Indexing and slicing a NumPy array

Q6.1.5Indexing and slicing a NumPy array

A $3\times 4 \times 4$ array is created with:

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.]

[[ 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.]]

Show solution

Indexing and slicing a NumPy array:

(a) a[1,0,3]

(b) a[0,2,:] (or just a[0,2])

(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.