Can somebody explain the output of this list operation?

Question

I got this in the Challanges:

a=[1,[2,3]]
b=a[:]
print(b)
a[0]=3
a[1][1]=5
print(b)

#Output is [1,[2,5]]

Why isn't the output [3, [2,5]]?

I understand th reason is in the semicolon, so that it's not the same to write b=a and b=a[:].
But this is what I don't understand :-)

Answer 1

a[:] makes a shallow copy. It takes all the references in a and puts them into a new list b.

So when you change an element of a, like in line 4, b will not know that, and not change.

In line 5, you change an element of the *inner* list - and this is just another reference to the same inner list in a.

I have recently written about the topic. I'd be happy if you take a look:

https://code.sololearn.com/c89ejW97QsTN/?ref=app

Answer 2

Thank you! 🙂

I think you got how it works, but only once again about the term 'shallow copy':

The opposite is 'deep copy'. This means, that however deep (multidimensional) an object might be, it gets actually copied down to the innermost item.

a = [[], []]
b = a[:]

This is a shallow copy because only the outermost object is copied, the outer list.

So although b is a separate list, both the inner lists are not copied, but are only other references to the very same inner lists as in a.

Now if instead I had written:

b = []
for list_ in a:
b.append(list_[:])

Now I have copied the outer list AND made copies from the inner lists and put them in there.

So here everything to the bottom of it is copied, making it a deep copy.

Answer 3

Thank you very much, HonFu!

So, if I understood correctly (and using your examples), the list a in the case above is something similar to this:

twins=["lucy","betty"]
a=["john",twins]

So, by doing b=a[:]

I create another list (b) which copies not the elements, but the references to the elements of (a): a string (shallow copy) and a list (not shallow copy).

Very clear article!

Answer 4

Great!
Thank you again, so much.