Python -将字典值列表缩减为更小的列表
Python -将字典值列表缩减为更小的列表
提问于 2018-07-18 05:33:30
我有一个字典，它的I是配方I，值是配料列表：
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，其中包含了我不想在一天中重复的成分：
non_repeatable_ingredients = ['egg', 'chicken', 'beef']现在，我遍历字典的每个值，然后遍历配料名称，将每个名称与non_repeatable_ingredients列表进行比较，然后创建一个共享单词列表。因此，我的缩小大小的字典将如下所示：
reduced_recipe_dictionary = { 134: ['chicken'],
523, ['egg'],
12: ['chicken']这个过程需要很长的时间，因为我真正的字典和配料列表很长。有没有比下面的更快的方法呢？
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我使用替换和相似比率，因为配料名称不像我的示例那样清晰；例如，我可以将鸡蛋或煮鸡蛋作为一种配料。
谢谢。
回答 3
Stack Overflow用户
回答已采纳
发布于 2018-07-18 05:59:50
使用正则表达式和defaultdict的组合，您可以确切地得到您正在寻找的东西。这种方法使用正则表达式来减少所需的for循环的数量。
注意，我已经调整了key 12，以显示它将获得两个匹配。
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']})Stack Overflow用户
发布于 2018-07-18 05:46:51
使用集合而不是列表怎么样，因为每个食谱都有独特的成分，顺序并不重要。
集合可以在O(1)固定时间内搜索，而列表可以在O(n)时间内搜索。
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'])您可以测试元素在集合中的存在情况，如下所示：
for ingredient in recipe_dictionary[134]:
if ingredient in non_repeatable_ingredients:
# do somethingStack Overflow用户
发布于 2018-07-18 06:12:42
>>> 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列表中的项目相匹配的干净配料，您可以使用fuzzywuzzy模块中的fuzz.partial_ratio来获取最匹配的配料(例如，比例大于80%的配料)。提前执行pip install fuzzywuzzy进行安装
>>> 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']}页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/51390499
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