How to group two lists and sort by number of matches?

Question

Django

There are two lists

first_names=
last_names=[[<User: username1>],[<User: username3>],[<User: username4>]]
intersection=[[<User: username1>],[<User: username3>]]
subtraction=[[<User: username2>],[<User: username4>]]

list_of_all.append(intersection)
list_of_all.append(subtraction)

list_of_all=[[<User: username1>],[<User: username3>],[<User: username2>],[<User: username4>]]

did something like this but didn't work

context["by_names"].append(list(set(first_names) & set(last_names)))
context["by_names"].append(list(set(last_names) - set(first_names)))

I need to sort the most appropriate option
and how to submit the list
Please help, @Kitaev))

Answer 1

from itertools import chain
from collections import Counter


list1 = [1, 2, 3]
list2 = [1, 3, 4]
list3 = [1, 4, 5]


counter = Counter(chain(list1, list2, list3))
ordered = sorted(
    set(list1).union(list2).union(list3), key=lambda k: counter[k], reverse=True
)

print(ordered)

Answer 2

The intersection of sets is unlikely to achieve adequate relevance. It can be told that an object has two or more occurrences (if the intersection of two lists, of course). Thus, objects with 10 and 2 occurrences will be on the same level.
Easier to count occurrences.