Python Question

Question

In this code (in the output), why is b[0] not equal to 3, but b[1][1] equal to 5?

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

Answer 1

a = [1, [2, 3]]
print(f"a = {a}") # Output: a = [1, [2, 3]]
b = a[:]
print(f"b = {b}") # Output: b = [1, [2, 3]]
a[0] = 3
print(f"a = {a}") # Output: a = [3, [2, 3]]
a[1][1] = 5
print(f"a = {a}") # Output: a = [3, [2, 5]]
print(f"b = {b}") # Output: b = [1, [2, 5]]
b[0] = 3
print(f"b = {b}") # Output: b = [3, [2, 5]]

So working through this it appears that 'b' shares references only to the nested list with 'a'
So b is still its own list. So you have to target the first interger of b to change it.

References to help explain it better.

https://www.geeksforgeeks.org/JUMP_LINK__&&__python__&&__JUMP_LINK-list-slicing/

Python Shallow Copy and Deep Copy

In Python, we use = operator to create a copy of an object. You may think that this creates a new object; it doesn't. It only creates a new variable that shares the reference of the original object.

https://www.programiz.com/python-programming/shallow-deep-copy

Answer 2

# Mohammadamin, when slicing, a new list is created, while the old list remains unchanged, respectively, object b receives its data storage address, but retains the cell addresses of the built-in list of object a.
# That is, if you change the second value in list b in the second cell, then it will also change in list a...😎:

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

Answer 3

I found this, let me know if any thing so far has been helpful.
https://www.sololearn.com/Discuss/2801058/?ref=app

Answer 4

a = [1, [2, 3]]
b = a[:] # Shallow copy of list a
a[0] = 3 # Modify the first element of a
a[1][1] = 5 # Modify the second element of the list within a
print(b) # Output: [1, [2, 5]]