Python：打印出没有特定项目的列表

Question

我正在尝试遍历列表并希望打印出除垃圾邮件之外的所有项目。 但是，我不断收到以下错误消息：预期类型“list[str]”（匹配的泛型类型“_T”）改为“str”。

menu = [
    ["egg", "bacon"],
    ["egg", "sausage", "bacon"],
    ["egg", "spam"],
    ["egg", "bacon", "spam"],
    ["egg", "bacon", "sausage", "spam"],
    ["spam", "bacon", "sausage", "spam"],
    ["spam", "sausage", "spam", "bacon", "spam", "tomato", "spam"],
    ["spam", "egg", "spam", "spam", "bacon", "spam"],
]
item = "-"
while item in menu != "spam":
    continue
if "spam" in menu:
    menu.remove("spam")
    print(menu)`

我还尝试了以下内容：

if item in menu == "spam"
    menu.remove("spam")

这导致了同样的问题。任何人都可以帮我解决这个问题吗？这让我发疯。

Answer 1

看起来问题出在 while 循环和 if 语句上。 这是您的代码的更正版本：

menu = [    ["egg", "bacon"],
    ["egg", "sausage", "bacon"],
    ["egg", "spam"],
    ["egg", "bacon", "spam"],
    ["egg", "bacon", "sausage", "spam"],
    ["spam", "bacon", "sausage", "spam"],
    ["spam", "sausage", "spam", "bacon", "spam", "tomato", "spam"],
    ["spam", "egg", "spam", "spam", "bacon", "spam"],
]
for sub_list in menu:
    if "spam" in sub_list:
        sub_list.remove("spam")
    print(sub_list)

由于菜单是列表的列表，因此您需要遍历每个子列表并检查其中是否包含“垃圾邮件”。然后您可以使用 remove 方法从该子列表中删除“垃圾邮件”。 此外，在 python 中，while 循环将运行直到满足某个条件，而 while 循环中的条件是 item in menu != "spam" 这始终为 True，这就是它不起作用的原因

编辑：问题是 remove 方法只从列表中删除第一次出现的项目。在最后三个列表中，“垃圾邮件”出现了多次，因此您需要使用 while 循环从每个子列表中删除所有出现的“垃圾邮件”。这是您如何做到这一点的示例：

menu = [    ["egg", "bacon"],
    ["egg", "sausage", "bacon"],
    ["egg", "spam"],
    ["egg", "bacon", "spam"],
    ["egg", "bacon", "sausage", "spam"],
    ["spam", "bacon", "sausage", "spam"],
    ["spam", "sausage", "spam", "bacon", "spam", "tomato", "spam"],
    ["spam", "egg", "spam", "spam", "bacon", "spam"],
]
for sub_list in menu:
    while "spam" in sub_list:
        sub_list.remove("spam")
    print(sub_list)

这将遍历每个子列表，当“垃圾邮件”仍在子列表中时，它将删除它。这样，它将从子列表中删除所有出现的“垃圾邮件”。