Why was p[:] designed to work differently in these two situations?

Question

p = [1,2,3]
print(p) # [1, 2, 3]

q=p[:]  # supposed to do a shallow copy
q[0]=11
print(q) #[11, 2, 3] 
print(p) #[1, 2, 3] 
# above confirms that q is not p, and is a d         


        

Answer 1

As others have stated; p[:] deletes all items in p; BUT will not affect q. To go into further detail the list docs refer to just this:

All slice operations return a new list containing the requested elements. This means that the following slice returns a new (shallow) copy of the list:

>>> squares = [1, 4, 9, 16, 25]
...
>>> squares[:]
[1, 4, 9, 16, 25]

So q=p[:] creates a (shallow) copy of p as a separate list but upon further inspection it does point to a completely separate location in memory.

>>> p = [1,2,3]
>>> q=p[:]
>>> id(q)
139646232329032
>>> id(p)
139646232627080

This is explained better in the copy module:

A shallow copy constructs a new compound object and then (to the extent possible) inserts references into it to the objects found in the original.

Although the del statement is performed recursively on lists/slices:

Deletion of a target list recursively deletes each target, from left to right.

So if we use del p[:] we are deleting the contents of p by iterating over each element, whereas q is not altered as stated earlier, it references a separate list although having the same items:

>>> del p[:]
>>> p
[]
>>> q
[1, 2, 3]

In fact this is also referenced in the list docs as well in the list.clear method:

list.copy()

Return a shallow copy of the list. Equivalent to a[:].

list.clear()

Remove all items from the list. Equivalent to del a[:].