accessing python list of list elements with only one index subscript

Question

So I have this function called replace_elem, written below:

def replace_elem(lst, index, elem):
    """Create and return a new list whose elements are the same as those in
    LST except at index INDEX, which should contain element ELEM instead.

    >>> old = [1, 2, 3, 4, 5, 6, 7]
    >>> new = replace_elem(old, 2, 8)
    >>> new
    [1, 2, 8, 4, 5, 6, 7]
    >>> new is old   # check that replace_elem outputs a new list
    False
    """
    assert index >= 0 and index < len(lst), 'Index is out of bounds'
    return [elem if i == lst[index] else i for i in lst]

I want to write this function below:

def put_piece(board, max_rows, column, player):
    """Puts PLAYER's piece in the bottommost empty spot in the given column of
    the board. Returns a tuple of two elements:

        1. The index of the row the piece ends up in, or -1 if the column
           is full.
        2. The new board

    >>> rows, columns = 2, 2
    >>> board = create_board(rows, columns)
    >>> row, new_board = put_piece(board, rows, 0, 'X')
    >>> row
    1
    >>> row, new_board = put_piece(new_board, rows, 0, 'O')
    >>> row
    0
    >>> row, new_board = put_piece(new_board, rows, 0, 'X')
    >>> row
    -1
    """

The hint was that I would use the replace_elem twice, but what I'm wondering is that replace_elem only takes in one index to give the location for what to replace so I'm curious as to how I could access lets say the first row and 3rd column index in python using only one subscript notation. Note I also have to return the whole board and not just the row

This isn't homework but self study as the material for this course is posted online for free and course has finished.

Your replace_elem() will replace all instances of the same value as lst[index] not just that one index, e.g. replace_elem([0,0,0], 1, 1) => [1,1,1] should be [0, 1, 0] — AChampion
– AChampion, Commented May 21, 2018 at 3:40
Why do you have a replace_elem function when you can just use list indexing? (e.g. lst[index] = elem)? — AAM111
– AAM111, Commented May 21, 2018 at 4:25
@OldBunny2800 it wasnt really my choice, its part of the lab im doing... im guessing its to have an abstraction so that the user doesn't know that the game board is implemented using a list. — JamesMat
– JamesMat, Commented May 21, 2018 at 5:10
Ok, I won’t question it but it doesn’t really make sense. — AAM111
– AAM111, Commented May 21, 2018 at 12:39

gyx-hh · Accepted Answer · 2018-05-21 12:22:34Z

I believe this what you're looking for. My assumption here is an empty spot in the board will be 0.

I also had to modify your replace_elem as you should be looking for the index and replacing that value with elem.

def replace_elem(lst, index, elem):
    assert index >= 0 and index < len(lst), 'Index out of bounds'
    return [elem if i == index else lst[i] for i in range(len(lst))]

def put_piece(board, max_rows, column, player):
    # return the column in board
    board_col = list(map(lambda x: x[column], board))
    try:
        # find an the last empty element - empty == 0
        row = len(board_col) - board_col[::-1].index(0) - 1
    except ValueError:
        return -1, board

    new_col = replace_elem(board_col, row, player)
    return row, [[board[r][c] if c != column else new_col[r] for c in range(len(board[r]))] for r in range(len(board))]

examples:

board = [[0, 0],[0,0]]
row, new_board = put_piece(board, 2, 0, 'X')
print('row: %s, board: %s' %(row, new_board))

Output: row: 1, board: [[0, 0], ['X', 0]]

row, new_board = put_piece(new_board, 2, 0, 'O')
print('row: %s, board: %s' %(row, new_board))

Output: row: 0, board: [['O', 0], ['X', 0]]

row, new_board = put_piece(new_board, 2, 0, 'X')
print('row: %s, board: %s' %(row, new_board))

Output: row: -1, board: [['O', 0], ['X', 0]]

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