Question

任何人都可以帮我这个功能吗？我不知道编写代码，我在函数体中编写的内容是错误的。

def get_quantities(table_to_foods: Dict[str, List[str]]) -> Dict[str, int]:

    """The table_to_foods dict has table names as keys (e.g., 't1', 't2', and
    so on) and each value is a list of foods ordered for that table.

    Return a dictionary where each key is a food from table_to_foods and each
    value is the quantity of that food that was ordered.

    >>> get_quantities({'t1': ['Vegetarian stew', 'Poutine', 'Vegetarian stew'],
    't3': ['Steak pie', 'Poutine', 'Vegetarian stew'], 't4': ['Steak pie', 'Steak pie']})
    {'Vegetarian stew': 3, 'Poutine': 2, 'Steak pie': 3}    
    """

    food_to_quantity = {}
    for t in table_to_foods:
        for i in table_to_foods[t]:
            if i in table_to_foods[t]:
                food_to_quantity[i] = food_to_quantity[i] + 1

    return food_to_quantity

Answer 1

如果您喜欢使用itertools.chain和collections.Counter：

，还有另一种做法

from itertools import chain
from collections import Counter

dict(Counter(chain.from_iterable(foods.values())))
#or Simply
dict(Counter(chain(*foods.values())))

#Output:
#{'apple': 3, 'banana': 4, 'grapes': 1, 'orange': 1}

Answer 2

计算没有库的项目的常用方法是使用python get()函数

foods = {
    't1': ['banana', 'apple', 'banana'],
    't2': ['orange', 'apple', 'banana'],
    't3': ['apple', 'grapes', 'banana']
    }

def get_quantities(foodLists):
    totals = {}
    for foodList in foodLists.values():
        for food in foodList:
            totals[food] = totals.get(food, 0) + 1
    return totals

print(get_quantities(foods))

打印哪些：

{'banana': 4, 'apple': 3, 'orange': 1, 'grapes': 1}

Answer 3

使用Counters

from collections import Counter

def get_quantities(table_to_foods: Dict[str, List[str]]) -> Dict[str, int]:
     return dict(Counter(x for v in table_to_foods.values() for x in v))

您可能不需要dict Counter Counter是dict的子类，但是我这样做，所以你的类型是相同

Answer 4

试试这个：

def get_quantities(table_to_foods):
    food_to_quantity = {}
    for table in table_to_foods.values():
        for food in table:
            food_to_quantity[food] = food_to_quantity.get(food, 0) + 1
    return food_to_quantity

您可以使用.values（）来获取字典中的值，然后遍历每个项目。如果食物在字典中，则在其值中加1，如果没有，则将食物添加为字典中的新项目。

get_quantities({
    't1': ['Vegetarian stew', 'Poutine', 'Vegetarian stew'],
    't2': ['Steak pie', 'Poutine', 'Vegetarian stew'],
    't3': ['Steak pie', 'Steak pie']
    })

如果打印，

应输出以下内容：

{'Poutine': 2, 'Steak pie': 3, 'Vegetarian stew': 3}

有关词典的更多信息： https://docs.python.org/3/tutorial/datastructures.html#dictionaries

输入dict：作为列表的值

4 个答案: