Python中的TicTacToe游戏

Question

我只想问一问，是否有办法使函数play_Game()或is_move_valid()变得更加高效。例如，使用某种类型的循环，我一直无法弄清楚如何并希望得到任何帮助，以使这些函数更短。

#TicTacToe
import sys
rows=[[0,1,2],[3,4,5],[6,7,8]]
columns =[[0,3,6],[1,4,7],[2,5,8]]
diagonals=[[0,4,8],[2,4,6]]
count=0

def display_Grid():
    print("\n")
    print(grid[0][0], "|" , grid[0][1], "|" , grid[0][2])
    print("---------")
    print(grid[1][0], "|", grid[1][1], "|", grid[1][2])
    print("---------")
    print(grid[2][0], "|", grid[2][1], "|", grid[2][2])

def check_turn():
    if count%2 == 0:
        player_1()
    else:
        player_2()

def game_over(shape):
    try:
        sys.exit()

    except SystemExit:
        print(shape," won")
        win = True
        quit()
def player_1():
    p_number = ("Player one")
    p_shape = p1
    play_Game(p_number, p_shape)

def player_2():
    p_number = ("Player two")
    p_shape = p2
    play_Game(p_number, p_shape)

def checkDraw():
    try:
        if count==10:
            sys.exit()

    except SystemExit:
        print("Draw")
        quit()

def check_win(shape):
    directions=[rows,columns,diagonals]
    for i in directions:
        for j in i:
            status = map(lambda val: True if grid[int(val/3)][val%3] == shape else False, j)
            if all(status):
                return True

    return False


def play_Game(p_number, p_shape):
    global count
    checkDraw()
    print(p_number, ",enter where you would like to move on the grid: ")
    move = int(input(""))
    if move in (1, 2, 0):
        valid = is_move_valid(move)
        if valid:
            count+=1
            grid[0][move] = p_shape
            display_Grid()
        else:
            check_turn()
    elif move in (3, 4, 5):
        valid = is_move_valid(move)
        if valid:
            count+=1
            move2 = move-3
            grid[1][move2] = p_shape
            display_Grid()
        else:
            check_turn()
    elif move in (6, 7, 8):
        valid = is_move_valid(move)
        if valid:
            count+=1
            move2 = move-6
            grid[2][move2] = p_shape
            display_Grid()
        else:
            check_turn()

    else:
        print("Enter one of the numbers visible on the grid: ")

    if check_win(p_shape):
        game_over(p_shape)
    else:
        None
    if count%2 == 0:
        player_1()
    else:
        player_2()

def is_move_valid(move):
    print(move)
    if move in(0, 1, 2):
        if grid[0][move] in ("X", "O"):
            print("This place has already been picked...")
            return False
        else:
            None
    elif move in(3, 4, 5):
        move2 = move-3
        if grid[1][move2] in ("X", "O"):
            print("This place has already been picked...")
            return False
        else:
              None
    elif move in(6, 7, 8):
        move2 = move-6
        if grid[2][move2] in ("X", "O"):
            print("This place has already been picked...")
            return False
        else:
              None

    return True

win = False    
grid = [[0,1,2],[3,4,5],[6,7,8]]
p_number = (" ")
display_Grid()
choose = input("Player one, would you like to be 'X' or 'O'?: ")
while win != True:
    try:
         if choose == "X":
             p2 = "O"
             p1 = choose
             player_1()
         elif choose == "O":
             p1 = "O"
             p2 = "X"
             player_1()
         else:
             print("Enter x or o: ")
             choose = input("Player 1, would you like to be 'X' or 'O'?: ")
    except ValueError:
        print("Please enter X OR O")

Answer 1

划分逻辑与用户交互

为什么要用一个函数来决定移动到屏幕上的print是否有效？如果你想改变这条信息呢？您会在逻辑函数中查看以更改用户界面消息吗？

我喜欢尽可能多地保持我的函数的“纯”，这样它们只会进行计算并返回结果，并在其上构建一个薄的交互皮肤。

长话短说，is_move_valid不应该打印任何东西。

Simplification

就简化而言，您只需将板压平，功能就变成：

def is_move_valid(move):
    return flatten(board)[move] not in ("X", "O")