我正在尝试用 python 构建一个简单的 Tic-Tac-Toe 游戏,为了检查是否获胜,我使用嵌套循环在嵌套列表中搜索匹配项。由于某种原因,我的代码只会搜索第一个嵌套列表,而不是像我预期的那样搜索其余列表。
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")
根据董事会预计 O 会获胜,这是我得到的输出:
Checking board for X win ...
[0, 1, 2]
Checking board for O win ...
[0, 1, 2]
有人知道为什么会发生这种情况吗?
最佳答案
无论这三个方 block 是否匹配,您都会从第一个嵌套列表测试中返回。相反,仅当 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,则上面的内容将继续到下一个嵌套列表以进行下一个测试。
更好的是,当 count
设置为 3
时返回,因为您知道在该阶段已找到匹配项:
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()
function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
all()
一旦生成器表达式中的某个测试失败,就会提前返回 False
。
最终版本添加了any()
一行完成测试:
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
关于python - 嵌套列表中的嵌套循环,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/33203038/