围绕两个列表之间的确定坐标进行迭代

Question

经典战舰游戏。我正在尝试制作一个 space（看起来更低 'o'），我们无法放置另一艘船。

我们有一个 5*5 的“O”数组：

board = [['O','O','O','O','O'],['O','O','O','O','O'],['O','O','O','O','O'],
         ['O','O','O','O','O'],['O','O','O','O','O']]

我已经放置了两格飞船并且知道它的坐标。

坐标是：

[(row, column),(row, column)]
[(0,0),(1,0)]

Board = [['S','O','O','O','O'],['S','O','O','O','O'],['O','O','O','O','O'],
         ['O','O','O','O','O'],['O','O','O','O','O']]

现在我必须找到两个坐标周围的所有 'O'-es 并将其更改为 'o'。如果有 'S' 不要更改它。

def mark_nearby_cells(board):  # function for mar
    """
    function for marking cells nearby to currently placed ship 0 -> o
    """
    ships_coordinates = []
    for i, row in enumerate(board):
        for j, ship in enumerate(row):
            if ship == "S":
                ship_coordinate = [row, column]
                ships_coordinates.append(ship_coordinate)

    for coordinate in ships_coordinates:
               #how to avoid simply iteration coordinate by adding and subtracting "1
            if ships_coordinates[ships_coordinates[0][0] - 1][ships_coordinates[0][1]] != 'S':
....
....
....

    return board

Answer 1

为了简单起见，我编写了一个附加函数来显示面板并将您的“mark_nearby_cells”函数分成两部分。

import numpy as np

def print_board(board):
    """Displays the board nicely."""
    print("\n".join("".join(row) for row in board))
    print()

# Create the board.
X = 5
Y = 5
board = [['O' for _ in range(X)] for _ in range(Y)]
print_board(board)
# OOOOO
# OOOOO
# OOOOO
# OOOOO
# OOOOO

# Place a ship.
for x, y in ((0, 0), (0, 1)):
    board[x][y] = "S"
print_board(board)
# SSOOO
# OOOOO
# OOOOO
# OOOOO
# OOOOO

def detect_ships(board):
    """Returns a list of all the cells coordinates containing "S"."""
    ships_coordinates = []
    for x, row in enumerate(board):
        for y, cell in enumerate(row):
            if cell == "S":
                ships_coordinates.append((x, y))
    return ships_coordinates


def mark_nearby_cells(board):
    """Puts a lower "o" in the cells near a ship."""
    for (x, y) in ships_coordinates:
        for dx in (-1, +1):
            for dy in (-1, +1):
                xdx = np.clip(0, x+dx, X-1) # We don't want the coordinates to loop around the
                ydy = np.clip(0, y+dy, Y-1) # board so we trim them to [0; X] and [0; Y].
                cell = board[xdx][ydy]
                if cell == "O":
                    board[xdx][ydy] = "o"
    return board

ships_coordinates = detect_ships(board)
board = mark_nearby_cells(board)
print_board(board)
# SSoOO
# oooOO
# OOOOO
# OOOOO
# OOOOO

围绕两个列表之间的确定坐标进行迭代

Iterations around definite coordinates between two lists

python

iteration

coordinates