scipy.sparse.linalg.cg#
- scipy.sparse.linalg.cg(A, b, x0=None, *, tol=<object object>, maxiter=None, M=None, callback=None, atol=0.0, rtol=1e-05)[source]#
Use Conjugate Gradient iteration to solve
Ax = b
.- Parameters:
- A{sparse matrix, ndarray, LinearOperator}
The real or complex N-by-N matrix of the linear system.
A
must represent a hermitian, positive definite matrix. Alternatively,A
can be a linear operator which can produceAx
using, e.g.,scipy.sparse.linalg.LinearOperator
.- bndarray
Right hand side of the linear system. Has shape (N,) or (N,1).
- x0ndarray
Starting guess for the solution.
- rtol, atolfloat, optional
Parameters for the convergence test. For convergence,
norm(b - A @ x) <= max(rtol*norm(b), atol)
should be satisfied. The default isatol=0.
andrtol=1e-5
.- maxiterinteger
Maximum number of iterations. Iteration will stop after maxiter steps even if the specified tolerance has not been achieved.
- M{sparse matrix, ndarray, LinearOperator}
Preconditioner for A. The preconditioner should approximate the inverse of A. Effective preconditioning dramatically improves the rate of convergence, which implies that fewer iterations are needed to reach a given error tolerance.
- callbackfunction
User-supplied function to call after each iteration. It is called as callback(xk), where xk is the current solution vector.
- tolfloat, optional, deprecated
Deprecated since version 1.12.0:
cg
keyword argumenttol
is deprecated in favor ofrtol
and will be removed in SciPy 1.14.0.
- Returns:
- xndarray
The converged solution.
- infointeger
- Provides convergence information:
0 : successful exit >0 : convergence to tolerance not achieved, number of iterations
Examples
>>> import numpy as np >>> from scipy.sparse import csc_matrix >>> from scipy.sparse.linalg import cg >>> P = np.array([[4, 0, 1, 0], ... [0, 5, 0, 0], ... [1, 0, 3, 2], ... [0, 0, 2, 4]]) >>> A = csc_matrix(P) >>> b = np.array([-1, -0.5, -1, 2]) >>> x, exit_code = cg(A, b, atol=1e-5) >>> print(exit_code) # 0 indicates successful convergence 0 >>> np.allclose(A.dot(x), b) True