can I split numpy array with mask?

You can generate the b as a broadcasted sum between the indices you want, and a shift-vector. Then you can broadcast again into a bigger size. Since the output in your examples are not depending on the a array, I'm disregarding that.

from numpy import array, broadcast_to, arange
from numpy.random import random

a = random((10,10,10)) # not used on the code at all.... don't understand what it is for...

b = [0,2,3]
b_array = array(b)
b_shifts = arange(3).reshape(-1,1)
c_cell= b+b_shifts # here, they are broadcasted toegether. one is a row-vector and one is a column-vector...
c = broadcast_to(c_cell,(10,10,3,3))

you might want to create b_shifts with some other method depending on step size and so on...


EDIT based on your comments, it seems a more accurate answer is:

from numpy import array, arange
a = arange(2*2*10).reshape((2,2,10)) # some example input 
b = array([0,2,3])                   # the 'template' to extract
shifts = arange(3).reshape(-1,1)     # 3 is the number of repeats
indexer = b+shifts                   # broadcasted sum makes a matrix
c = a[:,:,indexer]                   # extract

This will take the b array as a kind of template, and repeat it with a certain shift. Finally, it will extract those entries from every array a[i,j,:] into c[i,j,:,:]. Otput from above is:

print(a)

[[[ 0  1  2  3  4  5  6  7  8  9]
  [10 11 12 13 14 15 16 17 18 19]]
 [[20 21 22 23 24 25 26 27 28 29]
  [30 31 32 33 34 35 36 37 38 39]]]

print(c)

[[[[ 0  2  3]
   [ 1  3  4]
   [ 2  4  5]]
  [[10 12 13]
   [11 13 14]
   [12 14 15]]]
 [[[20 22 23]
   [21 23 24]
   [22 24 25]]
  [[30 32 33]
   [31 33 34]
   [32 34 35]]]]

A moving windows approach using as_strided:

In [1]: a = np.arange(6)                                                                       
In [2]: a                                                                                      
Out[2]: array([0, 1, 2, 3, 4, 5])
In [3]: as_strided = np.lib.stride_tricks.as_strided                                           

For this one-shift the strides parameter is easy. The shape requires more thinking - how many rows we expect, and the maximum index:

In [5]: b = as_strided(a, shape=(3,4), strides=(8,8))                                          
In [6]: b                                                                                      
Out[6]: 
array([[0, 1, 2, 3],
       [1, 2, 3, 4],
       [2, 3, 4, 5]])

Then select columns:

In [8]: b[:,[0,2,3]]                                                                           
Out[8]: 
array([[0, 2, 3],
       [1, 3, 4],
       [2, 4, 5]])

To extend it to the 3d case, I'll work from https://stackoverflow.com/a/60881930/901925, LudvigH's answer

In [10]: a = np.arange(2*2*10).reshape((2,2,10)) # some example input  
    ...: b = np.array([0,2,3])                                                                 
In [11]: a                                                                                     
Out[11]: 
array([[[ 0,  1,  2,  3,  4,  5,  6,  7,  8,  9],
        [10, 11, 12, 13, 14, 15, 16, 17, 18, 19]],

       [[20, 21, 22, 23, 24, 25, 26, 27, 28, 29],
        [30, 31, 32, 33, 34, 35, 36, 37, 38, 39]]])
In [12]: a.shape                                                                               
Out[12]: (2, 2, 10)
In [13]: a.strides                                                                             
Out[13]: (160, 80, 8)
In [18]: a1 = as_strided(a, shape=(2,2,3,4), strides=(160,80,8,8))                             
In [19]: a1                                                                                    
Out[19]: 
array([[[[ 0,  1,  2,  3],
         [ 1,  2,  3,  4],
         [ 2,  3,  4,  5]],

        [[10, 11, 12, 13],
         [11, 12, 13, 14],
         [12, 13, 14, 15]]],


       [[[20, 21, 22, 23],
         [21, 22, 23, 24],
         [22, 23, 24, 25]],

        [[30, 31, 32, 33],
         [31, 32, 33, 34],
         [32, 33, 34, 35]]]])

This is just an extension of the first case, with the first 2 dimensions just going-along for the ride. It's the last dimensions that expanded into the 2d window.

Again selecting a subset of the columns:

In [20]: a1[:,:,:,b]                                                                           
Out[20]: 
array([[[[ 0,  2,  3],
         [ 1,  3,  4],
         [ 2,  4,  5]],

        [[10, 12, 13],
         [11, 13, 14],
         [12, 14, 15]]],


       [[[20, 22, 23],
         [21, 23, 24],
         [22, 24, 25]],

        [[30, 32, 33],
         [31, 33, 34],
         [32, 34, 35]]]])

The as_strided step is efficient, creating a view. But the indexing will make a copy. We'd have to do some timing to test it against LudvigH's shifted index approach.