将对角矩阵提高到负功率1/2

Question

我正在尝试计算具有以下方程式的矩阵。

S = (D^?1/2) * W * (D^?1/2)

其中D是这种形式的对角矩阵：

array([[ 0.59484625,  0.        ,  0.        ,  0.        ],
       [ 0.        ,  0.58563893,  0.        ,  0.        ],
       [ 0.        ,  0.        ,  0.58280472,  0.        ],
       [ 0.        ,  0.        ,  0.        ,  0.58216725]])

和W：

array([[ 0.        ,  0.92311635,  0.94700586,  0.95599748],
       [ 0.92311635,  0.        ,  0.997553  ,  0.99501248],
       [ 0.94700586,  0.997553  ,  0.        ,  0.9995501 ],
       [ 0.95599748,  0.99501248,  0.9995501 ,  0.        ]])

我试图D^-1/2通过使用numpyfunction linalg.matrix_power(D,-1/2)和numpy.power(D,-1/2)and matrix_power函数引发TypeError: exponent must be an integer以及numpy.power函数引发进行计算RuntimeWarning: divide by zero encountered in power。

如何计算对角矩阵的负功率-1/2。请帮忙。

Answer 1

如果您可以更新D（例如您自己的答案），则只需更新其对角线索引处的项目，然后致电np.dot：

>>> D[np.diag_indices(4)] = 1/ (D.diagonal()**0.5)
>>> np.dot(D, W).dot(D)
array([[ 0.        ,  0.32158153,  0.32830723,  0.33106193],
       [ 0.32158153,  0.        ,  0.34047794,  0.33923936],
       [ 0.32830723,  0.34047794,  0.        ,  0.33913717],
       [ 0.33106193,  0.33923936,  0.33913717,  0.        ]])

或创建一个新的零点数组，然后用对角元素填充1/ (D.diagonal()**0.5)：

>>> arr = np.zeros(D.shape)
>>> np.fill_diagonal(arr, 1/ (D.diagonal()**0.5))
>>> np.dot(arr, W).dot(arr)
array([[ 0.        ,  0.32158153,  0.32830723,  0.33106193],
       [ 0.32158153,  0.        ,  0.34047794,  0.33923936],
       [ 0.32830723,  0.34047794,  0.        ,  0.33913717],
       [ 0.33106193,  0.33923936,  0.33913717,  0.        ]])