返回 NonType 的 Python 子例程

你的逻辑到达末尾而不返回任何内容，设置为 when you do .prizecheck()changeNonechange = prizecheck(...)

尝试在那里放置一个调试语句，看看什么时候会发生这种情况。

尝试在每次更改时打印值，以便可以看到模式。change

养成总是在你期望从中获得返回值的函数末尾的习惯。return

Python 不是静态类型语言，因此不能强制值为浮点数。如果可以（使用另一种语言），则必须指定函数的返回类型，并在过程中出现编译器错误。您可以提供类型提示来检查此类错误，但 Python 变量始终可以设置为 。None

我认为这应该有效：

import random

money = 1.00
finish = False

print("You have £", money)


def prizecheck(tocheck, money):
    winnings = 0  # Initialize winnings to 0
    if wheel1 == tocheck and wheel2 == tocheck and wheel3 == tocheck:
        if tocheck == "Skull":
            print("The house wins, All money lost")
        elif tocheck in ["Cherry", "Lemon", "Orange", "Star"]:
            print("You win £1")
            winnings = 1
        elif tocheck == "Bell":
            print("You win £5")
            winnings = 5
    elif (wheel1 == tocheck) + (wheel2 == tocheck) + (wheel3 == tocheck) == 2:
        if tocheck == "Skull":
            print("The house wins, you lost £1")
            winnings = -1
        elif tocheck in ["Cherry", "Lemon", "Orange", "Star", "Bell"]:
            print("You win 50p")
            winnings = 0.5
    return money + winnings


def spin():
    picture = random.randint(1, 6)
    if picture == 1:
        wheel = "Cherry"
    elif picture == 2:
        wheel = "Bell"
    elif picture == 3:
        wheel = "Lemon"
    elif picture == 4:
        wheel = "Orange"
    elif picture == 5:
        wheel = "Star"
    else:
        wheel = "Skull"
    return wheel

while money >= 0.2 and not finish:
    money -= 0.2
    wheel1 = spin()
    wheel2 = spin()
    wheel3 = spin()
    print(wheel1, "|", wheel2, "|", wheel3)

    money = prizecheck("Skull", money)
    money = prizecheck("Cherry", money)
    money = prizecheck("Bell", money)
    money = prizecheck("Lemon", money)
    money = prizecheck("Orange", money)
    money = prizecheck("Star", money)

    print("You have £", money)

    endcheck = input("Spin again (1), or quit(2)?")
    if endcheck == "2":
        finish = True

    if money < 0.20:
        print("Sorry, you have run out of money")

• return 语句之后的任何 print 语句都不会被执行，所以我移动了它们
• 我还简化了你的陈述，更少的嵌套使它不那么混乱if
• 每次你要求不同的水果时，你都在覆盖。这意味着只有最后一个果实的结果才会存储在 中。您需要累积所有水果的变化，然后进行相应的更新changeprizecheckchangemoney
• 在函数中，您使用的是在函数外部定义的名为变量。这会导致您未正确更新 money 变量的问题。您需要传递给函数并返回更新后的值prizecheckwinningsmoney

通过查看最频繁的轮子值的计数来考虑如何确定奖金可能会更容易。例如，我相信所有这些都会产生相同的“胜利”，对吧？

["Cherry", "Cherry", "Lemon"]
["Cherry", "Lemon", "Cherry"]
["Lemon", "Cherry", "Cherry"]

如果我们找到最常找到的项目计数，在所有情况下，我们都会得到： 使其更容易检查和处理。("Cherry", 2)

我建议：

def prize_check(wheels, bet_amount):
    ## ---------------------
    ## find the details of the most common item
    ## ---------------------
    most_common_value, most_common_count = collections.Counter(wheels).most_common(1)[0]
    ## ---------------------

    ## ---------------------
    ## All items different
    ## ---------------------
    if most_common_count == 1:
        return bet_amount  # a guess by me at what to return here
    ## ---------------------

    ## ---------------------
    ## Two of the items are the same
    ## ---------------------
    if most_common_count == 2:
        if most_common_value == "Skull":
            return bet_amount - 1.0
        return bet_amount + 0.5
    ## ---------------------

    ## ---------------------
    ## All three the same
    ## ---------------------
    if most_common_value == "Skull":
        return 0

    if most_common_value == "Bell":
        return bet_amount + 5.0

    return bet_amount + 1.0
    ## ---------------------

然后，我们不再多次调用这个新方法，而是调用一次并传入轮盘值和赌注。prizecheck()

money = prize_check([wheel1, wheel2, wheel3], money)