LU分解以numpy为中心

Question

我找不到从此处的伪代码进行部分透视实现LU分解的尝试的错误之处（第6页） 。 我的代码粘贴在下面。 谁能发现问题？

def LUPdecomposition(matrix):
n, m = matrix.shape
assert n == m, "LU decomposition is only applicable to square matrices"
L = np.identity(n)
U = matrix.copy()
P = np.identity(n)

for k in range(n-1):

    # pivoting: get index of maximum in k-th column on diagonal or below
    index = np.argmax(abs(U[k:,k]))
    index += k

    # pivoting: permute rows
    if index != k:
        U[[index,k],k:n] = U[[k,index],k:n]
        P[[index,k]] = P[[k,index]]
        if index > 0:
            L[[index,k],0:(k-1)] = L[[k,index],0:(k-1)]

    # calculating next column in L and modifing rows in U
    for j in range(k+1,n):
        L[j,k] = U[j,k] / U[k,k]
        U[j,k:] -= L[j,k]*U[k,k:]

return L,U,P

Answer 1

好了，下面的代码解决了这个问题。 比较scipy结果和您提供的链接是可以的。 但是，scipy库转置了P矩阵。

import numpy as np

def LU_partial_decomposition(matrix):
    n, m = matrix.shape
    P = np.identity(n)
    L = np.identity(n)
    U = matrix.copy()
    PF = np.identity(n)
    LF = np.zeros((n,n))
    for k in range(0, n - 1):
        index = np.argmax(abs(U[k:,k]))
        index = index + k 
        if index != k:
            P = np.identity(n)
            P[[index,k],k:n] = P[[k,index],k:n]
            U[[index,k],k:n] = U[[k,index],k:n] 
            PF = np.dot(P,PF)
            LF = np.dot(P,LF)
        L = np.identity(n)
        for j in range(k+1,n):
            L[j,k] = -(U[j,k] / U[k,k])
            LF[j,k] = (U[j,k] / U[k,k])
        U = np.dot(L,U)
    np.fill_diagonal(LF, 1)
    return PF, LF, U

# Usage
A = [[2, 1, 1, 0], [4, 3, 3, 1], [8, 7, 9, 5], [6, 7, 9, 8]]
A = np.array(A)
P1, L1, U1 = LU_partial_decomposition(A)

# Scipy
import scipy
import scipy.linalg

P2, L2, U2 = scipy.linalg.lu(A) # P2 = P2.T