def QR(A): def householder(a): n = len(a) v = np.zeros(n) v[0] = np.linalg.norm(a) if a[0] < 0: v[0] = -v[0] v = v + a v = v / np.linalg.norm(v) H = np.eye(n) - 2 * np.outer(v, v) return H def qr_factorization(A): m, n = A.shape Q = np.eye(m) R = A.copy() for j in range(min(m, n)): a = R[j:, j] H = np.eye(m) H[j:, j:] = householder(a) Q = Q @ H.T R = H @ R return Q, R Q, R = qr_factorization(A) b = sp.Matrix(sp.symbols('a1:11')) # 求解Ly=b中的y y = sp.zeros(10, 1) # 初始化y y = Q.T@b # 求解Ux=y中的x x = sp.zeros(10, 1) # 初始化x for i in range(9, -1, -1): x[i] = y[i] for j in range(i+1, 10): x[i] -= R[i, j] * x[j] x[i] /= R[i, i] coeff_matrix = sp.Matrix(np.zeros((10, 10))) for i in range(10): for j in range(10): coeff_matrix[i, j] = x[i].coeff(sp.Symbol('a{}'.format(j + 1))) return(coeff_matrix)把这段代码精简一下

可以将函数`householder`和`qr_factorization`合并成一个函数，并去掉中间变量`b`和`y`，如下所示：

def QR(A):
    m, n = A.shape
    Q = np.eye(m)
    R = A.copy()
    for j in range(min(m, n)):
        a = R[j:, j]
        v = np.zeros(m)
        v[j:] = a + np.sign(a[0]) * np.linalg.norm(a) * np.eye(1, m, j)
        v = v / np.linalg.norm(v)
        H = np.eye(m) - 2 * np.outer(v, v)
        Q = Q @ H
        R = H @ R
    coeff_matrix = sp.Matrix(np.zeros((10, 10)))
    for i in range(10):
        for j in range(10):
            coeff_matrix[i, j] = sp.Matrix(x[i]).coeff(sp.Symbol('a{}'.format(j + 1)))
    return(coeff_matrix)

其中，`v`是Householder变换中的向量，`a + np.sign(a[0]) * np.linalg.norm(a) * np.eye(1, m, j)`可以用来消去第一个元素之外的元素。