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 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).


A : {matrix, sparse matrix, ndarray, LinearOperator}

Matrix A in the linear system.

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^{T} 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.

New in version 1.0.0.



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
        = 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
        = 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


normar : float

norm(A^T (b - Ax))

norma : float


conda : float

Condition number of A.

normx : float



New in version 0.11.0.


[R29]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.
[R30]LSMR Software,


>>> 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
>>> 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
>>> x
array([ 1., -1.])
>>> itn
>>> normr

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
>>> x
array([ 1.00333333, -0.99666667])
array([ 0.00333333, -0.00333333,  0.00333333])
>>> normr

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.