bundles / scipy 2.0.0.dev0+git20260907.41eeb59 / scipy / sparse / linalg / _isolve / lsmr / lsmr

function

`scipy.sparse.linalg._isolve.lsmr:lsmr`

source: build-install/usr/lib/python3.14/site-packages/scipy/sparse/linalg/_isolve/lsmr.py:29

Signature

def lsmr(A, b, damp = 0.0, atol = 1e-06, btol = 1e-06, conlim = 100000000.0, maxiter = None, show = False, x0 = None)

Summary

Iterative solver for least-squares problems.

Extended Summary

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 array, 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.

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

Examples

import numpy as np
from scipy.sparse import csc_array
from scipy.sparse.linalg import lsmr
A = csc_array([[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
itn
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

✓

x
A.dot(x)-b

✗

array([ 0.00333, -0.00333,  0.00333])

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.

Aliases

scipy.sparse.linalg.lsmr