考虑以下数据：

data = np.array([[i for i in range(3)] for _ in range(9)])
print(data)
print(f'data has shape {data.shape}')

[[0 1 2]
 [0 1 2]
 [0 1 2]
 [0 1 2]
 [0 1 2]
 [0 1 2]
 [0 1 2]
 [0 1 2]
 [0 1 2]]
data has shape (9, 3)

还有一个参数，我们称之为history。history的功能是，它在第一维上堆叠history许多数组[0 1 2]。作为一个示例，考虑使用history=2对该过程进行一次迭代

history = 2
data = np.array([[[0, 1, 2], [0, 1, 2]]])
print(f'data has now shape {data.shape}')
data has now shape (1, 2, 3)

现在，让我们考虑两个迭代：

history = 2
data = np.array([[[0, 1, 2], [0, 1, 2]],[[0, 1, 2], [0, 1, 2]]])
print(f'data has now shape {data.shape}')
data has now shape (2, 2, 3)

应重复此过程，直到数据完全处理。这意味着，我们可能会在最后丢失一些数据，因为data.shape[0]/history % 2 != 0。因此，history=2的最终结果为

([[[0, 1, 2],
        [0, 1, 2]],

       [[0, 1, 2],
        [0, 1, 2]],

       [[0, 1, 2],
        [0, 1, 2]],

       [[0, 1, 2],
        [0, 1, 2]]])

如何才能做到这一点的性能？