如何对列表信息的某些部分进行分组？

Question

I have a list each element of list is like this:

 list[0]={'Keywords': 'program manager',
 'result': {'categoryId': '2712',
  'categoryName': 'program manager',
  'score': '0.9506622791290285'},
 'categoryId': '2712'}
 {'Keywords': 'technicalfunctional consultant', 'result': []}

I need to collect all keywords whose have the same categoryName . I did the following:

output1 = defaultdict(set)
for entry in list:
    kwds = entry['Keywords'].strip().split(' ')
    for word in kwds:
        output1[entry.get('categoryId', None)].add(word)

But it split all words and I don't want it. is there any way to collect all keywords with the same categoryName ?

Answer 1

You can find all the existent categoryIds in the dataset, then build a dict that for each one of them contains all the keywords associated.

EDIT: Changed the code to have a MRE and changed the name of the list to list1

Code:

list1=[]
list1.append({'Keywords': 'program manager',
'result': {'categoryId': '2712',
'categoryName': 'program manager',
'score': '0.9506622791290285'},
'categoryId': '2712'})

cat_set = set([elem["result"]["categoryName"] for elem in list1])

cat_dict = {}

for cat_name in cat_set:
    cat_dict[cat_name] = [elem["Keywords"] for elem in list1 if elem["result"]["categoryName"] == cat_name]

Answer 2

mylist=[{'Keywords': 'scrum master',
  'result': {'categoryId': '3193',
   'categoryName': 'agile coach',
   'score': '1.0'},
  'categoryId': '3193'},
 {'Keywords': 'principal consultant',
  'result': {'categoryId': '2655',
   'categoryName': 'principal consultant',
   'score': '1.045369052886963'},
  'categoryId': '2655'},{'Keywords': 'technicalfunctional consultant', 'result': []}]
def keywordcollector(yourlist):
    yourlist=list(filter(lambda x:type(x['result'])==type({}),yourlist))
    categories=set(x['result']['categoryName'] for x in yourlist)
    keywordslist=[]
    for category in categories:
        temp_list=list(filter(lambda x:x['result']['categoryName']==category,yourlist))
        temp_keywords=list(map(lambda x:x['Keywords'],temp_list))
        keywordslist.append({category:temp_keywords})
    return keywordslist
print(keywordcollector(mylist))

The code above gives you dictionary object with keywords for every categoryname. The code outputs

[{'principal consultant': ['principal consultant']}, {'agile coach': ['scrum master']}]