在Python上不返回任何內容

Question

我的學生正在上課模擬岩石，紙張，剪刀。 一組有一個錯誤，其中一個函數不會返回任何東西。 我檢查了一下，看看是否所有分支都具有return語句，並且確實如此。 我也嘗試過使用可視化工具，但功能只是停止，我也不知道為什么。

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']))

這應該是打印“ s”，但它打印“無”。

Answer 1

如果任何內部if失敗，它將返回None ，因為采用了外部if / elif，因此將不會執行else:塊。

要明確的是，它們的整個功能等效於：

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"

因為一旦采取了第一個if ，就不能采用所有其他elif 。 由於它們具有相同的條件，因此將永遠無法達到所有elif 。

如果您將else:取出，並且僅默認return "p" ，它將保證返回值：

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"

但這不能解決根本問題。

另外，他們可以結合條件：

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

這也將避免出現此問題，因為沒有內部條件可以跳過收益，而else:將按預期工作。

---

正如其他人提到的那樣，整個外部if塊是多余的。 對於我的女孩（我教11-13歲），建議他們將默認返回值移到頂部...然后，您不必檢查每個列表是否足夠大：

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'

如果出現問題，請在末尾添加一個例外：

raise AssertionError, "a combination not covered was encountered"

然后，這將確保我們知道是否忘記了可能性。

---

取決於其能力，可能不應該討論的其他樣式點：

通過嵌套其條件，它們可以減少重復，盡管如果要調整其機器人，這是一個障礙，而不是收益。

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語句處於錯誤的級別， 如@SiongThyeGoh所述 。 盡管效果不佳，但以下內容仍然可以正常工作。

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'
        ...

但這將是我最喜歡的解決方案：

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'

查看前三行，如果滿足條件的第一行但不滿足第二條，則您不返回任何內容，即得到None。

同樣，如果此條件通過：

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

那么它將進入前三行，否則，其他elif語句將被忽略，因為它是相同條件，並且只會到達else部分並返回p。

它只能返回p或無。