Unexpected list behavior in Python

Question

I wanted to reverse a list, and I managed to do so, but in the middle of the work I noticed something strange. The following program works as expected but uncommeting line list_reversed[i]=list[len(list)-1-i] and print(list[i]) (commenting the last line of course) leads to a change in list. What am I not seeing? My Python version is 3.3.3. Thank you in advance.

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

list_reversed=list

for i in range(0,len(list)):

    #list_reversed[i]=list[len(list)-1-i]
    #print(list[i])

    print(list[len(list)-1-i])

Answer 1

The following:

list_reversed = list 

makes the two variables refer to the same list. When you change one, they both change.

To make a copy, use

list_reversed = list[:]

Better still, use the builtin function instead of writing your own:

list_reversed = reversed(list)

P.S. I'd recommend against using list as a variable name, since it shadows the builtin.

Answer 2

When you do:

list_reversed = list

You don't create a copy of list, instead, you create a new name (variable) which references the same list you had before. You can see this by adding:

print(id(list_reversed), id(list))  # Notice, the same value!!

Answer 3

list_reversed = list does not make a copy of list. It just makes list_reversed a new name pointing to the same list. You can see any number of other questions about this on this site, some list in the related questions to the right.

Answer 4

list and reversed_list are the same list. Therefore, changing one also changes the other.

What you should do is this:

reversed_list = list[::-1]

This reverses and copies the list in one fell swoop.

Answer 5

This is about general behaviour of lists in python. Doing:

list_reversed = list

doesn't copy the list, but a reference to it. You can run:

print(id(list_reversed))
print(id(list))

Both would output the same, meaning they are the same object. You can copy lists by:

a = [1,2]
b = a.copy()

or

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