在python中实现快速的Walsh-hadamard变换

Question

使用这里https://github.com/dingluo/fwht/blob/master/FWHT.py给定的代码，我编写了一个变体，该变体适用于矩阵的每一列，并以Hadamard顺序返回转换。问题甚至是原始的，与简单的Fortran实现相比，我的变体至少慢了两个数量级。在改善性能（速度）方面的任何帮助将不胜感激。这是我的变体。

def fwhtvec(x, s=0):
    """ Fast Walsh-Hadamard Transform
    Based on mex function written by Chengbo Li@Rice Uni
    for his TVAL3 algorithm. His code is according to the
    K.G. Beauchamp's book -- Applications of Walsh and
    Related Functions. Modiefied by me to get it in hadamard order and
    now works for all the vectors in a matrix. For s = 0 it is
    usual Hadamard with matrix [[1, 1],[1, -1]]/sqrt(2), for s = 1
    it is [[1, -i],[i, 1]]/sqrt(2)
    """
    a = (-1j) ** s
    N, nc = x.shape
    x = x.T.reshape(nc, N, 1)
    G = N / 2  # Number of Groups
    M = 2  # Number of Members in Each Group

    # First stage
    y = np.zeros((nc, N / 2, 2), dtype=np.complex128)
    y[0:nc, 0:G, 0] = x[0:nc, 0:G, 0] + a * x[0:nc, G:N, 0]
    y[0:nc, 0:G, M - 1] = np.conj(a) * x[0:nc, 0:G, 0] - x[0:nc, G:N, 0]
    x = y.copy()
    # Second and further stage
    for nStage in xrange(2, int(log(N, 2)) + 1):
        y = np.zeros((nc, G / 2, M * 2), dtype=np.complex128)
        y[0:nc, 0:G / 2, 0:M * 2:4] = x[0:nc, 0:G // 2, 0:M:2] + a * x[0:nc, G // 2:G, 0:M:2]
        y[0:nc, 0:G / 2, 1:M * 2:4] = np.conj(a) * x[0:nc, 0:G // 2, 0:M:2] - x[0:nc, G // 2:G, 0:M:2]
        y[0:nc, 0:G / 2, 2:M * 2:4] = x[0:nc, 0:G // 2, 1:M:2] + a * x[0:nc, G // 2:G, 1:M:2]
        y[0:nc, 0:G / 2, 3:M * 2:4] = np.conj(a) * x[0:nc, 0:G // 2, 1:M:2] - x[0:nc, G // 2:G, 1:M:2]
        x = y.copy()
        G = G / 2
        M = M * 2
    x = y[:, 0, :].T / (np.sqrt(2) ** int(log(N, 2)))
    return x