计算字典中列表中单词的频率

Question

I have a list of common keywords:

common_keywords = ['dog', 'person', 'cat']

And a list of dictionaries, containing keywords and sometimes the common_keywords listed above:

people = [{'name':'Bob', 'keywords': ['dog', 'dog', 'car', 'trampoline']},
          {'name':'Kate', 'keywords': ['cat', 'jog', 'tree', 'flower']},
           {'name':'Sasha', 'keywords': ['cooking', 'stove', 'person', 'cat']}]

I would like to count the frequency of the common_keywords for each person, so the desired output would look something like:

counts = [{'name': 'Bob', 'counts': [{'dog': 2}]}, 
          {'name': 'Kate', 'counts': [{'cat': 1}]}, 
          {'name': 'Sasha', 'counts': [{'person':1}, {'cat': 1}]]

I am able to use dict(Counter()) to count the keywords and filter them if they appear in the common_keywords but I am struggling with linking these counts back to the original name as shown in the desired output: counts .

Current code (I think I am slowly getting there):

freq_dict = {}
for p in people:
    name = p['name']
    for c in p['keywords']:
        if c not in freq_dict:
            freq_dict[name] = {c: 1}
        else: 
            if c not in freq_dict[name]:
                freq_dict[c] = 1
            else:
                freq_dict[c] +=1

Answer 1

You can use a list-comprehension along with collections.Counter which does exactly what you want with the nested list. -

from collections import Counter

[{'name':i.get('name'),
  'keywords':[dict(Counter([j for j in i.get('keywords') 
                            if j in common_keywords]))]} for i in people]

[{'name': 'Bob', 'keywords': [{'dog': 2}]},
 {'name': 'Kate', 'keywords': [{'cat': 1}]},
 {'name': 'Sasha', 'keywords': [{'person': 1, 'cat': 1}]}]

---

1. First, with the list comprehension you want to reconstruct the original list of dicts with keys separately defined along with i.get('key') . This will let to work with the nested list value for keywords.
2. Iterate over the list and filter only the ones in common_keywords
3. Pass this list into collections.Counter to get your dict
4. Return it as a list with a single dict inside as you expect it to be