List index changes multiple elements

Question

I couldn't find anything matching my problem, so hopefully this wasn't already mentioned somewhere and I'm just too stupid to find it.

thelist = []
a = [0]
for i in range(5):
    thelist.append(a)
print(thelist)

At this point, the program returns [[0], [0], [0], [0], [0]]

thelist[0].append(1)
print(thelist)

After appending this I would expect it to return the same but with the first element modified, like this:

[[0, 1], [0], [0], [0], [0]]

What actually happens, is that every element was modified in the same way and I get the following.

[[0, 1], [0, 1], [0, 1], [0, 1], [0, 1]]

I have found out, that if I replace the a in line 2 with the value of a, everything works fine. But why does this not work when I append a variable multiple times??

Answer 1

When creating thelist, it's saving a reference to the same object (a) in each position. When you append to the first element, it's actually appending to a in all the "thelist" slots. What you need to do is a deep copy on the object during construction:

thelist = []
a = [0]
for i in range(5):
    thelist.append(list(a))
print(thelist)

Then the following append will work as desired.

Answer 2

a is an object that every item in the list is pointing to.

If you want every item in the list to be a different object (with the same value), you should create a new object before assigning it. the easiest way would be to do:

for i in range(5):
    thelist.append(list(a))

The list function receives a list and return a new instance with the same inner values. So you won't be changing the entire array every time.

Answer 3

thelist and a both are lists. So, inside the for loop you are appending a list into another list

Case 1: a is a list

thelist = []
a = [0]
for i in range(5):
    thelist.append(a)
print(thelist)

[[0], [0], [0], [0], [0]]

Case 2: a is not a list but variable

thelist = []
a = 0
for i in range(5):
    thelist.append(a)
print(thelist)


[0, 0, 0, 0, 0]