匹配元组列表中的元素并删除列表中的元组元素

Question

recipes = [('Peanut Butter','200g salt','500g sugar'),('Chocolate Brownie', '500g flour')]

def remove_recipe(recipe_name,recipes):
    recipe_dict = dict(recipes)
    size = recipe_dict.get(recipe_name)
    if size is not None:
        recipe_dict.pop(recipe_name)
        updated_recipes = [(k, v) for k, v in recipe_dict.items()]
        return updated_recipes

    else:
        print("There is no such recipe.")

我想达到什么目的？ 您好我是 python 的新手，我正在尝试从元组列表中删除一个项目。我将食谱列表转换为字典，并将大小用于特定键的值。一旦键不是 None 它将从字典中删除该项目并将其转换回列表。

收到错误： ValueError：字典更新序列元素#0 的长度为 3；需要 2 个

Answer 1

我会说不需要转换为字典——假设元组的第一个元素是食谱的名称——只需使用列表理解并比较大小，:

def remove_recipe(recipe_name,recipes):    
    res = [r for r in recipes if r[0] != recipe_name] # filter out by the recipe name
    
    if len(res) != len(recipes):
        print("Recipe removed.")
    else:
        print("There is no such recipe.")

    return res

Answer 2

字典是具有键值对的结构。所以你不能将这个元组转换为字典，因为它不知道它应该做什么。你实际上不需要字典来做你想做的事情。

recipes = [('Peanut Butter','200g salt','500g sugar'),('Chocolate Brownie', '500g flour')]

def remove_recipe(recipe_name,recipes):
    for recipe in recipes:
        if recipe_name in recipe:
            recipes.remove(recipe)
            print("Recipe removed.")
            break

    else:
        print("There is no such recipe.")

字典可以构造你的食谱，但你必须重构你的食谱：

{
    "name": 'Peanut Butter',
    "ingredients": ['200g salt', '500g sugar']
}