向量化在ndarray的子​​阵列上操作的函数

Question

我在3D阵列的每个2D切片上都有一个功能。如何对函数进行向量化以避免循环以提高性能？例如：

def interp_2d(x0,y0,z0,x1,y1):
    # x0, y0 and z0 are 2D array
    # x1 and y1 are 2D array
    # peform 2D interpolation
    return z1

# now I want to call the interp_2d for each 2D slice of z0_3d as following:
for k in range(z0_3d.shape[2]):
    z1_3d[:,:,k]=interp_2d(x0, y0, z0_3d[:,:,k], x1, y1)

Answer 1

如果不重新实现interp_2d，则无法进行矢量化。但是，假设interp_2d是某种类型的插值，那么操作可能是线性的。那是lambda z0: interp_2d(x0, y0, z0, x1, y1)可能等同于np.dot(M, z0)，其中M是一些（可能是稀疏的）矩阵，依赖于x0，y0，x1和y1。现在，通过调用interp_2d函数，您可以在每次调用时隐式重新计算此矩阵，即使每次调用都相同。找出该矩阵的次数并将其重新应用到新的z0次数会更有效。

这是一个非常简单的1D插值示例：

x0 = [0., 1.]
x1 = 0.3
z0_2d = "some very long array with shape=(2, n)"

def interp_1d(x0, z0, x1):
    """x0 and z0 are length 2, 1D arrays, x1 is a float between x0[0] and x0[1]."""

    delta_x = x0[1] - x0[0]
    w0 = (x1 - x0[0]) / delta_x
    w1 = (x0[1] - x1) / delta_x
    return w0 * z0[0] + w1 * z0[1]

# The slow way.
for i in range(n):
     z1_2d[i] = interp_1d(x0, z0_2d[:,i], x1)
# Notice that the intermediate products w1 and w2 are the same on each
# iteration but we recalculate them anyway.

# The fast way.
def interp_1d_weights(x0, x1):
    delta_x = x0[1] - x0[0]
    w0 = (x1 - x0[0]) / delta_x
    w1 = (x0[1] - x1) / delta_x
    return w0, w1

 w0, w1 = interp_1d_weights(x0, x1)
 z1_2d = w0 * z0_2d[0,:] + w1 * z0_2d[1:0]

如果n非常大，预计速度会超过100倍。