Match set of dictionaries. Most elegant solution. Python

Question

Given two lists of dictionaries, new one and old one. Dictionaries represent the same objects in both lists.
I need to find differences and produce new list of dictionaries where will be objects from new dictionaries only and updated attributes from old dictionaries.
Example:

   list_new=[
             { 'id':1,
               'name':'bob',
               'desc': 'cool guy'
              },
             
             { 'id':2,
               'name':'Bill',
               'desc': 'bad guy'
              },

              { 'id':3,
               'name':'Vasya',
               'desc': None
              },
        ]

    list_old=[
             { 'id':1,
               'name':'boby',
               'desc': 'cool guy',
                'some_data' : '12345'
              },
             { 'id':2,
               'name':'Bill',
               'desc': 'cool guy',
               'some_data' : '12345'

              },
              { 'id':3,
               'name':'vasya',
               'desc': 'the man',
               'some_data' : '12345'
              },
              { 'id':4,
               'name':'Elvis',
               'desc': 'singer',
               'some_data' : '12345'
              },
            ]
            

In that example I want produce new list where will be only new guys from list_new with updated data. Matched by id. So Bob will become Boby, Bill will become coll guy, Vasya become - the man. End Elvis have to be absent.

Give me an elegant solution. With less amount of iteration loops.

There is way, I resolve that. Which is not the best:

 def match_dict(new_list, old_list)
    ids_new=[]
    for item in new_list:
            ids_new.append(item['id'])
    result=[] 
    for item_old in old_medias:
        if item_old['id'] in ids_new:
            for item_new in new_list:
                if item_new['id']=item_old['id']
                    item_new['some_data']=item_old['some_data']
                    result.append(item_new)
    return result

The reason why I'm doubt, because there is loop inside loop. If there will be lists of 2000 items the process would take same time.

Answer 1

Can't quite get it to one line, but here's a simpler version:

def match_new(new_list, old_list) :
    ids = dict((item['id'], item) for item in new_list)
    return [ids[item['id']] for item in old_list if item['id'] in ids]