slinalg.py 文件源码

def perform(self, node, inputs, outputs):
        # Kalbfleisch and Lawless, J. Am. Stat. Assoc. 80 (1985) Equation 3.4
        # Kind of... You need to do some algebra from there to arrive at
        # this expression.
        (A, gA) = inputs
        (out,) = outputs
        w, V = scipy.linalg.eig(A, right=True)
        U = scipy.linalg.inv(V).T

        exp_w = numpy.exp(w)
        X = numpy.subtract.outer(exp_w, exp_w) / numpy.subtract.outer(w, w)
        numpy.fill_diagonal(X, exp_w)
        Y = U.dot(V.T.dot(gA).dot(U) * X).dot(V.T)

        with warnings.catch_warnings():
            warnings.simplefilter("ignore", numpy.ComplexWarning)
            out[0] = Y.astype(A.dtype)