How to add an extra column to a NumPy array

Question

Let’s say I have a NumPy array, a:

a = np.array([
    [1, 2, 3],
    [2, 3, 4]
    ])

And I would like to add a column of ze

Answer 1

I find the following most elegant:

b = np.insert(a, 3, values=0, axis=1) # Insert values before column 3

An advantage of insert is that it also allows you to insert columns (or rows) at other places inside the array. Also instead of inserting a single value you can easily insert a whole vector, for instance duplicate the last column:

b = np.insert(a, insert_index, values=a[:,2], axis=1)

Which leads to:

array([[1, 2, 3, 3],
       [2, 3, 4, 4]])

For the timing, insert might be slower than JoshAdel's solution:

In [1]: N = 10

In [2]: a = np.random.rand(N,N)

In [3]: %timeit b = np.hstack((a, np.zeros((a.shape[0], 1))))
100000 loops, best of 3: 7.5 µs per loop

In [4]: %timeit b = np.zeros((a.shape[0], a.shape[1]+1)); b[:,:-1] = a
100000 loops, best of 3: 2.17 µs per loop

In [5]: %timeit b = np.insert(a, 3, values=0, axis=1)
100000 loops, best of 3: 10.2 µs per loop

Answer 2

I think a more straightforward solution and faster to boot is to do the following:

import numpy as np
N = 10
a = np.random.rand(N,N)
b = np.zeros((N,N+1))
b[:,:-1] = a

And timings:

In [23]: N = 10

In [24]: a = np.random.rand(N,N)

In [25]: %timeit b = np.hstack((a,np.zeros((a.shape[0],1))))
10000 loops, best of 3: 19.6 us per loop

In [27]: %timeit b = np.zeros((a.shape[0],a.shape[1]+1)); b[:,:-1] = a
100000 loops, best of 3: 5.62 us per loop

Answer 3

np.concatenate also works

>>> a = np.array([[1,2,3],[2,3,4]])
>>> a
array([[1, 2, 3],
       [2, 3, 4]])
>>> z = np.zeros((2,1))
>>> z
array([[ 0.],
       [ 0.]])
>>> np.concatenate((a, z), axis=1)
array([[ 1.,  2.,  3.,  0.],
       [ 2.,  3.,  4.,  0.]])

Answer 4

I like JoshAdel's answer because of the focus on performance. A minor performance improvement is to avoid the overhead of initializing with zeros, only to be overwritten. This has a measurable difference when N is large, empty is used instead of zeros, and the column of zeros is written as a separate step:

In [1]: import numpy as np

In [2]: N = 10000

In [3]: a = np.ones((N,N))

In [4]: %timeit b = np.zeros((a.shape[0],a.shape[1]+1)); b[:,:-1] = a
1 loops, best of 3: 492 ms per loop

In [5]: %timeit b = np.empty((a.shape[0],a.shape[1]+1)); b[:,:-1] = a; b[:,-1] = np.zeros((a.shape[0],))
1 loops, best of 3: 407 ms per loop

Answer 5

One way, using hstack, is:

b = np.hstack((a, np.zeros((a.shape[0], 1), dtype=a.dtype)))