如何用RandomCard()函数替换AceOfSpades()中的卡牌生成代码？

现有以下Python代码：

def RandomCard():
    x=number[ran.randint(0,12)]
    y=suit[ran.randint(0,3)]
    print("Your card is the:", x, "of", y)


def AceOfSpades():
    count = 1
    x=number[ran.randint(0,12)]
    y=suit[ran.randint(0,3)]
    while x != 'Ace' or y != 'Spades':
        x=number[ran.randint(0,12)]
        y=suit[ran.randint(0,3)]
        count += 1
    else: 
        print("Your card is the:", x, "of", y)
        print("It took", count, "tries to get the Ace of Spades!")

想将AceOfSpades()函数中生成随机牌的代码：

x=number[ran.randint(0,12)]
y=suit[ran.randint(0,3)]

替换为调用RandomCard()，但直接替换后无法达到预期效果，该如何操作？

直接替换无效的原因是原RandomCard()仅打印结果，没有返回生成的牌面和花色值，导致AceOfSpades()无法获取x和y来判断是否为黑桃A。需要修改RandomCard()让它返回牌面和花色，再在AceOfSpades()中接收返回值。

修改后的代码 #

1. 调整RandomCard()函数，使其返回生成的牌面与花色：

def RandomCard():
    x = number[ran.randint(0,12)]
    y = suit[ran.randint(0,3)]
    # 若需保留打印功能，可保留下方print语句；不需要则删除
    # print("Your card is the:", x, "of", y)
    return x, y

2. 修改AceOfSpades()，调用RandomCard()并接收返回值：

def AceOfSpades():
    count = 1
    x, y = RandomCard()
    while x != 'Ace' or y != 'Spades':
        x, y = RandomCard()
        count += 1
    print("Your card is the:", x, "of", y)
    print("It took", count, "tries to get the Ace of Spades!")

说明 #

修改后，RandomCard()生成牌面和花色后会将其返回，AceOfSpades()通过x, y = RandomCard()获取这些值，就能正常进行循环判断，直到生成黑桃A为止。

内容的提问来源于stack exchange，提问作者ConquestAce