Module « scipy.linalg »

Fonction solve_banded - module scipy.linalg

Signature de la fonction solve_banded

def solve_banded(l_and_u, ab, b, overwrite_ab=False, overwrite_b=False, debug=None, check_finite=True) 

Description

solve_banded.__doc__

    Solve the equation a x = b for x, assuming a is banded matrix.

    The matrix a is stored in `ab` using the matrix diagonal ordered form::

        ab[u + i - j, j] == a[i,j]

    Example of `ab` (shape of a is (6,6), `u` =1, `l` =2)::

        *    a01  a12  a23  a34  a45
        a00  a11  a22  a33  a44  a55
        a10  a21  a32  a43  a54   *
        a20  a31  a42  a53   *    *

    Parameters
    ----------
    (l, u) : (integer, integer)
        Number of non-zero lower and upper diagonals
    ab : (`l` + `u` + 1, M) array_like
        Banded matrix
    b : (M,) or (M, K) array_like
        Right-hand side
    overwrite_ab : bool, optional
        Discard data in `ab` (may enhance performance)
    overwrite_b : bool, optional
        Discard data in `b` (may enhance performance)
    check_finite : bool, optional
        Whether to check that the input matrices contain only finite numbers.
        Disabling may give a performance gain, but may result in problems
        (crashes, non-termination) if the inputs do contain infinities or NaNs.

    Returns
    -------
    x : (M,) or (M, K) ndarray
        The solution to the system a x = b. Returned shape depends on the
        shape of `b`.

    Examples
    --------
    Solve the banded system a x = b, where::

            [5  2 -1  0  0]       [0]
            [1  4  2 -1  0]       [1]
        a = [0  1  3  2 -1]   b = [2]
            [0  0  1  2  2]       [2]
            [0  0  0  1  1]       [3]

    There is one nonzero diagonal below the main diagonal (l = 1), and
    two above (u = 2). The diagonal banded form of the matrix is::

             [*  * -1 -1 -1]
        ab = [*  2  2  2  2]
             [5  4  3  2  1]
             [1  1  1  1  *]

    >>> from scipy.linalg import solve_banded
    >>> ab = np.array([[0,  0, -1, -1, -1],
    ...                [0,  2,  2,  2,  2],
    ...                [5,  4,  3,  2,  1],
    ...                [1,  1,  1,  1,  0]])
    >>> b = np.array([0, 1, 2, 2, 3])
    >>> x = solve_banded((1, 2), ab, b)
    >>> x
    array([-2.37288136,  3.93220339, -4.        ,  4.3559322 , -1.3559322 ])