Module « scipy.sparse.linalg »

Fonction spsolve - module scipy.sparse.linalg

Signature de la fonction spsolve

def spsolve(A, b, permc_spec=None, use_umfpack=True) 

Description

spsolve.__doc__

Solve the sparse linear system Ax=b, where b may be a vector or a matrix.

    Parameters
    ----------
    A : ndarray or sparse matrix
        The square matrix A will be converted into CSC or CSR form
    b : ndarray or sparse matrix
        The matrix or vector representing the right hand side of the equation.
        If a vector, b.shape must be (n,) or (n, 1).
    permc_spec : str, optional
        How to permute the columns of the matrix for sparsity preservation.
        (default: 'COLAMD')

        - ``NATURAL``: natural ordering.
        - ``MMD_ATA``: minimum degree ordering on the structure of A^T A.
        - ``MMD_AT_PLUS_A``: minimum degree ordering on the structure of A^T+A.
        - ``COLAMD``: approximate minimum degree column ordering
    use_umfpack : bool, optional
        if True (default) then use umfpack for the solution.  This is
        only referenced if b is a vector and ``scikit-umfpack`` is installed.

    Returns
    -------
    x : ndarray or sparse matrix
        the solution of the sparse linear equation.
        If b is a vector, then x is a vector of size A.shape[1]
        If b is a matrix, then x is a matrix of size (A.shape[1], b.shape[1])

    Notes
    -----
    For solving the matrix expression AX = B, this solver assumes the resulting
    matrix X is sparse, as is often the case for very sparse inputs.  If the
    resulting X is dense, the construction of this sparse result will be
    relatively expensive.  In that case, consider converting A to a dense
    matrix and using scipy.linalg.solve or its variants.

    Examples
    --------
    >>> from scipy.sparse import csc_matrix
    >>> from scipy.sparse.linalg import spsolve
    >>> A = csc_matrix([[3, 2, 0], [1, -1, 0], [0, 5, 1]], dtype=float)
    >>> B = csc_matrix([[2, 0], [-1, 0], [2, 0]], dtype=float)
    >>> x = spsolve(A, B)
    >>> np.allclose(A.dot(x).todense(), B.todense())
    True