python - 即使嵌套 if 条件失败后如何继续执行剩余部分

Question

我只是想验证一个 tic tac toe 板，其值在列表列表中。

board = ([1, 2, 0], [2, 1, 0], [2, 1, 2])


def print_winner(variable):
    if variable == 1:
        return print("Winner of this board is Player1")
    else:
        return print("Winner of this board is Player2")


def interpret(game_board):
    if game_board[0][0] != 0:
        if game_board[0][1] and game_board[0][2] == game_board[0][0]:
            print(print_winner(game_board[0][0]))
        elif game_board[1][0] and game_board[2][0] == game_board[0][0]:
            print(print_winner(game_board[0][0]))
        elif game_board[1][1] and game_board[2][2] == game_board[0][0]:
            print(print_winner(game_board[0][0]))
    elif game_board[0][1] != 0:
        if game_board[1][1] and game_board[2][1] == game_board[0][1]:
            print(print_winner(game_board[0][0]))
    elif game_board[0][2] != 0:
        if game_board[1][2] and game_board[2][2] == game_board[0][2]:
            print(print_winner(game_board[0][0]))
        elif game_board[1][2] and game_board[2][0]:
            print(print_winner(game_board[0][0]))
    elif game_board[1][0] != 0:
        if game_board[1][1] and game_board[1][2] == game_board[1][0]:
            print(print_winner(game_board[0][0]))
    elif game_board[2][0] != 0:
        if game_board[2][1] and game_board[2][2] == game_board[2][0]:
            print(print_winner(game_board[0][0]))
    else:
        print("No Winner")


interpret(board)

上面代码的问题是，一旦所有第一个嵌套 if 语句失败，我不知道如何指示我的程序继续执行剩余的代码。因此我的代码没有给出任何输出。预期输出为“No Winner”。

Answer 1

第一个问题是您错误地使用了和。您需要在 和 的任一操作数上指定条件。例如:

if game_board[0][1] == game_board[0][0] and game_board[0][2] == game_board[0][0]  

或者

if game_board[0][1] == game_board[0][2] == game_board[0][0] 

而不是:

if game_board[0][1] and game_board[0][2] == game_board[0][0]

但是，您也可以使用集合并检查它是否只包含一个数字:

if len({game_board[0][0], game_board[0][1], game_board[0][2]}) == 1:

实际问题

how to instruct my program to continue with the remaining piece of code once all my first nested if statements failed

其实很简单。使用普通的 if 并在满足一个条件时return。

加上一些其他改进(您总是打印 game_board[0][0])，它看起来像这样:

def interpret(game_board):
    if game_board[0][0] != 0:
        if len({game_board[0][0], game_board[0][1], game_board[0][2]}) == 1:
            return print(print_winner(game_board[0][0]))
        elif len({game_board[0][0], game_board[1][0], game_board[2][0]}) == 1:
            return print(print_winner(game_board[0][0]))
        elif len({game_board[0][0], game_board[1][1], game_board[2][2]}) == 1:
            return print(print_winner(game_board[0][0]))
    if game_board[0][1] != 0:
        if len({game_board[0][1], game_board[1][1], game_board[2][1]}) == 1:
            return print(print_winner(game_board[0][1]))
    if game_board[0][2] != 0:
        if len({game_board[0][2], game_board[1][2], game_board[2][2]}) == 1:
            return print(print_winner(game_board[0][2]))
        elif len({game_board[0][2], game_board[1][1], game_board[2][0]}) == 1:
            return print(print_winner(game_board[0][2]))
    if game_board[1][0] != 0:
        if len({game_board[1][0], game_board[1][1], game_board[1][2]}) == 1:
            return print(print_winner(game_board[1][0]))
    if game_board[2][0] != 0:
        if len({game_board[2][0], game_board[2][1], game_board[2][2]}) == 1:
            return print(print_winner(game_board[2][0]))
    return print("No Winner")

这是有效的，因为 print 返回 None 并打印到 stdout。