Why do I get zero here? Assigning array elements-CodePudding

Why does this

import numpy as np

a = np.array([[1,2,3],[4,5,6]])
a_old = a[0]
a[0] = a[0]   np.array([1,1,1])
print(a[0] - a_old)

give

[0 0 0]

and not

[1 1 1]

? But on the other hand

b = 2
b_old = b
b = b   1
print(b - b_old)

gives indeed

as I would expect. I suppose there is a difference between assigning something to np.array elements, and assigning something to a variable. But I don't quite see through it. Thanks for help!

CodePudding user response：

This can be tricky in numpy! It's because that when you do a_old = a[0], you are creating a view of the data in a, not a copy of the data. To quote the numpy indexing documentation:

It must be noted that the returned array is a view, i.e., it is not a copy of the original, but points to the same values in memory as does the original array.

So when you add to the original array: a[0] = a[0] np.array([1,1,1]), the "value" of a_old changes as well since a_old points to the same location in memory. If you want to create a copy, you can simply edit your code:

a_old = a[0].copy()

In your floating point example, b_old = b copies the value of b to a new point in memory, so b and b_old can be changed without affecting one another.

CodePudding user response：

This happens because in Python, as well as other programming languages, lists are Reference type. So, the value stored in a[0], isn't the list but it's a memory address where the list is stored.

So, the instruction a_old = a[0], copy the address of a[0] in a_old, and both the variables will reference the same list.

Instead, doing b_old = b, you're copying the value 2 directly.

To notice this, try running print(a_old is a[0]), you should see True.

Now, to solve your problem, you can run a_old = a[0].copy(). After this, print(a_old is a[0]) will output False, because the reference of the two variables are different.