如何在Python中使用部分透视实现LU分解?
我想实现我自己的LU分解p,L,U=my_LU(A),这样给定一个矩阵A,就可以用部分枢轴计算LU分解。但我只知道如何不旋转。 有人能帮忙做部分旋转吗如何在Python中使用部分透视实现LU分解?,python,matrix,decomposition,Python,Matrix,Decomposition,我想实现我自己的LU分解p,L,U=my_LU(A),这样给定一个矩阵A,就可以用部分枢轴计算LU分解。但我只知道如何不旋转。 有人能帮忙做部分旋转吗 def lu(A): import numpy as np # Return an error if matrix is not square if not A.shape[0]==A.shape[1]: raise ValueError("Input matrix must be square")
def lu(A):
import numpy as np
# Return an error if matrix is not square
if not A.shape[0]==A.shape[1]:
raise ValueError("Input matrix must be square")
n = A.shape[0]
L = np.zeros((n,n),dtype='float64')
U = np.zeros((n,n),dtype='float64')
U[:] = A
np.fill_diagonal(L,1) # fill the diagonal of L with 1
for i in range(n-1):
for j in range(i+1,n):
L[j,i] = U[j,i]/U[i,i]
U[j,i:] = U[j,i:]-L[j,i]*U[i,i:]
U[j,i] = 0
return (L,U)
为此,可以使用Scipy的Scipy.linalg.lu 请同时查看此链接: 使用单位正向和反向代换求解x(带和不带部分枢轴):
# No partial pivoting
LU = naive_lu_factor(A)
y = ufsub( LU, b )
x = bsub( LU, y )
这不是问得更充分吗?来自Trefethen和Bau,带有清晰的伪代码
def lu_factor(A):
"""
LU factorization with partial pivorting
Overwrite A with:
U (upper triangular) and (unit Lower triangular) L
Return [LU,piv]
Where piv is 1d numpy array with row swap indices
"""
n = A.shape[0]
piv = np.arange(0,n)
for k in range(n-1):
# piv
max_row_index = np.argmax(abs(A[k:n,k])) + k
piv[[k,max_row_index]] = piv[[max_row_index,k]]
A[[k,max_row_index]] = A[[max_row_index,k]]
# LU
for i in range(k+1,n):
A[i,k] = A[i,k]/A[k,k]
for j in range(k+1,n):
A[i,j] -= A[i,k]*A[k,j]
return [A,piv]
def ufsub(L,b):
""" Unit row oriented forward substitution """
for i in range(L.shape[0]):
for j in range(i):
b[i] -= L[i,j]*b[j]
return b
def bsub(U,y):
""" Row oriented backward substitution """
for i in range(U.shape[0]-1,-1,-1):
for j in range(i+1, U.shape[1]):
y[i] -= U[i,j]*y[j]
y[i] = y[i]/U[i,i]
return y
# No partial pivoting
LU = naive_lu_factor(A)
y = ufsub( LU, b )
x = bsub( LU, y )
# Partial pivoting
LU, piv = lu_factor(A)
b = b[piv]
y = ufsub( LU, b )
x = bsub( LU, y )