[英]Nested loops in nested lists
我正在尝试在python中构建一个简单的井字游戏,以检查是否有获胜。我正在使用嵌套循环在嵌套列表中搜索匹配项。 由于某种原因,我的代码将只搜索第一个嵌套列表,而不搜索我期望的其余列表。
board = [ 'O', 'X', ' ', 'O', ' ', 'X', 'O', 'X', 'X' ]
wins = [[0,1,2],[3,4,5],[6,7,8],[0,3,6],[1,4,7],[2,5,8],[0,4,8],[2,4,6]]
def checkWin(player):
win = False
for test in wins:
print (test)
count = 0
for squares in test:
if board[squares] == player:
count = count + 1
if count == 3:
win = True
return win
if __name__ == '__main__':
print ("\nChecking board for X win ...\n")
if checkWin("X"):
print ("Game over, X wins!")
print ("\nChecking board for O win ...\n")
if checkWin("O"):
print ("Game over, O wins")
基于提供O获胜的董事会,这是我得到的输出:
Checking board for X win ...
[0, 1, 2]
Checking board for O win ...
[0, 1, 2]
有谁知道为什么会这样吗?
无论这三个正方形是否匹配,您都将从第一个嵌套列表测试中返回。 相反,只有在win
为true时才返回:
def checkWin(player):
win = False
for test in wins:
count = 0
for squares in test:
if board[squares] == player:
count = count + 1
if count == 3:
win = True
if win:
return True
return False
如果win
为false,则上面的内容继续到下一个嵌套列表,以进行下一个测试。
更好的是,在count
设置为3
时返回即可,因为您知道在该阶段找到了一个匹配项:
def checkWin(player):
for test in wins:
count = 0
for squares in test:
if board[squares] == player:
count = count + 1
if count == 3:
return True
return False
您可以使用all()
函数来代替计数:
def checkWin(player):
for test in wins:
if all(board[square] == player for square in test):
return True
return False
生成器表达式中的测试之一失败后, all()
尽早返回False
。
最终版本添加了any()
可以在一行中完成测试:
def checkWin(player):
return any(all(board[square] == player for square in test)
for test in wins)
演示:
>>> board = [ 'O', 'X', ' ', 'O', ' ', 'X', 'O', 'X', 'X' ]
>>> wins = [[0,1,2],[3,4,5],[6,7,8],[0,3,6],[1,4,7],[2,5,8],[0,4,8],[2,4,6]]
>>> def checkWin(player):
... return any(all(board[square] == player for square in test)
... for test in wins)
...
>>> checkWin('X')
False
>>> checkWin('O')
True
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.