solve_banded#
- scipy.linalg.solve_banded(l_and_u, ab, b, overwrite_ab=False, overwrite_b=False, check_finite=True)[source]#
- Solve the equation - a @ x = bfor- x, where- ais the banded matrix defined by ab.- 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_abbool, optional
- Discard data in ab (may enhance performance) 
- overwrite_bbool, optional
- Discard data in b (may enhance performance) 
- check_finitebool, 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 *] - >>> import numpy as np >>> 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 ])