Question

我的学生正在为课堂制作摇滚，纸张，剪刀模拟。一个组有一个错误，其中一个函数不会返回任何东西。我已检查过它，看看所有分支机构是否都有退货声明，而且确实如此。我也试过了一个可视化工具，功能就此停止了，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＆＃39;，但它打印无。

Answer 1

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

要明确的是，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"

因为一旦获得第一个if，就无法获取所有其他elif。由于它们具有相同的条件，因此永远不会达到所有elif。

如果您取消else:，只是默认return "p"，则会保证返回值：

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'

这也可以避免这个问题，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/年），我建议他们将默认返回到顶部...然后你不必检查每个人是否足够大：

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，最后会添加一个例外：

raise AssertionError, "a combination not covered was encountered"

然后会确保我们知道我们是否忘记了这种可能性。

根据他们的能力，可能会或可能不值得讨论的其他风格点：

通过嵌套他们的条件，他们可以减少重复，但如果他们想要调整他们的机器人这是一个障碍而不是一个好处。

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语句处于错误的级别as explained by @SiongThyeGoh。以下工作正常，尽管不是很优秀。

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'

看看前三行，如果第一个if条件满足但第二个不满足，则不返回任何内容，即得到None。

如果这种情况通过：

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

然后它将进入前三行，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。

在Python上返回None

3 个答案: