Confusion in array operation in numpy

Question

I generally use MATLAB and Octave, and i recently switching to python numpy. In numpy when I define an array like this

>>> a = np.array([[2,3],[4,5]])

it works great and size of the array is

>>> a.shape
(2, 2)

which is also same as MATLAB But when i extract the first entire column and see the size

>>> b = a[:,0]
>>> b.shape
(2,)

I get size (2,), what is this? I expect the size to be (2,1). Perhaps i misunderstood the basic concept. Can anyone make me clear about this??

Answer 1

A 1D numpy array* is literally 1D - it has no size in any second dimension, whereas in MATLAB, a '1D' array is actually 2D, with a size of 1 in its second dimension.

If you want your array to have size 1 in its second dimension you can use its .reshape() method:

a = np.zeros(5,)
print(a.shape)
# (5,)

# explicitly reshape to (5, 1)
print(a.reshape(5, 1).shape)
# (5, 1)

# or use -1 in the first dimension, so that its size in that dimension is 
# inferred from its total length
print(a.reshape(-1, 1).shape)
# (5, 1)

Edit

As Akavall pointed out, I should also mention np.newaxis as another method for adding a new axis to an array. Although I personally find it a bit less intuitive, one advantage of np.newaxis over .reshape() is that it allows you to add multiple new axes in an arbitrary order without explicitly specifying the shape of the output array, which is not possible with the .reshape(-1, ...) trick:

a = np.zeros((3, 4, 5))
print(a[np.newaxis, :, np.newaxis, ..., np.newaxis].shape)
# (1, 3, 1, 4, 5, 1)

np.newaxis is just an alias of None, so you could do the same thing a bit more compactly using a[None, :, None, ..., None].

---

* An np.matrix, on the other hand, is always 2D, and will give you the indexing behavior you are familiar with from MATLAB:

a = np.matrix([[2, 3], [4, 5]])
print(a[:, 0].shape)
# (2, 1)

For more info on the differences between arrays and matrices, see here.