Question

我有一本字典，其ID是配方ID，值是成分列表：

recipe_dictionary  = { 134: ['salt', 'chicken', 'tomato paste canned'],
                       523: ['toast whole grain', 'feta cheese' 'egg', 'salt'], 
                       12: ['chicken', 'rice', 'parsley']}

我还有一个静态列表，其中包含我白天不想重复的成分：

non_repeatable_ingredients = ['egg', 'chicken', 'beef']

现在，我遍历字典的每个值，然后遍历成分名称，将每个名称与 non_repeatable_ingredients 列表进行比较，然后创建共享单词的列表。因此，我的缩小字典将如下所示：

   reduced_recipe_dictionary  = { 134: ['chicken'],
                                  523, ['egg'], 
                                  12: ['chicken']

此过程需要很长时间，因为我的实际字典和成分列表很长。是否有比以下方法更快的方法？

这是 get_reduced_meal_plans_dictionry 方法：

reduced_meal_plans_dictionary = {}

# For each recipe
for recipe in meal_plans_dictionary:

    # Temp list for overlapp ingredients found for each recipe
    overlapped_ingredients_list = []

    # For each complete name of ingredient in the recipe
    for ingredient_complete_name in meal_plans_dictionary[recipe]:

        # Clean up the ingredient name as it sometimes involves comma, parentheses or spaces
        ingredient_string = ingredient_complete_name.replace(',', '').replace('(', '').replace(')', '').lower().strip()

        # Compare each ingredient name against the list of ingredients that shall not repeated in a day
        for each in PROTEIN_TAGS:

            # Compute the partial similarity
            partial_similarity = fuzz.partial_ratio(ingredient_string, each.lower())

            # If above 90, means one of the ingredients in the PROTEIN_TAGS exists in this recipe
            if partial_similarity > 90:
                # Make a list of such ingredients for this recipe
                overlapped_ingredients_list.append(each.lower())

    # Place the recipe ID as the key and the reduced overlapped list as the value
    reduced_meal_plans_dictionary[recipe] = overlapped_ingredients_list

我使用替换率和相似率是因为成分名称并不干净。例如，我可以将鸡蛋或煮鸡蛋作为一种成分。

谢谢。

Answer 1

由于每个食谱都有独特的成分并且顺序无关紧要，所以使用集合而不是列表怎么样？

可以在O（1）个恒定时间内搜索集合，而列表可以在O（n）个恒定时间内搜索。

Here are some examples。

例如：

recipe_dictionary = { 
    134: set(['salt', 'chicken', 'tomato paste canned']),
    523: set(['toast whole grain', 'feta cheese' 'egg', 'salt']), 
    12: set(['chicken', 'rice', 'parsley'])
}

non_repeatable_ingredients = set(['egg', 'chicken', 'beef'])

您可以在这样的集合中测试元素的存在：

for ingredient in recipe_dictionary[134]:
    if ingredient in non_repeatable_ingredients:
        # do something

Answer 2

使用正则表达式和defaultdict的组合，您可以确切地找到所需的内容。这种方法使用正则表达式来减少所需的for循环数。

请注意，我已经调整了键12以表明它将同时获得两个匹配项。

recipe_dictionary  = { 134: ['salt', 'chicken', 'tomato paste canned'],
                        523: ['toast whole grain', 'feta cheese', 'egg', 'salt'],
                        12: ['whole chicken', 'rice', 'parsley', 'egg']}
non_repeatable_ingredients = ['egg', 'chicken', 'beef']
non_repeat = '(' + '|'.join(non_repeatable_ingredients) + ')'

d = defaultdict(list)
for k, j in recipe_dictionary.items():
     for i in j:
            m = re.search(non_repeat, i)
            if m:
                d[k].append(m.groups()[0])
d
defaultdict(list, {134: ['chicken'], 523: ['egg'], 12: ['chicken', 'egg']})

Answer 3

>>> reduced_recipe_dictionary = {k: list(filter(lambda x: x in non_repeatable_ingredients, v)) for k,v in recipe_dictionary.items()}
>>> reduced_recipe_dictionary
{134: ['chicken'], 523: ['egg'], 12: ['egg']}
>>>

如果您没有与non_repeatable_ingredients列表中的项目匹配的干净配料，则可以使用fuzz.partial_ratio模块中的fuzzywuzzy来获得最匹配的配料（比率更大的配料）比说80％）。手动进行pip install fuzzywuzzy安装

>>> from fuzzywuzzy import fuzz
>>> reduced_recipe_dictionary = {k: list(filter(lambda x: fuzz.partial_ratio(v,x) >80, non_repeatable_ingredients)) for k,v in recipe_dictionary.items()}
>>> reduced_recipe_dictionary
{134: ['chicken'], 523: ['egg'], 12: ['chicken']}

Python-将字典值列表缩小为较小的列表

3 个答案: