使用numpy数组对函数进行矢量化

Question

我正在尝试加快我编写的一些代码的速度，但在这样做时遇到了很大的困难。我知道，能够删除for循环和使用numpy可以帮助做到这一点，所以这是我一直尝试的，但收效甚微

没有任何加速的工作功能是

def acf(x, y, z, cutoff=0):
    steps = x.shape[1]
    natoms = x.shape[0]

    z_x = np.zeros((steps,natoms))
    z_y, z_z = np.zeros_like(z_x), np.zeros_like(z_x)

    xmean = np.mean(x, axis=1)
    ymean = np.mean(y, axis=1)
    zmean = np.mean(z, axis=1)

    for k in range(steps-cutoff): # x.shape[1]
        xtemp, ytemp, ztemp = [], [], []
        for i in range(x.shape[0]): # natoms
            xtop, ytop, ztop = 0.0, 0.0, 0.0
            xbot, ybot, zbot = 0.0, 0.0, 0.0
            for j in range(steps-k): # x.shape[1]-k
                xtop += (x[i][j] - xmean[i]) * (x[i][j+k] - xmean[i])
                ytop += (y[i][j] - ymean[i]) * (y[i][j+k] - ymean[i])
                ztop += (z[i][j] - zmean[i]) * (z[i][j+k] - zmean[i])
                xbot += (x[i][j] - xmean[i])**2
                ybot += (y[i][j] - ymean[i])**2
                zbot += (z[i][j] - zmean[i])**2
            xtemp.append(xtop/xbot)
            ytemp.append(ytop/ybot)
            ztemp.append(ztop/zbot)
        z_x[k] = xtemp
        z_y[k] = ytemp
        z_z[k] = ztemp

    z_x = np.mean(np.array(z_x), axis=1)
    z_y = np.mean(np.array(z_y), axis=1)
    z_z = np.mean(np.array(z_z), axis=1)

    return z_x, z_y, z_z

此函数的输入x、y和z是相同尺寸的numpy数组。x（或y或z）的一个例子是：

x = np.array([[1,2,3],[4,5,6]])

到目前为止，我所能做的是

def acf_quick(x, y, z, cutoff=0):
    steps = x.shape[1]
    natoms = x.shape[0]

    z_x = np.zeros((steps,natoms))
    z_y, z_z = np.zeros_like(z_x), np.zeros_like(z_x)

    x -= np.mean(x, axis=1, keepdims=True)
    y -= np.mean(y, axis=1, keepdims=True)
    z -= np.mean(z, axis=1, keepdims=True)

    for k in range(steps-cutoff): # x.shape[1]
        for i in range(natoms):
            xtop, ytop, ztop = 0.0, 0.0, 0.0
            xbot, ybot, zbot = 0.0, 0.0, 0.0
            for j in range(steps-k): # x.shape[1]-k
                xtop += (x[i][j]) * (x[i][j+k])
                ytop += (y[i][j]) * (y[i][j+k])
                ztop += (z[i][j]) * (z[i][j+k])
                xbot += (x[i][j])**2
                ybot += (y[i][j])**2
                zbot += (z[i][j])**2
            z_x[k][i] = xtop/xbot
            z_y[k][i] = ytop/xbot
            z_z[k][i] = ztop/xbot

    z_x = np.mean(np.array(z_x), axis=1)
    z_y = np.mean(np.array(z_y), axis=1)
    z_z = np.mean(np.array(z_z), axis=1)

    return z_x, z_y, z_z

这将使其速度提高约33%，但我相信有一种方法可以使用类似于x[:][j]的东西来删除for i in range(natoms)。到目前为止，我一直没有成功，任何帮助都将不胜感激

在任何人询问之前，我知道这是一个自相关函数，numpy、scipy等中内置了一些函数，但我需要编写自己的函数

Answer 1

以下是循环的矢量化形式：

def acf_quick_new(x, y, z, cutoff=0):
    steps = x.shape[1]
    natoms = x.shape[0]

    lst_inputs = [x.copy(),y.copy(),z.copy()]
    lst_outputs = []
    for x_ in lst_inputs:

        z_x_ = np.zeros((steps,natoms))

        x_ -= np.mean(x_, axis=1, keepdims=True)

        x_top = np.diag(np.dot(x_,x_.T))
        x_bot = np.sum(x_**2, axis=1)

        z_x_[0,:] = np.divide(x_top, x_bot)


        for k in range(1,steps-cutoff): # x.shape[1]

            x_top = np.diag(np.dot(x_[:,:-k],x_.T[k:,:]))
            x_bot = np.sum(x_[:,:-k]**2, axis=1)

            z_x_[k,:] = np.divide(x_top, x_bot)


        z_x_ = np.mean(np.array(z_x_), axis=1)
        lst_outputs.append(z_x_)    

    return lst_outputs

请注意，在您的_快速函数中有一个小错误：您总是用xbot来除，而不是用xbot、ybot和zbot。此外，我的建议可以写得更好一点，但它应该可以解决您的问题，并大大加快计算速度：）