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Module « scipy.sparse.linalg »

Fonction spsolve_triangular - module scipy.sparse.linalg

Signature de la fonction spsolve_triangular

def spsolve_triangular(A, b, lower=True, overwrite_A=False, overwrite_b=False, unit_diagonal=False) 

Description

help(scipy.sparse.linalg.spsolve_triangular)

Solve the equation ``A x = b`` for `x`, assuming A is a triangular matrix.

Parameters
----------
A : (M, M) sparse array or matrix
    A sparse square triangular matrix. Should be in CSR or CSC format.
b : (M,) or (M, N) array_like
    Right-hand side matrix in ``A x = b``
lower : bool, optional
    Whether `A` is a lower or upper triangular matrix.
    Default is lower triangular matrix.
overwrite_A : bool, optional
    Allow changing `A`.
    Enabling gives a performance gain. Default is False.
overwrite_b : bool, optional
    Allow overwriting data in `b`.
    Enabling gives a performance gain. Default is False.
    If `overwrite_b` is True, it should be ensured that
    `b` has an appropriate dtype to be able to store the result.
unit_diagonal : bool, optional
    If True, diagonal elements of `a` are assumed to be 1.

    .. versionadded:: 1.4.0

Returns
-------
x : (M,) or (M, N) ndarray
    Solution to the system ``A x = b``. Shape of return matches shape
    of `b`.

Raises
------
LinAlgError
    If `A` is singular or not triangular.
ValueError
    If shape of `A` or shape of `b` do not match the requirements.

Notes
-----
.. versionadded:: 0.19.0

Examples
--------
>>> import numpy as np
>>> from scipy.sparse import csc_array
>>> from scipy.sparse.linalg import spsolve_triangular
>>> A = csc_array([[3, 0, 0], [1, -1, 0], [2, 0, 1]], dtype=float)
>>> B = np.array([[2, 0], [-1, 0], [2, 0]], dtype=float)
>>> x = spsolve_triangular(A, B)
>>> np.allclose(A.dot(x), B)
True

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