import numpy as np def lu_decomposition(A): n = A.shape[0] L = np.zeros_like(A) U = np.zeros_like(A) for i in range(n): L[i, i] = 1 for j in range(i, n): U[i, j] = A[i, j] - np.dot(L[i, :i], U[:i, j]) for j in range(i + 1, n): L[j, i...
We can solve the linear system using the matrix LU Decomposition in scipy with the help of the functions scipy.linalg.lu() and scipy.linalg.lu_solve(). In this example we are solving AX = B for a given A and B −import numpy as np from scipy.linalg import lu_factor, lu_solve #...
LU分解(LU decomposition)是一种将一个方阵分解为一个下三角矩阵(Lower triangular matrix)和一个上三角矩阵(Upper triangular matrix)的乘积的方法。这种分解通常用于解决线性方程组、计算行列式以及求解矩阵的逆。 给定一个n阶方阵A,LU分解的目标是找到一个n阶下三角矩阵L和一个n阶上三角矩阵U,使得A=LU。 高斯消...
http://www.quantstart.com/articles/LU-Decomposition-in-Python-and-NumPy http://www.quantstart.com/articles/LU-Decomposition-in-Python-and-NumPy
python import numpy as np def lu_decomposition(A): n = A.shape[0] L = np.zeros_like(A) U = np.zeros_like(A) for i in range(n): L[i, i] = 1 for j in range(i, n): U[i, j] = A[i, j] - np.dot(L[i, :i], U[:i, j]) for j in range(i + 1, n): L...
eye(m, min(m, n)) U = A for i in range(min(m, n)): for j in range(i + 1, m): L[j, i] = U[j, i] / U[i, i] U[j, i:] -= L[j, i] * U[i, i:] return L, U[:min(m,n), :] PLU 分解,稍微复杂一点 def plu_decomposition(A): n = A.shape[0] L =...
deflu_decomposition(matrix):"""计算矩阵的 LU 分解"""check_square_matrix(matrix)n=matrix.shape[0]L,U=initialize_matrices(matrix)foriinrange(n):# 上三角矩阵 Uforjinrange(i,n):U[i,j]=matrix[i,j]# 下三角矩阵 Lforkinrange(i):U[i,j]-=L[i,k]*U[k,j]L[i,i]=1# L 的对角线...
1. 计算矩阵A的LU分解,使用`LU_decomposition`函数: ```python L, U = LU_decomposition(A) ``` 其中,`LU_decomposition`函数用于计算矩阵A的LU分解,返回L和U矩阵。 2. 使用L矩阵求解Ly=b: ```python y = np.zeros_like(b) n = A.shape(0) # Solve Ly = b for i in range(n): y(i) ...
def LU_decomposition(A): n = len(A[0]) L = np.zeros([n, n]) U = np.zeros([n, n]) for i in range(n): L[i][i] = 1 U[0][0] = A[0][0] if i == 0: for j in range(1, n): U[0][j] = A[0][j] L[j][0] = A[j][0] / U[0][0] else: for j ...
LU分解(python) MagicBones 人微言轻 来自专栏 · 数值模拟 import numpy as np def LUdecomposition(a): n=len(a) L1=np.zeros((n,n)) U1=np.zeros((n,n)) for k in range(n): # 0 1 2 3 L1[k,k]=1 for j in range(k,n): sum1=0 for r in range(0,k): # matlab是1:k-1...