA, condA)
|
||
|
|
print(''.join([str1, str2, str3, str4]))
|
||
|
|
|
||
|
|
if istop > 0:
|
||
|
|
break
|
||
|
|
|
||
|
|
# Print the stopping condition.
|
||
|
|
|
||
|
|
if show:
|
||
|
|
print(' ')
|
||
|
|
print('LSMR finished')
|
||
|
|
print(msg[istop])
|
||
|
|
print('istop =%8g normr =%8.1e' % (istop, normr))
|
||
|
|
print(' normA =%8.1e normAr =%8.1e' % (normA, normar))
|
||
|
|
print('itn =%8g condA =%8.1e' % (itn, condA))
|
||
|
|
print(' normx =%8.1e' % (normx))
|
||
|
|
print(str1, str2)
|
||
|
|
print(str3, str4)
|
||
|
|
|
||
|
|
return x, istop, itn, normr, normar, normA, condA, normx
|