idiom for getting contiguous copies

Question

In the help of numpy.broadcst-array, an idiom is introduced. However, the idiom give exactly the same output as original command. Waht is the meaning of "getting contiguous copies instead of non-contiguous views."?

https://docs.scipy.org/doc/numpy/reference/generated/numpy.broadcast_arrays.html

x = np.array([[1,2,3]])
y = np.array([[1],[2],[3]])

np.broadcast_arrays(x, y)
[array([[1, 2, 3],
       [1, 2, 3],
       [1, 2, 3]]), array([[1, 1, 1],
       [2, 2, 2],
       [3, 3, 3]])]

Here is a useful idiom for getting contiguous copies instead of non-contiguous views.

[np.array(a) for a in np.broadcast_arrays(x, y)]

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

Answer 1

To understand the difference try writing into the new arrays:

Let's begin with the contiguous copies.

>>> import numpy as np
>>> x = np.array([[1,2,3]])
>>> y = np.array([[1],[2],[3]])
>>> 
>>> xc, yc = [np.array(a) for a in np.broadcast_arrays(x, y)]
>>> xc
array([[1, 2, 3],
       [1, 2, 3],
       [1, 2, 3]])

We can modify an element and nothing unexpected will happen.

>>> xc[0, 0] = 0
>>> xc
array([[0, 2, 3],
       [1, 2, 3],                                                                                                   
       [1, 2, 3]])
>>> x
array([[1, 2, 3]])

Now, let's try the same with the broadcasted arrays:

>>> xb, yb = np.broadcast_arrays(x, y)
>>> xb
array([[1, 2, 3],
       [1, 2, 3],
       [1, 2, 3]])

Although we only write to the top left element ...

>>> xb[0, 0] = 0

... the entire left column will change ...

>>> xb
array([[0, 2, 3],
       [0, 2, 3],
       [0, 2, 3]])

... and also the input array.

>>> x
array([[0, 2, 3]])