Python 如何阻止这些变量重置，同时仍能循环它们？
以下是我的部分代码：Python 如何阻止这些变量重置，同时仍能循环它们？,python,Python,以下是我的部分代码： def dice_game(): dice_window = Tk() dice_window.geometry('450x450') player1_score = 0 player2_score = 0 def player1_turn(): player1_num1 = random.randint(1, 6) player1_num2 = random.randint(1, 6)
def dice_game():
dice_window = Tk()
dice_window.geometry('450x450')
player1_score = 0
player2_score = 0
def player1_turn():
player1_num1 = random.randint(1, 6)
player1_num2 = random.randint(1, 6)
player1_score = player1_num1 + player1_num2
player1_total = player1_num1 + player1_num2
if (player1_total % 2) == 0:
player1_score = player1_score + 10
else:
player1_score = player1_score - 5
player1_result = Label(dice_window, text = ( "Player 1 got", player1_num1, "and", player1_num2, "and their total score is:", player1_score))
player1_result.pack()
for x in range(1, 6):
player1_turn()
我曾尝试将循环放入player1_turn（）函数和dice_game（）函数中，但结果总是一样的。如何保持两名玩家的分数，而不在本节每次循环5次时重置它们？好吧，代码本身在骰子游戏（）函数中有for循环。 如您所见，您正在该函数中将分数声明为零。因此，要调用for循环，需要调用dice_game（），它会再次将分数重置为零 也许您应该尝试在函数dice_game（）之外使用该for循环，最后，分数将被保留。或者你也可以接受声明
player1_score = 0
player2_score = 0
在函数外部，作为全局变量。好吧，代码本身在dice\u game（）函数中有for循环。 如您所见，您正在该函数中将分数声明为零。因此，要调用for循环，需要调用dice_game（），它会再次将分数重置为零 也许您应该尝试在函数dice_game（）之外使用该for循环，最后，分数将被保留。或者你也可以接受声明
player1_score = 0
player2_score = 0
在函数外部，作为全局变量。由于您在
player1\u得分player2\u得分player1\u得分和player2\u得分player1\u得分python3nonlocalplayer1\u分数player2\u分数def dice_game():
dice_window = Tk()
dice_window.geometry('450x450')
player1_score = 0
player2_score = 0
def player1_turn():
nonlocal player1_score, player2_score
player1_num1 = random.randint(1, 6)
player1_num2 = random.randint(1, 6)
player1_score = player1_num1 + player1_num2
player1_total = player1_num1 + player1_num2
if (player1_total % 2) == 0:
player1_score = player1_score + 10
else:
player1_score = player1_score - 5
player1_result = Label(dice_window, text = ( "Player 1 got", player1_num1, "and", player1_num2, "and their total score is:", player1_score))
player1_result.pack()
for x in range(1, 6):
player1_turn()
# END
对于Python2，有一种解决方法，您可以将
player1_分数和player2_分数
存储在类似score={'player1_分数：0'，player2_分数：0}
的dict中，因为您正在player1_分数
和player2_分数
分配到函数体中，它使函数的player1\u分数和player2\u分数局部变量player1\u-turn（）
因此，如果您使用的是python3
，则可以使用nonlocal
关键字将player1\u分数
和player2\u分数
定义为非局部变量
代码如下所示
def dice_game():
dice_window = Tk()
dice_window.geometry('450x450')
player1_score = 0
player2_score = 0
def player1_turn():
nonlocal player1_score, player2_score
player1_num1 = random.randint(1, 6)
player1_num2 = random.randint(1, 6)
player1_score = player1_num1 + player1_num2
player1_total = player1_num1 + player1_num2
if (player1_total % 2) == 0:
player1_score = player1_score + 10
else:
player1_score = player1_score - 5
player1_result = Label(dice_window, text = ( "Player 1 got", player1_num1, "and", player1_num2, "and their total score is:", player1_score))
player1_result.pack()
for x in range(1, 6):
player1_turn()
# END
对于Python2，有一种变通方法，您可以将player1\u分数和player2\u分数存储在一个dict中，比如score={'player1\u分数：0，'player2\u分数：0}
让我们把它分解一下。我可以看到player1\u turn（）
是嵌套的，因为您想访问dice\u窗口。但是有更好的办法
首先，将功能分开player1_turn
需要一个骰子窗口，并且您想要保存的东西似乎是player1_score
，因此我们将返回该值
def player1_turn(dice_window):
player1_num1 = random.randint(1, 6)
player1_num2 = random.randint(1, 6)
player1_score = player1_num1 + player1_num2
player1_total = player1_num1 + player1_num2
if (player1_total % 2) == 0:
player1_score = player1_score + 10
else:
player1_score = player1_score - 5
player1_result = Label(dice_window, text = ( "Player 1 got", player1_num1, "and", player1_num2, "and their total score is:", player1_score))
player1_result.pack()
return player1_score
现在进入游戏：
def dice_game():
dice_window = Tk()
dice_window.geometry('450x450')
# we're not using x, so we should make it clear that this is a loop of 5
scores = []
total = 0
for x in range(5):
player1_score = player1_turn(dice_window)
# now what to do with the score?
scores.append(player1_score) # save it?
print(player1_score) # print it?
total += player1_score # add it up
print(scores)
print(total)
让我们来分析一下。我可以看到player1\u turn（）
是嵌套的，因为您想访问dice\u窗口。但是有更好的办法
首先，将功能分开player1_turn
需要一个骰子窗口，并且您想要保存的东西似乎是player1_score
，因此我们将返回该值
def player1_turn(dice_window):
player1_num1 = random.randint(1, 6)
player1_num2 = random.randint(1, 6)
player1_score = player1_num1 + player1_num2
player1_total = player1_num1 + player1_num2
if (player1_total % 2) == 0:
player1_score = player1_score + 10
else:
player1_score = player1_score - 5
player1_result = Label(dice_window, text = ( "Player 1 got", player1_num1, "and", player1_num2, "and their total score is:", player1_score))
player1_result.pack()
return player1_score
现在进入游戏：
def dice_game():
dice_window = Tk()
dice_window.geometry('450x450')
# we're not using x, so we should make it clear that this is a loop of 5
scores = []
total = 0
for x in range(5):
player1_score = player1_turn(dice_window)
# now what to do with the score?
scores.append(player1_score) # save it?
print(player1_score) # print it?
total += player1_score # add it up
print(scores)
print(total)
查找有关嵌套函数作用域的更多信息。然后，你必须增加分数，而不是每次都重新初始化。这将重置该值。创建一个并将其添加到您的帖子中。查找有关嵌套函数范围的更多信息。然后，你必须增加分数，而不是每次都重新初始化。这将重置该值。创建一个并将其添加到您的帖子中。那么我应该如何将其添加到我的代码中？就让他们一个接一个吗？是的，一个接一个。如果“现在开始游戏：”已经从我的帖子中删除了。它有点有效。我会稍微调整一下，但你给我的已经解决了我的核心问题。谢谢！那么，我应该如何把这些放在我的代码中呢？就让他们一个接一个吗？是的，一个接一个。如果“现在开始游戏：”已经从我的帖子中删除了。它有点有效。我会稍微调整一下，但你给我的已经解决了我的核心问题。谢谢！