如何确定元素是否已经在某个位置的列表中？

Question

我制作了一个 5x5 矩阵游戏，由 25 个 0 组成。玩家 1 可以将任何 0 更改为 1，玩家 2 可以将任何 0 更改为 2。我只是在弄清楚如何验证板上的位置时遇到了麻烦，这样如果玩家已经放置了他们的号码，他们就需要输入一个不同的号码，而这个号码没有被占用。例如：Player 1 | Please enter a number between 1-25: 30 0 3 0 00 0 0 0 00 0 0 0 00 0 0 0 00 0 0 0 0Player 2 | Please enter a number between 1-25: 3This position is already taken! Please enter a different position:另外，我将如何对游戏进行编程以确定板上是否不再有任何 0？因为那将是平局。代码：def player1_turn():    player1_option = int(input("Player 1 | Please enter a number between 1-25: "))    if player1_option <= 0:        print("You can only enter a number between 1 and 25")        player1_turn()    elif player1_option > 25:        print("You can only enter a number between 1 and 25")        player1_turn()    player1 = (player1_option - 1) #Counter-acts the elements from starting at 0    grid[int(player1) // 5][int(player1) % 5] = 1 #places a 1 in inputted position    for row in grid: #for each row in the grid        print(*row, sep=" ")    print()    for y in range(0,4):        for x in range(0,4):            if grid[y][x] == grid[y][x+1] == grid[y+1][x] == grid[y+1][x+1] >0: #if there is a 2x2 of same number in grid:                print("Player",grid[y][x],"has won!")                exit()

Answer 1

您只需要在更新之前检查您正在更新的位置是否为 0

def player1_turn():

    player1_option = int(input("Player 1 | Please enter a number between 1-25: "))

    if player1_option <= 0:

        print("You can only enter a number between 1 and 25")

        player1_turn()

    elif player1_option > 25:

        print("You can only enter a number between 1 and 25")

        player1_turn()

    player1 = (player1_option - 1) #Counter-acts the elements from starting at 0

    #You must check if the position is a 0

    if (grid[int(player1) // 5][int(player1) % 5] == 0):

        grid[int(player1) // 5][int(player1) % 5] = 1 #places a 1 in inputted position

    else:

        #Do something else here instead of updating it 

    for row in grid: #for each row in the grid

        print(*row, sep=" ")

    print()

至于填满的板子，你可以使用常用的方式（双循环）或更直观的方式使用列表理解来检查它

def board_filled():

    for i in grid:

        for j in i:

            if j == 0:

                return False

    return True

def board_filled():

    return (sum([0 in i for i in grid]) == 0)

Answer 2

如果您为此使用二维数组，您可以检查一个地方是否充满，

filled_array = np.array([[0, 0, 0, 0, 0],

            [0, 0, 0, 0, 0],

            [0, 0, 0, 0, 0],

            [0, 0, 0, 0, 0],

            [0, 0, 0, 0, 0]])

selected_pos = 23

# Get row

row = np.floor(selected_pos)

# Get pos in row

pos = selected_pos % 5 - 1

# Check if pos is not 0

if(filled_array[row][pos] != 0):

    raise ["Place is filled"]

编辑：使用它来检查是否没有剩余的 0

if(np.count_nonzero(filled_array) == 25):

     raise ["No 0's left"]

忘了说：

import numpy as np

让它工作