List manipulation (mutable or not)?

Question

I have this snippet of code:

list1 = [1, 2, 3]
list2 = list1
list1 = [4, 5, 6]

print(list2)
print(list1)

Which results in the following output:

[1, 2, 3]
[4, 5, 6]

Why is list2 not still pointed to [4, 5, 6]? I was under the impression that, since lists are mutable, the change would affect both list1 and list2, since in RAM both lists are pointed to the same sequence of items.

Any explanation would be appreciated.

Related: http://stackoverflow.com/q/575196/395760 In fact it pretty much answers the question if you replace "pass to a function" with "assign to a variable". — , Mar 22 '15 at 13:40

Bhargav Rao · Answer 1 · 2015-03-24T14:29:33.623

4

I have added the reason as comments in the code :

# list1 variable points to the memory location containing [1, 2, 3]
list1 = [1, 2, 3] 
# list2 variable made to point to the memory location pointed to by list1 
list2 = list1    
# list1 variable made to point to the memory location containing 
# a new list [4, 5, 6], list2 still pointing to [1, 2, 3] 
list1 = [4, 5, 6] 

print(list2) # list2 prints [1, 2, 3]
print(list1) # list1 prints [4, 5, 6]

edited Mar 24 '15 at 14:29

answered Mar 22 '15 at 13:36

Bhargav Rao

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2

Perhaps, I shouldn't have gone with inline comments, they are tough to read! – Bhargav Rao Mar 22 '15 at 13:44
1

@PadraicCunningham Thanks a lot Paddy, Your edit saved ma day! 3 votes in 6 mins (none from 35 mins) Seems impossible? Not anymore .... Thanks again!!! – Bhargav Rao Mar 22 '15 at 14:20
1

No worries, the invoice is in the post ;) – Padraic Cunningham Mar 22 '15 at 14:38

score 3 · Answer 2 · answered Mar 22 '15 at 13:36

I'll go through the lines one by one:

# Define a list [1, 2, 3] and save it into list1 variable
list1 = [1, 2, 3]

# Define list2 to be equal to list1 (that is, list2 == list1 == [1, 2, 3])
list2 = list1

# Create a new list [4, 5, 6] and save it into list1 variable
# Notice, that this replaces the existing list1!!
list1 = [4, 5, 6]

# Print list2, which still points to the original list [1, 2, 3]
print(list2)

# Print the new list1, [4, 5, 6] that is
print(list1)

However, this:

list1 = [1, 2, 3]
list2 = list1
list1.append(4)

print(list2)
print(list1)

Will output:

[1, 2, 3, 4]
[1, 2, 3, 4]

Since we are editing the list1 (and therefore list2, they are mutable), not creating a new list and saving it under the variable name list1

The keyword here is create a new list, so you're not editing list1 in your example, you're actually changing the name list1 to point to a whole different list.

score 3 · Answer 3 · answered Mar 22 '15 at 13:37

Lists are mutable. However, the line:

list1 = [4, 5, 6]

does not mutate the list object previously referenced by list1, it creates a brand new list object, and switches the list1 identifier to reference the new one. You can see this by looking at object IDs:

>>> list1 = [1, 2, 3]
>>> list2 = list1
>>> id(list1)
4379272472
>>> id(list2)
4379272472  # both reference same object
>>> list1 = [4, 5, 6]
>>> id(list1)
4379279016  # list1 now references new object
>>> id(list2)
4379272472  # list2 still references previous object

List manipulation (mutable or not)?

3 Answers3