Numpy slicing vs list slicing

Question

My list:

a=[1,2,3,4,5]

b=a[:]

id(a)
>>2181314756864

id(b)
>>2181314855232 (different as slicing creates a new object)

id(a[0])
>>140734633334432

id(b[0])
>>140734633334432 (same)

b[0]=-1

b
>>[-1,2,3,4,5]

a
>>[1,2,3,4,5]  --perfectly fine

But in case of numpy:

import numpy as np

l=np.array([1,2,3,4,5])

p=l[:]

id(l)
>>2181315005136

id(p)
>>2181315019840 (different as it creates a new object, which is fine)

id(l[0])

>>2181314995440

id(p[0])
>>2181314995952 (different)

But:

p[0]=-1

p
>>array([-1,2,3,4,5])

l

>>array([-1,2,3,4,5])

Even though the memory addresses of first elements of numpy arrays are different, l is also being updated. Can anyone explain the concept behind this?

Answer 1

Talking about id as a memory address is a bit of a red herring. Yes, in CPython, it happens to be a memory address, but the address of what? Not , as it happens, the actual numeric data, but the Python object that describes the numeric data!

Slicing in numpy creates a new Python object (hence a different id), but the new object shares the storage for the actual array with the old one.

If id(A) == id(B) , then, for example, A.shape could not possibly be different from B.shape !