Changing list elements in shallow copy

Question

I have one question about list shallow copy.

In both examples, I modified one element of the list, but in example 1, list b changed, while in example 2,

Answer 1

[1. = copies the reference of object, hence any changes in either list, reflects in another

2. b=list(a) or b=a.copy() -> do the same work. That is it copies the reference of only the individual objects i.e like b[0]=a[0] and b2=a2 and so on. With int, string etc, it's like if x = 10 and y = x and changing the value of 'x' or 'y' won't affect the other. This is what happens for the remaining elements of the a and b when you do a shallow copy.

So as in your question when doing b=list(a) and a[1]=0 using a shallow copy behaves as explained above and hence the changes are not reflected in both the list . But the nested listed acts as list assignment like a=[1,2,3] and b=a and making a2=3 will change b2 to 3 as well, i.e.changes in a or b effect both (same as in case 1 above). So this is why in case of a list with in a list any changes reflects in both the list. As in your example doing d=list(c) (here when copying d[2]=c[2] this is similar to list assignment i.e. the reference is copied and in case of list assignment changes are reflected in both so changes to d2 or c2 is reflected in both list) so doing c[2][0] = 0 will also change d[2][0] to zero.

Try the code at http://www.pythontutor.com/visualize.html#mode=edit to understand better

a=[1,2,"hello",[3,4],5]
b=a
c=a.copy()
a[0]=2
a[3][0]=6