列出索引超出范围，我不知道该怎么办

Question

我一直在尝试制作一个带有 2 个列表的程序。第一个列表是问题，我必须将它们与用户输入进行比较。如果用户输入与列表中的项目(问题(完全相同，则打印第二个列表中的其他数据。 例如：

Questions=["hello","yellow","horse"]
Ans=["world","I pref red","I pref dog"]
# now if input of user is something from the Questions list, it will print # from Ans
# if input --> yellow
# then --> print(Ans[1])

我写的代码是这样的：

x = len(Questions)
leng = int(x / 2)

quest = str(input('Which your question: '))
while(quest!='@'):
counter = 0
if(quest == Questions[counter]):
print(Ans[counter])
else:
counter+=1
while(quest != Questions[counter] and counter<x):
counter+=1
print(Ans[counter])
quest = str(input('Which your question: '))

出于某种原因，我正在接受此错误：列出索引超出范围第 244 行，在 而(任务！=问题[计数器]和计数器

Answer 1

如果问题不在您的列表中，它将遍历列表，然后在尝试访问列表中不存在的元素时将索引抛出范围。

你能用字典吗？

questions = {
"hello": "world",
"yellow": "I pref red",
"horse": "I pref dog"
}
quest = str(input('Which your question: '))
while (quest != '@'):
if quest in questions:
print(questions[quest])
else:
print("invalid input")
quest = str(input('Which your question: '))

建议通读一些关于字典的文档：

https://docs.python.org/3/tutorial/datastructures.html#dictionaries

https://realpython.com/python-dicts/

Answer 2

如果答案不在列表中，则counter==x(指的是不存在的元素(。

通常，应该使用并行列表，因为它们难以操作和维护。更好的解决方案是使用字典：

qAndA = {"hello" : "world", "yellow" : "I pref red", 
"horse": "I pref dog"}
if quest in qAndA:
print(qAndA[quest]) # Otherwise, repeat

Answer 3

您可以使用字典结构：

questions = {
"hello": "world",
"yellow": "I pref red",
"horse": "I pref dog"
}
quest = ""
while quest != '@':
quest = str(input('Which your question: '))
answer = questions.get(quest, "I have no answer")
print(answer)