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: 3

0 0 3 0 0

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

Player 2 | Please enter a number between 1-25: 3

This 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 0def player1_turn():&nbsp; &nbsp; player1_option = int(input("Player 1 | Please enter a number between 1-25: "))&nbsp; &nbsp; if player1_option <= 0:&nbsp; &nbsp; &nbsp; &nbsp; print("You can only enter a number between 1 and 25")&nbsp; &nbsp; &nbsp; &nbsp; player1_turn()&nbsp; &nbsp; elif player1_option > 25:&nbsp; &nbsp; &nbsp; &nbsp; print("You can only enter a number between 1 and 25")&nbsp; &nbsp; &nbsp; &nbsp; player1_turn()&nbsp; &nbsp; player1 = (player1_option - 1) #Counter-acts the elements from starting at 0&nbsp; &nbsp; #You must check if the position is a 0&nbsp; &nbsp; if (grid[int(player1) // 5][int(player1) % 5] == 0):&nbsp; &nbsp; &nbsp; &nbsp; grid[int(player1) // 5][int(player1) % 5] = 1 #places a 1 in inputted position&nbsp; &nbsp; else:&nbsp; &nbsp; &nbsp; &nbsp; #Do something else here instead of updating it&nbsp;&nbsp; &nbsp; for row in grid: #for each row in the grid&nbsp; &nbsp; &nbsp; &nbsp; print(*row, sep=" ")&nbsp; &nbsp; 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():&nbsp; &nbsp; for i in grid:&nbsp; &nbsp; &nbsp; &nbsp; for j in i:&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if j == 0:&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; return False&nbsp; &nbsp; return Truedef board_filled():&nbsp; &nbsp; 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],&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; [0, 0, 0, 0, 0],&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; [0, 0, 0, 0, 0],&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; [0, 0, 0, 0, 0],&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; [0, 0, 0, 0, 0]])selected_pos = 23# Get rowrow = np.floor(selected_pos)# Get pos in rowpos = selected_pos % 5 - 1# Check if pos is not 0if(filled_array[row][pos] != 0):&nbsp; &nbsp; raise ["Place is filled"]编辑：使用它来检查是否没有剩余的 0if(np.count_nonzero(filled_array) == 25):&nbsp; &nbsp; &nbsp;raise ["No 0's left"]忘了说：import numpy as np让它工作