Question

所以我需要做一个＆＃34;游戏＆＃34;用户询问是否要选择卡的情况，如果是，则返回随机卡。像这样的东西：

>>>selectCard()
Would you like to draw a random card? (yes/no)
The card you got is: 8 of hearts
Would you like to draw a random card? (yes/no)
>>>yes
The card you got is: Ace of diamonds
Would you like to draw a random card? (yes/no)
>>>no

到目前为止，我有这个，但我被困住了，不知道如何完成它

def selectCards():
cardranks = ['Ace',1,2,3,4,5,6,7,8,9,'Jack','Queen','King','Ace',1,2,3,4,5,6,7,8,9,'Jack','Queen','King','Ace',1,2,3,4,5,6,7,8,9,'Jack','Queen','King','Ace',1,2,3,4,5,6,7,8,9,'Jack','Queen','King']
cardsuits = ['of Clubs','of Clubs','of Clubs','of Clubs','of Clubs','of Clubs','of Clubs','of Clubs','of Clubs','of Clubs','of Clubs','of Clubs','of Clubs','of Diamonds','of Diamonds','of Diamonds','of Diamonds','of Diamonds','of Diamonds','of Diamonds','of Diamonds','of Diamonds','of Diamonds','of Diamonds','of Diamonds','of Diamonds','of Hearts','of Hearts','of Hearts','of Hearts','of Hearts','of Hearts','of Hearts','of Hearts','of Hearts','of Hearts','of Hearts','of Hearts','of Hearts','of Spades','of Spades','of Spades','of Spades','of Spades','of Spades','of Spades','of Spades','of Spades','of Spades','of Spades','of Spades','of Spades']
shuffle(cardranks)
shuffle(cardsuits)
for a,b in zip(cardranks,cardsuits):
    response = (input ('Would you like to draw a random card? (yes/no)'))
    if response == yes:
        return (a,b)

当我运行这个程序时，它说肯定没有定义，因此我不知道如何继续。

谢谢

Answer 1

输入传递一个字符串，因此您必须检查是否收到字符串'yes'而不仅仅是。是的不是价值或变量。

最佳实践的细节：我喜欢你给出好的变量名，但是响应用s拼写;）

Answer 2

问题在于，当您检查响应（拼写错误的响应错误）是否等于是时，您将它与未定义的变量“是”进行比较，您需要做的是：

#The '.upper()' bit changes the response string to uppercase.
if response.upper() == 'YES': #Yes must be in quotes to be a string comparison.
    print('Its Working!') #Prints out Its Working! to the Python IDE.

Answer 3

您需要将yes括在引号中：'yes' - 因为现在您正在告诉Python解释器将其与变量中包含的值进行比较 yes而不是文字字符串 'yes'。通过将response转换为小写来使检查不区分大小写也是一个好主意：

response = input ('Would you like to draw a random card? (yes/no)')
if response.lower() == 'yes':
    # Return the card

然后您的代码还有一些其他问题：

如果用户不想绘制卡片，for循环将继续运行。这意味着您必须在循环结束前输入 no 52次，并且函数返回时不会绘制卡片。

zip(cardranks,cardsuits)

cardranks

cardsuits不会给你一个有效的，正常的套牌 - 你最终可能会得到一个有4个黑桃王牌而没有任何其他王牌的套牌适合例如。

更好的解决方案是：

# All possible ranks
cardranks = ['Ace','1','2','3','4','5','6','7','8','9','Jack','Queen','King']
# All possible suits
cardsuits = ['of Clubs','of Diamonds','of Hearts','of Spades']

# The complete deck
deck = list(itertools.product(cardranks, cardsuits))

# Shuffle the deck
random.shuffle(deck)

# Draw a random card from the deck (the deck doesn't need to be shuffled before)
rank, suit = random.choice(deck)

从python中的列表中选择一张随机卡

3 个答案: