允许用户使用索引位置从列表中进行选择答案

【问题标题】：Allowing user to select from a list using the index location允许用户使用索引位置从列表中进行选择
【发布时间】：2021-09-04 12:45:41
【问题描述】：

如果我没有正确地问这个问题，我很抱歉，这是我第一次在这里问。

我有一个当前可以运行的脚本，但我正在努力改进它。

我创建了一个星期几的字典，并根据每个值允许用户输入他们想在那天吃哪顿饭。但是，目前我要求用户从列表中键入他们选择的选项，该选项必须完全匹配。例如“四川炒猪肉”——这里很容易打错。而不是管理错字，我想让用户更容易从列表中进行选择，例如通过从字典中选择一个键或从列表中选择一个索引位置 - 我无法让其中任何一个工作。

我的代码现在看起来像这样：

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

我尝试通过执行以下操作来创建字典，但我无法弄清楚如何将每个项目的键设置为递增 1：

week_meals_dict = {}

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

但是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。这甚至可能吗？

这让我想到让用户在列表中输入索引位置可能更容易，但我也想不通。

【问题讨论】：

标签： python list dictionary input

【解决方案1】：

我想通了……

不太确定我是否完全理解它的工作原理，但这似乎是：

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

我什至可以把它清理干净，但我现在很满意

【讨论】：

请注意，您可以使用 some_dict.keys() 遍历字典键列表。因此，将来如果您想并行遍历键和列表，您可以使用：for i,j in zip(som_list, some_dict.keys()):

【解决方案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

【讨论】：

【解决方案3】：

首先，don't modify a list during iteration。

您似乎想为一周中的每一天匹配一顿饭。您可以迭代这些日子：

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!")

【讨论】：