将2D数组转换为具有重叠步幅的3D数组

Question

I would convert the 2d array into 3d with previous rows by using NumPy or native functions.

Input:

[[1,2,3],
 [4,5,6],
 [7,8,9],
 [10,11,12],
 [13,14,15]]

Output:

[[[7,8,9],    [4,5,6],    [1,2,3]],
 [[10,11,12], [7,8,9],    [4,5,6]],
 [[13,14,15], [10,11,12], [7,8,9]]]

Any one can help? I have searched online for a while, but cannot got the answer.

Answer 1

Approach #1

One approach with np.lib.stride_tricks.as_strided that gives us a view into the input 2D array and as such doesn't occupy anymore of the memory space -

L = 3  # window length for sliding along the first axis
s0,s1 = a.strides

shp = a.shape
out_shp = shp[0] - L + 1, L, shp[1]
strided = np.lib.stride_tricks.as_strided
out = strided(a[L-1:], shape=out_shp, strides=(s0,-s0,s1))

Sample input, output -

In [43]: a
Out[43]: 
array([[ 1,  2,  3],
       [ 4,  5,  6],
       [ 7,  8,  9],
       [10, 11, 12],
       [13, 14, 15]])

In [44]: out
Out[44]: 
array([[[ 7,  8,  9],
        [ 4,  5,  6],
        [ 1,  2,  3]],

       [[10, 11, 12],
        [ 7,  8,  9],
        [ 4,  5,  6]],

       [[13, 14, 15],
        [10, 11, 12],
        [ 7,  8,  9]]])

---

Approach #2

Alternatively, a bit easier one with broadcasting upon generating all of row indices -

In [56]: a[range(L-1,-1,-1) + np.arange(shp[0]-L+1)[:,None]]
Out[56]: 
array([[[ 7,  8,  9],
        [ 4,  5,  6],
        [ 1,  2,  3]],

       [[10, 11, 12],
        [ 7,  8,  9],
        [ 4,  5,  6]],

       [[13, 14, 15],
        [10, 11, 12],
        [ 7,  8,  9]]])

Answer 2

How about a list comprehension?

In [144]: np.array([l[i:i + 3][::-1] for i in range(0, len(l) - 2)])
Out[144]: 
array([[[ 7,  8,  9],
        [ 4,  5,  6],
        [ 1,  2,  3]],

       [[10, 11, 12],
        [ 7,  8,  9],
        [ 4,  5,  6]],

       [[13, 14, 15],
        [10, 11, 12],
        [ 7,  8,  9]]])