scipy.sparse.linalg.spsolve_triangular#
- scipy.sparse.linalg.spsolve_triangular(A, b, lower=True, overwrite_A=False, overwrite_b=False, unit_diagonal=False)[source]#
Solve the equation
A x = b
for x, assuming A is a triangular matrix.- Parameters
- A(M, M) sparse matrix
A sparse square triangular matrix. Should be in CSR format.
- b(M,) or (M, N) array_like
Right-hand side matrix in
A x = b
- lowerbool, optional
Whether A is a lower or upper triangular matrix. Default is lower triangular matrix.
- overwrite_Abool, optional
Allow changing A. The indices of A are going to be sorted and zero entries are going to be removed. Enabling gives a performance gain. Default is False.
- overwrite_bbool, 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_diagonalbool, optional
If True, diagonal elements of a are assumed to be 1 and will not be referenced.
New in version 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
New in version 0.19.0.
Examples
>>> from scipy.sparse import csr_matrix >>> from scipy.sparse.linalg import spsolve_triangular >>> A = csr_matrix([[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