Question

这是卡片游戏（项目）的一个示例我试图创建

p1 = []
p2 = []
player_list = [p1,p2]

def hit(self):
    self.append(random.choice(Deck))
    print(self)

for player in player_list:
    print('Player')
    play = input('Hit or Stay >> ')
    if play != 'hit' or play != 'stay': continue
    if play == 'hit':
        hit(player)
    print(player)
    if play == 'stay':
        print(player)
    print (player_list)

当您运行代码时，它不会从hit（）函数更新播放器。谁能告诉我为什么收到这个结果？

Answer 1

这个条件

if play != 'hit' or play != 'stay': continue

总是如此：总是要么不被击中＆＃39;是否留下＆＃39;

应该是＆＃39;和＆＃39;：

if play != 'hit' and play != 'stay': continue

更多Pythonic表达式是：

if play not in ('hit', 'stay'):
    continue

您还应该考虑诸如“HIT＆＃39;”之类的变体。我更喜欢将输入转换为大写。例如，

if play.upper() not in ('HIT', 'STAY'):
    continue

Answer 2

像往常一样，将代码分解为小的，单一用途的函数，事情变得更容易理解。

def ask():
    result = None
    while result not in ('hit', 'stay'):
        result = raw_input('hit or stay? >> ')
    return result

def process(player, command):
    if command == 'hit':
        hit(player)

for player in players:
    command = ask()
    process(player, command)
    print player

现在您可以独立确保每个都有效。 ask中的错误比原始代码的整个交错逻辑更容易检测。

从函数更新列表

2 个答案: