在 Python 上返回 None

Question

我的学生正在为课堂制作石头、纸、剪刀模拟。一组有一个错误，他们的一个函数不会返回任何东西。我已经检查过它是否所有分支都有返回语句，它们确实有。我也试过可视化工具，但功能停止了，我不知道为什么。

def Win_Loss(my_history, their_history):
    if len(my_history) > 1 and len(their_history) > 1:
        if (their_history[-1] == 's' and my_history[-1] == 'r'):
            return 'p'

    elif len(my_history) > 1 and len(their_history) > 1:
        if (their_history[-1] == 'r' and my_history[-1] == 'p'):
            return 's'

    elif len(my_history) > 1 and len(their_history) > 1:
        if (their_history[-1] == 'p' and my_history[-1] == 's'):
            return 'r'

    elif len(my_history) > 1 and len(their_history) > 1:
        if (their_history[-1] == 's' and my_history[-1] == 'p'):
            return 'r'

    elif len(my_history) > 1 and len(their_history) > 1:
        if (their_history[-1] == 'r' and my_history[-1] == 's'):
            return 'p'

    elif len(my_history) > 1 and len(their_history) > 1:
        if (their_history[-1] == 'p' and my_history[-1] == 'r'):
            return 's'

    elif len(my_history) > 1 and len(their_history) > 1:
        if (their_history[-1] == 's' and my_history[-1] == 's'):
            return 'r'

    elif len(my_history) > 1 and len(their_history) > 1:
        if (their_history[-1] == 'r' and my_history[-1] == 'r'):
            return 'p'

    elif len(my_history) > 1 and len(their_history) > 1:
        if (their_history[-1] == 'p' and my_history[-1] == 'p'):
            return 's'
    else:
        return "p"
print(Win_Loss(['s','p','p'],['s','p','p']))

这应该是打印's'，但它打印的是None。

Answer 1

如果任何内部ifs 失败，它将返回None，因为外部if/elif 被采用，因此else: 块将不会被执行。

要清楚，它们的整个功能相当于：

def Win_Loss(my_history, their_history):
    if len(my_history) > 1 and len(their_history) > 1:
        if (their_history[-1] == 's' and my_history[-1] == 'r'):
            return 'p'
    else:
        return "p"

因为一旦第一个if 被占用，其他所有elifs 就不能被占用。由于它们具有相同的条件，因此永远不会到达所有elifs。

如果你把else:取出来，只是默认return "p"，它会保证一个返回值：

elif len(my_history) > 1 and len(their_history) > 1:
    if (their_history[-1] == 'p' and my_history[-1] == 'p'):
        return 's'
# If anything else happens
return "p"

但这并不能解决根本问题。

或者，他们可以结合条件：

if len(my_history) > 1 and len(their_history) > 1 and
    (their_history[-1] == 's' and my_history[-1] == 'r'):
        return 'p'

这也将避免该问题，因为没有内部条件可以跳过返回，else: 将按预期工作。

---

正如其他人所提到的，整个外部 if 块是多余的。对于我的女孩（我教 11-13 岁/os），我建议他们将默认返回移到顶部......然后你不必检查每个列表是否足够大：

if not (len(my_history) > 1 and len(their_history) > 1):
    return "p"  # default value
elif their_history[-1] == 's' and my_history[-1] == 'r':
    return 'p'
elif their_history[-1] == 'r' and my_history[-1] == 'p':
    return 's'
elif their_history[-1] == 'p' and my_history[-1] == 's':
    return 'r'
elif their_history[-1] == 's' and my_history[-1] == 'p':
    return 'r'
elif their_history[-1] == 'r' and my_history[-1] == 's':
    return 'p'
elif their_history[-1] == 'p' and my_history[-1] == 'r':
    return 's'
elif their_history[-1] == 's' and my_history[-1] == 's':
    return 'r'
elif their_history[-1] == 'r' and my_history[-1] == 'r':
    return 'p'
elif their_history[-1] == 'p' and my_history[-1] == 'p':
    return 's'

如果我们失败了，在最后添加一个异常：

raise AssertionError, "a combination not covered was encountered"

这将确保我们知道我们是否忘记了一种可能性。

---

其他风格点可能值得讨论，也可能不值得讨论，具体取决于他们的能力：

通过嵌套条件，他们可以减少重复，但如果他们想调整他们的机器人，这是一个障碍而不是好处。

if not len(my_history) > 1 and len(their_history) > 1:
    return "p"  # default value

elif their_history[-1] == 's':  # if they threw scissors
    if my_history[-1] == 'r':       # and I threw rock
        return 'p'                      # throw paper
    return 'r'                      # otherwise throw rock

elif their_history[-1] == 'r':
    if my_history[-1] == 'p':
        return 's'
    return 'p'

elif their_history[-1] == 'p': 
    if my_history[-1] == 's':
        return 'r'
    return 's'

Answer 2

您的elif 语句处于错误级别，as explained by @SiongThyeGoh。下面的工作正常，虽然不优雅。

def Win_Loss(my_history, their_history):
    if len(my_history) > 1 and len(their_history) > 1:
        if (their_history[-1] == 's' and my_history[-1] == 'r'):
            return 'p'
        elif (their_history[-1] == 'r' and my_history[-1] == 'p'):
            return 's'
        ...

但这将是我最喜欢的解决方案：

def Win_Loss(my_history, their_history):

    d = {frozenset({'s', 'r'}): 'p',
         frozenset({'r', 'p'}): 's',
         frozenset({'s', 'p'}): 'r',
         frozenset({'s', 's'}): 'r',
         frozenset({'r', 'r'}): 'p',
         frozenset({'p', 'p'}): 's'}

    if len(my_history) > 1 and len(their_history) > 1:
        return d.get(frozenset({their_history[-1], my_history[-1]}))

    else:
        return "p"

print(Win_Loss(['s','p','p'],['s','p','p']))

Answer 3

if len(my_history) > 1 and len(their_history) > 1:
        if (their_history[-1] == 's' and my_history[-1] == 'r'):
            return 'p'

看前三行，如果第一个 if 条件满足而第二个不满足，那么你什么都不返回，也就是你得到 None。

如果这个条件通过：

if len(my_history) > 1 and len(their_history) > 1:

那么它将进入前三行，否则，另一个elif语句将被忽略，因为它是相同的条件，它只会到达else部分并返回p。

它只能返回 p 或 None。