【PLA】基于Python实现的线性代数算法库之lu分解
算法包下载链接:https://download.csdn.net/download/qq_42629529/79481514
from .Matrix import Matrix
from .Vector import Vector
from ._globals import is_zero#判断是否为0
#方阵的lu分解
def lu(matrix):
assert matrix.row_num() == matrix.col_num(), "matrix must be a square matrix"
n = matrix.row_num()#行数
A = [matrix.row_vector(i) for i in range(n)]#矩阵的行向量,上三角矩阵
L = [[1.0 if i == j else 0.0 for i in range(n)] for j in range(n)]
#高斯消元的过程
for i in range(n):
# 看A[i][i]位置是否可以是主元
if is_zero(A[i][i]):
return None, None#不能分解两个都空
else:
for j in range(i + 1, n):
p = A[j][i] / A[i][i]
A[j] = A[j] - p * A[i]
L[j][i] = p
return Matrix(L), Matrix([A[i].underlying_list() for i in range(n)])