Python 3.3中函数内部For循环的问题

时间:2013-10-04 21:03:04

标签: python function for-loop

好吧,所以我正在编写一个tic tac toe游戏,并且遇到了一个我似乎无法解决的非常糟糕的错误。我已经创建了一个函数,如果玩家即将获胜,计算机将阻止玩家,但是,如果成功阻止一次,即使条件满足,它也不会再触发。该函数的代码是:

def block():
for t in range(0, 9, 3):
        if slot[t] == user_team and slot[t+1] == user_team and (slot[t+2] \
           != user_team) and (slot[t+2] != computer_team):
            slot[int(t+2)] = computer_team
            return
        elif slot[t+1] == user_team and slot[t+2] == user_team and (slot[t] \
             != user_team) and (slot[t] != computer_team):
            slot[int(t)] = computer_team
            return
        elif slot[t] == user_team and slot[t+2] == user_team and (slot[t+1] \
             != user_team) and (slot[t+1] != computer_team):
            slot[int(t+1)] = computer_team
            return

for t in range(3):
        if slot[t] == user_team and slot[t+3] == user_team and (slot[t + 6] \
           != user_team) and (slot[t+6] != computer_team):
            slot[int(t+6)] = computer_team
            return
        elif slot[t+3] == user_team and slot[t+6] == user_team and (slot[t] \
             != user_team) and (slot[t] != computer_team):
            slot[int(t)] = computer_team
            return
        elif slot[t] == user_team and slot[t+6] == user_team and (slot[t+3] \
             != user_team) and (slot[t+3] != computer_team):
            slot[int(t+3)] = computer_team

此外,user_team和computer_team会返回该播放器是X还是O,而slot[int()] = computer_team用于将移动放置在电路板上。 下面是调用函数的地方(以防我搞砸了。):

else:
    draw_board()
    '''win()'''
    block()
    cmove()
    turn = "user"
    if end_game() == True:
        computer_win += 1
        draw_board()
        print ("The computer has won! But... We already knew that would happen. (:")
        playing = False
    elif end_game() == "Tie":
        tie_win += 1
        draw_board()
        print ("The game is a tie. You're going to have to try harder " \
               + "\n" + "if you wish to beat the computer!" + "\n")
        playing = False
    else:
        pass

如果你们中的任何一个人能告诉我哪里出错了,那么那将是我的一天。 C:

Board(打印是缩进的,它只是不想在这里。)

def draw_board():
'''Opted to use lists so that the numbers can be replaced with either
    X or O later on and so that testing whether the game is over is simpler'''
print (" " + str(slot[0]) + " | " + str(slot[1]) + " | " + str(slot[2]))
print ("-----------")
print (" " + str(slot[3]) + " | " + str(slot[4]) + " | " + str(slot[5]))
print ("-----------")
print (" " + str(slot[6]) + " | " + str(slot[7]) + " | " + str(slot[8]))
print ("\n")

新错误:

这是我进入第4步后的我的董事会

X | O | X

O | 4 | 5

X | 7 | X

计算机主板移动4(两次移动,并替换x)

X | O | X

O | 4 | 0

O | 7 | X

1 个答案:

答案 0 :(得分:2)

我相信您的问题符合block函数的逻辑。

这是你的董事会:

0 1 2
3 4 5
6 7 8

遍历第一对嵌套for循环,让我们看看你的代码做了什么:

for t in range(0,9,3):
    for y in range(1, 9, 3):

这将为您提供以下t, y对:(0,1),(0,4),(0,7),(3,1),(3,4),(3, 7),(6,1),(6,4)和(6,7)。马上,我认为这不是你想要的。据我所知,你试图检查玩家是否连续两个标记。

这个问题很容易解决 - 您不需要两个for循环。相反,只需使用tt+1t+2即可。

接下来,考虑一行:

0 1 2

要检查三个条件,玩家的标记为0和1,标记为0和2,或者标记为1和2.您只检查其中两个条件 - 0和1,以及1和2。

此外,if语句没有按照您的想法执行:

if ... and slot[y+1] != user_team and computer_team:

这相当于:

if ... and (slot[y+1] != user_team) and computer_team:

我认为computer_team'x''o',在这种情况下,python会在True语句中使用它与if相同。你想要的是这个:

if ... and (slot[y+1] != user_team) and (slot[y+1] != computer_team):

这也可能是您的代码只能运行一次的原因 - 下次评估之前找到的同一行或列时,if语句将再次评估为True,并且将再次设置相同的空间,看起来好像什么也没做。

检查列的代码具有相同的问题。希望我已经指出的问题足以让您弄清楚如何修复代码。祝你好运!