允许用户使用索引位置从列表中进行选择

Question

如果我没有问对这个问题，我先道歉，这是我第一次在这里问。

我有一个脚本，目前工作，但我正在努力改进它。

我创建了一个包含一周中的天数的字典，并允许用户根据每个值输入他们想在当天吃哪顿饭。但是，目前我要求用户从列表中输入他们选择的选项，这些选项必须完全匹配。例如"四川炒肉";这里很容易打错字。而不是管理错字，我想让它更容易使用，使他们的选择从列表中，例如，通过选择一个关键字从字典或从列表中的索引位置-我有麻烦，让这些工作虽然。

我的代码现在是这样的:

for d in week_meals:
try:
answer = input(f"What would you like to eat on {d}? options are {week_meals_list}")
week_meals_list.remove(answer)
week_meals[d] = answer
except ValueError:
print("That isn't an option!")
answer = input(f"What would you like to eat on {d}? options are {week_meals_list} type {week_meals_list.index")
week_meals_list.remove(answer)
week_meals[d] = answer

我尝试通过做下面的事情来创建一个字典，但我不知道如何设置每个项的键以递增1:

week_meals_dict = {}
for k in range(int(days)):
week_meals_dict[k] = None

，但我真的无法找到一种方法来遍历每个键，同时并行遍历列表。这可能吗?

这让我想到，让用户在列表中输入索引位置可能更容易，但我也不知道。

Answer 1

I figure out…

不太确定我完全理解为什么它工作，但这似乎:

for k in range(int(days)):
week_meals_dict[k] = week_meals_list[k]

我甚至可以把它清理一下但是我现在很满意

Answer 2

week_meals = {}
week_meals_list = ['meal1', 'meal2',]
for d in range(1, 8):
meal = None
while not meal:
answer = input(f"What would you like to eat on {d}? options are {week_meals_list}")
try:
meal = week_meals_list[int(answer)]
week_meals_list.remove(meal)
week_meals[d] = meal
except (IndexError, ValueError):
pass

Answer 3

首先，不要在迭代期间修改列表。

看来你是在试着为一周中的每一天搭配一顿饭。您可以按日期进行迭代:

week_meals_options = {1: "Chicken", 2: "Soup"} # Your dict containing choices mapped to the meal options
week_meals = {}
for day in week_days:
while True:
answer = input(f"What would you like to eat on {day}? options are {week_meals_options.keys()}")
if answer in week_meals_options:
week_meals[d] = week_meals_options.pop(answer) # Remove the option from the dict, and assign it to the day
break
else:
print("That isn't an option!")