Python 我想了解我的3on1函数有什么问题
因此,如果我们使用:Python 我想了解我的3on1函数有什么问题,python,Python,因此,如果我们使用: def vencedor_linha(t): if (t[0]=='X' and t[1] and t[2]=='X') or (t[3]=='X' and t[4]=='X' and t[5]=='X') or (t[6]=='X' and t[7]=='X' and t[8]=='X'): print("'X'") return True elif (t[0]=='O' and t[1] and t[2]=='O') o
def vencedor_linha(t):
if (t[0]=='X' and t[1] and t[2]=='X') or (t[3]=='X' and t[4]=='X' and t[5]=='X') or (t[6]=='X' and t[7]=='X' and t[8]=='X'):
print("'X'")
return True
elif (t[0]=='O' and t[1] and t[2]=='O') or (t[3]=='O' and t[4]=='O' and t[5]=='O') or (t[6]=='O' and t[7]=='O' and t[8]=='O'):
print("'O'")
return True
else:
return False
def vencedor_coluna(t):
if (t[0]=='X' and t[3] and t[6]=='X') or (t[1]=='X' and t[4]=='X' and t[7]=='X') or (t[2]=='X' and t[5]=='X' and t[8]=='X'):
print("'X'")
return True
elif (t[0]=='O' and t[3] and t[6]=='O') or (t[1]=='X' and t[4]=='O' and t[7]=='O') or (t[2]=='O' and t[5]=='O' and t[8]=='X'):
print("'O'")
return True
else:
return False
def vencedor_diagonal(t):
if (t[0]=='X' and t[4] and t[8]=='X') or (t[6]=='X' and t[4]=='X' and t[2]=='X'):
print("'X'")
return True
elif (t[0]=='O' and t[4] and t[8]=='O') or (t[6]=='X' and t[4]=='O' and t[2]=='O'):
print("'O'")
return True
else:
return False
def determina_vencedor(t):
if vencedor_coluna(t):
return vencedor_coluna
elif vencedor_linha(t):
return vencedor_linha
elif vencedor_diagonal(t):
return vencedor_diagonal
else:
return False
及
它应该返回我,determina\u vencedor(test1)
->'O'
和determina\u vencedor(test2)
->False
对于test2
,一切正常,问题是它在test2处给了我“'O',函数vencedor_对角线0x0000000002C29A48”
怎么了?!帮我弄清楚 您希望返回函数的结果,而不是函数本身。或者更确切地说,由于所有函数都返回True
或False
,因此如果函数返回True
,则需要返回True
:
test2=(' ',' ',' ','X',' ',' ',' ',' ',' ')
这可以组合成一个循环:
def determina_vencedor(t):
if vencedor_coluna(t):
return True
if vencedor_linha(t)
return True
if vencedor_diagonal(t)
return True
return False
或者,您可以使用来测试所有功能:
def determina_vencedor(t):
for func in (vencedor_coluna, vencedor_linha, vencedor_diagonal):
if func(t):
return True
return False
它起作用了!非常感谢,现在我只有一个问题。我希望determina_vencedor(test1)返回'O'而不是'O'和True@wankinglemur:您的函数打印
'O'
,但不返回它。在这种情况下,返回'O'
而不是True
,并调整determina\u vencedor()
以返回第一个返回True的函数的返回值。非常感谢,现在我对python有了更多的了解,因为我知道了u。竖起大拇指!
def determina_vencedor(t):
for func in (vencedor_coluna, vencedor_linha, vencedor_diagonal):
if func(t):
return True
return False
def determina_vencedor(t):
return any(f(t) for f in (vencedor_coluna, vencedor_linha, vencedor_diagonal))