Python list updating elements when appending

Question

I have a python list that I want to append a list to. The list was declared like this:

data = []

Then I append the list with:

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

After that I want to append another list:

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

Then when I print out the entire list it shows:

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

When I am trying to have it print out:

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

Where it updated the first element of the array when I just appended it like this:

data.append(prev_board)

And prev_board stores the list to append to the list. My entire code for this is:

def gather_training_data(games):
global data
global game_won

for each_game in range(games):
    game = Game()
    while True:
        pygame.event.get()
        game.create_board()
        if not game_won:
            prev_board = game.board
            move = game.random_move()
            data.append(prev_board)
            print data
            time.sleep(5)
        else:
            break
    game_won = False
    time.sleep(1)
    window.fill((255, 255, 255))

return data

So why are the first elements of the list updating when I am only appending a list to the list?

Answer 1

You must be using the same list over and mutating it - something like this

>>> a = [1,2,3,4,5]
>>> b = []
>>> b.append(a)
>>> b
[[1, 2, 3, 4, 5]]

Now change the contents of a

>>> a[:4] = [9,9,9,9]
>>> a
[9, 9, 9, 9, 5]
>>> b
[[9, 9, 9, 9, 5]]

>>> b.append(a)
>>> b
[[9, 9, 9, 9, 5], [9, 9, 9, 9, 5]]

Now b contains a twice

>>> id(b[0]), id(b[1])
(50336328, 50336328)
>>> b[0] is b[1]
True
>>> id(a)
50336328
>>> a is b[0] is b[1]
True
>>> 

---

To prevent this from happening, create new lists to append instead of mutating the same list or append copies of the mutated list if it cannot be helped.
copy — Shallow and deep copy operations
How to clone or copy a list?