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python - 创建一个新字典的更简单方法，其中另一个字典的值作为键，出现次数作为值

问题描述

我将首先为我的代码提供一些上下文，而不是进一步解释，这些代码有效但似乎效率很低：

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}    

    >>> get_quantities({'t1': ['pie'],
    't2': ['orange pie'], 't3': ['pie']})

    {'pie': 2, 'orange pie': 1} 
    """

    food_to_quantity = {}
    
    # Accumulate the food information here.

    # Creating a dictionary with the new keys as values from the other
    for j in table_to_foods.values():
      for a in j:
        food_to_quantity[a] = 0
    
    # Increment based on number of occurrences
    for j in table_to_foods.values():
      for a in j:
        food_to_quantity[a] += 1

    return food_to_quantity

必须有一种更简单的方法来创建一个新字典，其中 table_to_foods 的值作为键，任何食物值的出现次数作为值。

标签： pythonlistdictionaryaccumulate

解决方案

这就是我会怎么做。

table_to_foods = {'t1': ['Vegetarian stew', 'Poutine', 'Vegetarian stew'],
't3': ['Steak pie', 'Poutine', 'Vegetarian stew'], 't4': ['Steak pie', 'Steak pie']}

food_to_quantity = {}
for foods in table_to_foods.values():
    for food in foods:
        if(food not in food_to_quantity):
            food_to_quantity[food]=1
        else:
            food_to_quantity[food]+=1

print(food_to_quantity)

输出：{'素食炖菜'：3，'Poutine'：2，'牛排馅饼'：3}

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