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带有IndexError消息的并行列表
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Stack Overflow用户
提问于 2021-08-05 19:28:10
回答 1查看 39关注 0票数 0

我几乎完成了一个项目,如果没有"max = IndexError“行的环,我似乎无法完成这一部分。如果有帮助,变量ring和player是常量变量,并标记为player = []和ring = []。这是最新的Python 3版本。谢谢!

代码语言:javascript
复制
def main():
    player = []
    ring = []
    menu()
    addNBA(player, ring)
    rankNBA(player, ring)

# yes or no input request to continue updating ranking
def addNBA(player, ring):
    choice = "y"
    while choice.lower() != 'y' and choice.lower() != "n":
        print("Invalid. Enter 'y' or 'n'.")
        playerName = playerEntry()
        player.append(playerName)
        nbaRing = ringEntry()
        ring.append(nbaRing)
        choice = input("Do you want to continue updating the ranking? Enter 'y' or 'n'. ")

# sorts list of players and rings from lowest to highest amount of rings
def rankNBA(player, ring):
    max = ring[0]
    bestPlayer = player[0]
    min = ring[0]
    worsePlayer = player[0]
    print(player)
    for i in range(len(ring)):
        print(ring[i])
        if max < ring[i]:
            max = ring[i]
            bestPlayer = player[i]
        if min > ring[i]:
            min = ring[i]
            worsePlayer = player[i]
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回答 1

Stack Overflow用户

发布于 2021-08-05 19:52:36

我认为你想要的是忽略错误并运行代码。因此,您应该根据您的代码,在出现错误的地方使用try except statement : Put try:

代码语言:javascript
复制
def rankNBA(player, ring):
    try:
        max = ring[0]
    except IndexError:
        pass
    bestPlayer = player[0]
    min = ring[0]
    worsePlayer = player[0]
    print(player)
    for i in range(len(ring)):
        print(ring[i])
        if max < ring[i]:
            max = ring[i]
            bestPlayer = player[i]
        if min > ring[i]:
            min = ring[i]
            worsePlayer = player[i]

如果min = ring或其他任何情况下再次出现同样的错误,只需使用try except。结论

我们使用try except语句只是为了告诉程序应该忽略错误并继续执行。

我希望这回答了你的问题。

票数 0
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页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/68672497

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