检查TicTacToe获胜条件的有效方法//编辑:如何检查平局

Question

所以我制作了一个TicTacToe程序，作为我在Python中的第一个小项目（使用3.4）

到目前为止，它是有效的，但我想知道是否有可能简化获胜条件检查

import os
clear = lambda: os.system('cls')

def playerChange(player):  #Function for easily swapping out players
    if player == "X":
        return "O"
    else:
        return "X"
player = "X"  #Setting initial player
tttfield = ["1","2","3","4","5","6","7","8","9"] #setting up tictactoe field
clear()
while True:
    print("", tttfield[0], "|", tttfield[1], "|", tttfield[2], "n",
          "---------", "n",
          tttfield[3], "|", tttfield[4], "|", tttfield[5], "n",
          "---------", "n",
          tttfield[6], "|", tttfield[7], "|", tttfield[8], "n")
    choice = 0
    choice = input("n%s, choose a slot: " % player)
    if choice in tttfield:
        tttfield[int(choice)-1] = player #marks space
        player = playerChange(player) #changes player
    else:
        input("Not a valid number! Choose again!")
    clear()
    #check for win condition
    if ((tttfield[0]==tttfield[1]==tttfield[2]) or
        (tttfield[3]==tttfield[4]==tttfield[5]) or
        (tttfield[6]==tttfield[7]==tttfield[8]) or
        (tttfield[0]==tttfield[3]==tttfield[6]) or
        (tttfield[1]==tttfield[4]==tttfield[7]) or
        (tttfield[2]==tttfield[5]==tttfield[8]) or
        (tttfield[0]==tttfield[4]==tttfield[8]) or
        (tttfield[6]==tttfield[4]==tttfield[2])) :
        clear()
        input("nn  %s wins!" % playerChange(player))
        break

由于所有的检查，获胜条件检查看起来相当笨拙。有没有办法压缩它？

编辑：刚刚注意到我的程序中有个错误。我没有任何领带检查，遇到领带的情况会让你陷入困境——我该如何检查领带？我不知道我该怎么做。

Answer 1

一种常见的方法是将获胜状态存储在紧凑的数据结构中，例如

winners = [[0, 1, 2], [3, 4, 5] ...]

然后循环通过它们，例如

for squares in winners:
    if all(tttfield[square]=='x' for square in squares):
        print "X wins!"

（您希望对X和O都运行此操作，即添加一个外部循环，并在内部使用其变量而不是文字X/O）

p.s.你不需要那些反斜杠。位于圆括号内就足以让Python知道表达式在继续。