Question

字典中的键由具有不同整数的不同元组组成，我需要用该整数打印字典中的所有值。

我试图使用dictionary.get（）和dictionary.values（）获取值结合不同的for循环。

#Vegetarian = 1
#Vegan = 2
#Gluten-free = 3
a = tuple([0])
# "Joe's Gourmet Burger"

b = tuple([1,3])
# "Main Street Pizza Company"

c = tuple([1,2,3])
# "Corner Cafe"

d = tuple([1])
# "Mama's Fine Italian"

e = tuple([1,2,3])
# "The Chef's Kitchen"

rest = {a:"Joe's Gourmet Burger",
        b:"Main Street Pizza Company",
        c:"Corner Cafe",
        d:"Mama's Fine Italian",
        e:"The Chef's Kitchen"}

food = set()
veg = input("does anyone need vegetarian? yes/no ")
if veg == "yes":
    food.add(1)

vegan = input("does anyone need vegan? yes/no ")
if vegan == "yes":
    food.add(2) 


glut =input("does anyone need gluten free? yes/no ")
if glut == "yes":
    food.add(3)

for num in food:
    print(num)
    print(rest[num])

这将为您提供所选过敏的价值。我希望字典能够识别元组中的键，然后打印出相应的餐厅。此外，如果用户选择了多个过敏症，则脚本应仅返回满足所有用户过敏症的餐馆。

我得到的是“ dict对象在0x000001ACC9DA4A68处的内置方法获取”

Answer 1

如果用户对任何问题的回答为“是”，您是否要打印出这些值？

如果是这样的话，这将无法完全解决您的问题，但这应该为您指明正确的方向：

for num in food:
    for k in rest.keys():
        if num in k:
            print(rest[k])

如果我的理解正确，则需要检查字典的每个键（即每个元组）中是否存在给定的数字。

@Michael Butscher的观点，这将不会很有用，因为您将拥有重复的键。相反，您可以将数据结构重组为包含每个键的餐厅列表的内容：

requirements = {
    0: "None",
    1: "Vegetarian",
    2: "Vegan",
    3: "Gluten-free"
}
requirement_restaurant_map = {
    0: ["Joe's Gourmet Burger"],
    1: ["Main Street Pizza Company", "Corner Cafe", "Mama's Fine Italian", "The Chef's Kitchen"],
    2: ["Corner Cafe", "The Chef's Kitchen"],
    3: ["Main Street Pizza Company", "Corner Cafe", "The Chef's Kitchen"]
}

food = set()
veg = input("does anyone need vegetarian? yes/no ")
if veg == "yes":
    food.add(1)

vegan = input("does anyone need vegan? yes/no ")
if vegan == "yes":
    food.add(2) 

glut = input("does anyone need gluten free? yes/no ")
if glut == "yes":
    food.add(3)

intersection = []
for num in food:
    for k in requirements:
        if num == k:
            if not intersection:
                intersection.extend(requirement_restaurant_map[k])
        else:
            intersection = list(set(intersection).intersection(set(requirement_restaurant_map[k])))

print(f"The following restaurants satisfy your requirements...\n {intersection}")

在元组中查找整数，这是字典中的关键

1 个答案: