# scipy.sparse.linalg.splu

scipy.sparse.linalg.splu(A, permc_spec=None, diag_pivot_thresh=None, relax=None, panel_size=None, options={})

  Compute the LU decomposition of a sparse, square matrix.

  Parameters:
    A sparse matrix

      Sparse matrix to factorize. Most efficient when provided in CSC format. Other formats will be converted to CSC before factorization.

    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

    diag_pivot_thresh float, optional

      Threshold used for a diagonal entry to be an acceptable pivot. See SuperLU user’s guide for details [1]

    relax int, optional

      Expert option for customizing the degree of relaxing supernodes. See SuperLU user’s guide for details [1]

    panel_size int, optional

      Expert option for customizing the panel size. See SuperLU user’s guide for details [1]

    options dict, optional

      Dictionary containing additional expert options to SuperLU. See SuperLU user guide [1] (section 2.4 on the ‘Options’ argument) for more details. For example, you can specify `options=dict(Equil=False, IterRefine='SINGLE'))` to turn equilibration off and perform a single iterative refinement.

  Returns:
    invA scipy.sparse.linalg.SuperLU

      Object, which has a `solve` method.

  See also

  `spilu`

    incomplete LU decomposition

  Notes

  This function uses the SuperLU library.

  References

  [1] (1,2,3,4)

  SuperLU https://portal.nersc.gov/project/sparse/superlu/

  Examples

      >>> import numpy as np
      >>> from scipy.sparse import csc_matrix
      >>> from scipy.sparse.linalg import splu
      >>> A = csc_matrix([[1., 0., 0.], [5., 0., 2.], [0., -1., 0.]], dtype=float)
      >>> B = splu(A)
      >>> x = np.array([1., 2., 3.], dtype=float)
      >>> B.solve(x)
      array([ 1. , -3. , -1.5])
      >>> A.dot(B.solve(x))
      array([ 1.,  2.,  3.])
      >>> B.solve(A.dot(x))
      array([ 1.,  2.,  3.])