Python variable question - how does this work

Question

a = [1,2,3]
b = a

b.append(5)
a.append(4)

print(a)
print(b)

Result
[1,2,3,5,4]
[1,2,3,5,4]

Why isn't a [1,2,3,4] and b [1,2,3,5]?

Edit:
And why is this different

a = "hello"
b = a

a += "world"
b += "!"

print(a)
print(b)

Result
Helloworld
Hello!

Answer 1

The reason is that b = a creates a shallow copy of a. That means that a and b refer to the same object, not to two identical objects.
For more information about shallow and deep copies:
https://realpython.com/copying-JUMP_LINK__&&__python__&&__JUMP_LINK-objects/

Answer 2

because b is referring to a when you give b=a.So if you change the value of b or a it will affect both.
If you want to create duplicate instead of reference, you can use list slicing .Like that:
b=a[:]
https://code.sololearn.com/c4i4vuKlD5J2/?ref=app

Answer 3

When you assign things in python they then reference to the same location . So any change to values stored in variable a or b will affect both .

Answer 4

Yvonne Rizkallah In Python, everything is an object and every initialized variable holds the reference to an object present in the heap. Here's what happens in your code:

# Create an array object and store its reference in variable 'a'.
a = [1, 2, 3]

# Assign this reference contained in variable 'a' to variable 'b'.
b = a

# Call the append method of the list object
b.append(5)
a.append(4)

# Since both 'a' and 'b' are the same (as mentioned earlier), printing 'a' and 'b' shows the same result.

Now, considering the edit:

# This creates a new instance (a new string object) of the str class and assigns its reference (i.e., the memory location that contains the data related to the object) to variable 'a'.
a = "hello"

# This copies the reference contained in variable 'a' to variable 'b'
b = a

# This creates 2 different str objects, the first one being the 'world' string and the second one being a + 'world', the concatenated version (note: a+="world" is the same as a = a + "world")
a += "world"

Continued...

Answer 5

# Do the same thing to variable 'b'
b += "!"

Clearly, both the variables contain different references to two different string objects, hence the differing outputs. I hope that this explanation helps.

Answer 6

Yvonne Rizkallah The assignment is the same. However, what happens next is not. In the first case, a method is called. Both the variables 'a' and 'b' refer to the same object. So, the method can be called in either one. But the second case is a bit different. When concatenating strings, a new string object is created and the reference to this is returned. Since strings are immutable, the original string can't be modified. The only way is to make a new string object. 'a + "world"' and 'b + "!"' create two new string objects, even though 'a' and 'b' were initially the same. Please read my previous comment to understand about this better.

Answer 7

The reason it didn't work for the list because a and b are pointing to the same object as what Simon Sauter said but, you can instead do:

a = [1, 2, 3]
b = a[:]

print(a)
print(b)

result:
[1, 2, 3, 4]
[1, 2, 3, 5]

Which will create a copy of the list stored in a and store it in b

https://code.sololearn.com/c6FI3vR1PcxV/?ref=app

Answer 8

Or you can just use the list.copy() method to make a distinct, but duplicated copy of the original list.

Note: Only the list object is copied, not the objects inside them.

Answer 9

Calvin Thomas Right!

Answer 10

Calvin Thomas why is the list assignment different than a string assignment

Answer 11

The reason is that you created a shallow copy as b=a when you assign anything then they refer the same location