如果你想弹出4个东西，那就数到4

for _ in range(4):
    popped_guest = guest.pop()
    print("I am sorry " + popped_guest + " I can no longer invite you to dinner")

for循环在第三次迭代后停止，因为这是弹出前一个元素后guest中剩余的元素数。您可以使用while循环来连续弹出元素，直到列表中只剩下2个元素。

while len(guest) > 2:
    popped_guest = guest.pop()
    ...

如前所述，您的代码根本没有做您认为它是什么，因为您正在从列表中弹出元素，同时积极地迭代它。我会说“一个更好的编码实践是复制要弹出的列表”，但这不是一个“更好”的实践-你的方式根本不像你想要的那样工作，它总是弹出列表的前半部分。
我会问自己“我如何指定在我当前的迭代中弹出的人”，以及“我在哪里设置在我当前的迭代中弹出的人的数量”。这两个问题的答案似乎都是“我不知道”。

这种行为的原因是，当你在Python中使用for循环时，它实际上是通过它的索引号遍历列表。因此，当你在一个列表上循环并同时改变它时，你可能会跳过一些元素。更好的方法是在循环原始列表的同时改变列表的副本。

invite = ['Pankaj', 'Deepak', 'Nischey', 'Ram', 'Martin', 'Gopal']

for item in invite:
    if invite.index(item) > 1:
        popped_guest = invite.pop()
        print("Sorry you are not invited " + popped_guest)
    else:
        print("you are invited " + item)

让2个值仍然在列表中。而当列表长度未知时。

guest = ['john', 'phil', 'andy']

except2 = (len(guest)) - 2

for i in range(except2):
    popped_guest = guest.pop()
    print("I am sorry " + popped_guest + " I can no longer invite you to dinner") 
    
else:
    for i in guest:
        print(f"You are invited  {i}")

I am sorry andy I can no longer invite you to dinner
You are invited  john
You are invited  phil

[Program finished]

您可以将list()和copy()用于numbers，以创建不同的复制numbers，如下所示：

guest = ['john', 'phil', 'andy', 'mark', 'frank', 'joe']
             # ↓ Here ↓
for people in list(guest):
    popped_guest = guest.pop()
    print("I am sorry " + popped_guest + " I can no longer invite you to dinner")

或者：

guest = ['john', 'phil', 'andy', 'mark', 'frank', 'joe']
            # ↓ ↓ Here ↓ ↓
for people in guest.copy():
  popped_guest = guest.pop()
  print("I am sorry " + popped_guest + " I can no longer invite you to dinner")

输出：

I am sorry joe I can no longer invite you to dinner
I am sorry frank I can no longer invite you to dinner
I am sorry mark I can no longer invite you to dinner
I am sorry andy I can no longer invite you to dinner
I am sorry phil I can no longer invite you to dinner
I am sorry john I can no longer invite you to dinner

我添加了一些代码，希望它可以帮助你。

text = ["hello", "world"]

while True:
    try:
        val = text.pop()
        print(val)
    except IndexError:
        return