A $3\times 4 \times 4$ array is created with:
In [x]: a = np.linspace(1,48,48).reshape(3,4,4)
Index or slice this array to obtain the following
(a)
20.0
(b)
[ 9. 10. 11. 12.]
(c) The $4 \times 4$ array:
[[ 33. 34. 35. 36.]
[ 37. 38. 39. 40.]
[ 41. 42. 43. 44.]
[ 45. 46. 47. 48.]]
(d) The $3 \times 2$ array:
[[ 5., 6.],
[ 21., 22.],
[ 37., 38.]]
(e) The $4 \times 2$ array:
[[ 36. 35.]
[ 40. 39.]
[ 44. 43.]
[ 48. 47.]]
(f) The $3 \times 4$ array:
[[ 13. 9. 5. 1.]
[ 29. 25. 21. 17.]
[ 45. 41. 37. 33.]]
(g) (Harder) Using an array of indexes, the $2 \times 2$ array:
[[ 1. 4.]
[ 45. 48.]]
Indexing and slicing a NumPy array:
(a) a[1,0,3]
(b) a[0,2,:]
(or just a[0,2]
)
(c) a[2,...]
(or a[2,:,:]
or a[2]
)
(d) a[:,1,:2]
(e) a[2,:,:1:-1]
("in the third block, for each row take the items in the middle two columns'').
(f) a[:,::-1,0]
("for each block, traverse the rows backwards and take the item in the first column of each'').
(g) Defining the three $2\times 2$ index arrays for the blocks, rows and columns locating our elements as follows:
ia = np.array([[0, 0], [2, 2]])
ja = np.array([[0, 0], [3, 3]])
ka = np.array([[0, 3], [0, 3]])
a[ia,ja,ka]
returns the desired result.