Popping items from a list using a loop

Question

I am trying to write a for loop in python to pop out all the items in a list but two, so I tried this:

guest = ['john', 'phil', 'andy', 'mark', 'frank', 'joe']

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

and this is what I get when I run it:

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

因此，它仅弹出3，但是否有办法使其弹出6中的4？我尝试添加一个if语句：

guest = ['john', 'phil', 'andy', 'mark', 'frank', 'joe']

for people in guest:
  if people > guest[1]:
    popped_guest = guest.pop()
    print("I am sorry " + popped_guest + " I can no longer invite you to dinner")

我以为'phil'是1，它将弹出最后4个字符，但是当我运行该程序时，它什么也不返回。那么可以在一个for循环中做吗？

Answer 1

如果您想弹出4件事，那么只需数4

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

Answer 2

您的for循环在第 3 次迭代后停止，因为这是guest弹出前一个元素后剩余的元素数量。您可以潜在地使用 while 循环来连续弹出元素，直到列表中只剩下 2 个。

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