scipy.sparse.linalg.lsmr¶
- scipy.sparse.linalg.lsmr(A, b, damp=0.0, atol=1e-06, btol=1e-06, conlim=100000000.0, maxiter=None, show=False, x0=None)[source]¶
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 problemmin ||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 produceAx
andA^H x
using, e.g.,scipy.sparse.linalg.LinearOperator
.- barray_like, shape (m,)
Vector
b
in the linear system.- dampfloat
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, btolfloat, optional
Stopping tolerances.
lsmr
continues iterations until a certain backward error estimate is smaller than some quantity depending on atol and btol. Letr = b - Ax
be the residual vector for the current approximate solutionx
. IfAx = b
seems to be consistent,lsmr
terminates whennorm(r) <= atol * norm(A) * norm(x) + btol * norm(b)
. Otherwise, lsmr terminates whennorm(A^H r) <= atol * norm(A) * norm(r)
. If both tolerances are 1.0e-6 (say), the finalnorm(r)
should be accurate to about 6 digits. (The finalx
will usually have fewer correct digits, depending oncond(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 ofA
andb
respectively. For example, if the entries ofA
have 7 correct digits, setatol = 1e-7
. This prevents the algorithm from doing unnecessary work beyond the uncertainty of the input data.- conlimfloat, optional
lsmr
terminates if an estimate ofcond(A)
exceeds conlim. For compatible systemsAx = 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 settingatol = btol = conlim = 0
, but the number of iterations may then be excessive.- maxiterint, optional
lsmr
terminates if the number of iterations reaches maxiter. The default ismaxiter = min(m, n)
. For ill-conditioned systems, a larger value of maxiter may be needed.- showbool, optional
Print iterations logs if
show=True
.- x0array_like, shape (n,), optional
Initial guess of
x
, if None zeros are used.New in version 1.0.0.
- Returns
- xndarray of float
Least-square solution returned.
- istopint
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.
- itnint
Number of iterations used.
- normrfloat
norm(b-Ax)
- normarfloat
norm(A^H (b - Ax))
- normafloat
norm(A)
- condafloat
Condition number of A.
- normxfloat
norm(x)
Notes
New in version 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
>>> 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.