np.concatenate a ND tensor/array with a 1D array

Question

I have two arrays a & b

a.shape
(5, 4, 3)
array([[[ 0.        ,  0.        ,  0.        ],
        [ 0.        ,  0.        ,  0.        ],
        [ 0.          


        

Answer 1

You can use np.repeat:

r = np.concatenate((a, b.reshape(1, 1, -1).repeat(a.shape[0], axis=0)), axis=1)

What this does, is first reshape your b array to match the dimensions of a, and then repeat its values as many times as needed according to a's first axis:

b3D = b.reshape(1, 1, -1).repeat(a.shape[0], axis=0)

array([[[1, 2, 3]],

       [[1, 2, 3]],

       [[1, 2, 3]],

       [[1, 2, 3]],

       [[1, 2, 3]]])

b3D.shape
(5, 1, 3)

This intermediate result is then concatenated with a -

r = np.concatenate((a, b3d), axis=0)

r.shape
(5, 5, 3)

This differs from your current answer mainly in the fact that the repetition of values is not hard-coded (i.e., it is taken care of by the repeat).

If you need to handle this for a different number of dimensions (not 3D arrays), then some changes are needed (mainly in how remove the hardcoded reshape of b).

---

Timings

a = np.random.randn(100, 99, 100)
b = np.random.randn(100)

# Tai's answer
%timeit np.insert(a, 4, b, axis=1)
100 loops, best of 3: 3.7 ms per loop

# Divakar's answer
%%timeit 
b3D = np.broadcast_to(b,(a.shape[0],1,len(b)))
np.concatenate((a,b3D),axis=1)

100 loops, best of 3: 3.67 ms per loop

# solution in this post
%timeit np.concatenate((a, b.reshape(1, 1, -1).repeat(a.shape[0], axis=0)), axis=1)
100 loops, best of 3: 3.62 ms per loop

These are all pretty competitive solutions. However, note that performance depends on your actual data, so make sure you test things first!