- scipy.sparse.linalg.onenormest(A, t=2, itmax=5, compute_v=False, compute_w=False)¶
Compute a lower bound of the 1-norm of a sparse matrix.
A : ndarray or other linear operator
A linear operator that can be transposed and that can produce matrix products.
t : int, optional
A positive parameter controlling the tradeoff between accuracy versus time and memory usage. Larger values take longer and use more memory but give more accurate output.
itmax : int, optional
Use at most this many iterations.
compute_v : bool, optional
Request a norm-maximizing linear operator input vector if True.
compute_w : bool, optional
Request a norm-maximizing linear operator output vector if True.
est : float
An underestimate of the 1-norm of the sparse matrix.
v : ndarray, optional
The vector such that ||Av||_1 == est*||v||_1. It can be thought of as an input to the linear operator that gives an output with particularly large norm.
w : ndarray, optional
The vector Av which has relatively large 1-norm. It can be thought of as an output of the linear operator that is relatively large in norm compared to the input.
This is algorithm 2.4 of .
In  it is described as follows. “This algorithm typically requires the evaluation of about 4t matrix-vector products and almost invariably produces a norm estimate (which is, in fact, a lower bound on the norm) correct to within a factor 3.”
New in version 0.13.0.
[R346] Nicholas J. Higham and Francoise Tisseur (2000), “A Block Algorithm for Matrix 1-Norm Estimation, with an Application to 1-Norm Pseudospectra.” SIAM J. Matrix Anal. Appl. Vol. 21, No. 4, pp. 1185-1201. [R347] Awad H. Al-Mohy and Nicholas J. Higham (2009), “A new scaling and squaring algorithm for the matrix exponential.” SIAM J. Matrix Anal. Appl. Vol. 31, No. 3, pp. 970-989.
>>> from scipy.sparse import csc_matrix >>> from scipy.sparse.linalg import onenormest >>> A = csc_matrix([[1., 0., 0.], [5., 8., 2.], [0., -1., 0.]], dtype=float) >>> A.todense() matrix([[ 1., 0., 0.], [ 5., 8., 2.], [ 0., -1., 0.]]) >>> onenormest(A) 9.0 >>> np.linalg.norm(A.todense(), ord=1) 9.0