Python实现Householder变换报列表索引不能为元组TypeError错误

问题原因 #

你传入函数的A是Python原生嵌套列表，不支持A[行切片, 列索引]这种逗号分隔的多维索引写法，Python会把(k + 1) :, k识别为一个元组作为索引，所以触发类型错误。

解决方案 #

1. 函数内部先将输入转换为numpy数组，保证支持多维索引操作
2. 修正代码缩进问题：原代码中第二个for l in ...循环及后续矩阵赋值代码，应缩进放在外层for k in ...循环内部，否则只会在k循环结束后执行一次，逻辑不符合Householder变换的迭代要求
3. 给householder函数添加返回值，否则最后打印new会得到None

修正后可运行代码 #

import numpy as np
from numpy import array,identity,diagonal
from math import sqrt,asin,acos,sin,cos,atan

def householder(A):
    # 转换输入为numpy数组，支持多维索引
    A = np.array(A, dtype=np.double)
    n = len(A)
    v = np.zeros(n, dtype=np.double)
    u = np.zeros(n, dtype=np.double)
    z = np.zeros(n, dtype=np.double)

    for k in range(0, n - 2):
        if np.isclose(A[k + 1][k], 0.0):
            α = -np.sqrt(np.sum(A[(k + 1) :, k] ** 2))
        else:
            α = -np.sign(A[k + 1][k]) * np.sqrt(np.sum(A[(k + 1) :, k] ** 2))

        two_r_squared = α ** 2 - α * A[k + 1][k]
        v[k] = 0.0
        v[k + 1] = A[k + 1][k] - α
        v[(k + 2) :] = A[(k + 2) :, k]
        u[k:] = 1.0 / two_r_squared * np.dot(A[k:, (k + 1) :], v[(k + 1) :])
        z[k:] = u[k:] - np.dot(u, v) / (2.0 * two_r_squared) * v[k:]

        # 以下逻辑移到k循环内部，保证每轮迭代都更新矩阵
        for l in range(k + 1, n - 1):
            A[(l + 1) :, l] = (
                A[(l + 1) :, l] - v[l] * z[(l + 1) :] - v[(l + 1) :] * z[l]
            )
            A[l, (l + 1) :] = A[(l + 1) :, l]
            A[l, l] = A[l, l] - 2 * v[l] * z[l]

        A[-1, -1] = A[-1, -1] - 2 * v[-1] * z[-1]
        A[k, (k + 2) :] = 0.0
        A[(k + 2) :, k] = 0.0

        A[k + 1][k] = A[k + 1][k] - v[k + 1] * z[k]
        A[k][k + 1] = A[k + 1][k]
    return A

A= [ [3,2,1,2] , [2,-1,1,2] , [1,1,4,30] , [2,2,3,1] ]
new= householder(A)
print(new)

内容的提问来源于stack exchange，提问作者Dries Gouws