How can you most efficiently stack N rows of a 3D array to form a 4D array, but only advance one row-CodePudding

I would like to stack N rows of a 3D array one after the other to form a 4D array. I would like to start with the first N rows, i.e. rows = 1, ..., N, and always progress one row, i.e. the next N rows would be rows = 2, ..., N 1. That means the first N rows of the 3D array become the first row of the 4D array. The next N rows of the 3D array become the second row of the 4D array, and so on.

I've already found a way to solve my problem. However, the code is inefficient and it takes forever to compute a large 3D array with many rows. Here is what I did:

import numpy as np

N = 3
X = (np.ones((5,3)).transpose()*np.array([1,2,3,4,5])).transpose()
X = np.dstack((X, 2*X))

Z = np.empty((0, N, X.shape[1], X.shape[2]))

for i in range(N-1, X.shape[0]):
    temp1 = X[i-N 1:i 1, :, :]
    temp1 = np.expand_dims(temp1, axis=0)
    Z = np.append(Z, temp1, axis=0)

I hope the code makes it clear what my goal is.

Thanks in advance for your help.

CodePudding user response：

The part that makes your solutions slow, probably is the np.append operation as internally memory for the array has to be (re-)allocated for each loop iteration. For large arrays this can be quite some overhead. From your code snippet I assume the length of Z is kown prior to the loop, so this modification could probably speed up your code:

import numpy as np

N = 3
X = (np.ones((5,3)).transpose()*np.array([1,2,3,4,5])).transpose()
X = np.dstack((X, 2*X))

n_z = X.shape[0] - N   1
Z = np.empty((n_z, N, X.shape[1], X.shape[2]))

for idx, i in enumerate(range(N-1, X.shape[0])):
    Z[idx] = X[i-N 1:i 1, :, :]