I tried to reproduce your example, I came up with a user schema like this…
schema "users" do
field :name, :string
field :password, :string, virtual: true
field :password_digest, :string
many_to_many :friends, User,
join_through: "friendships",
join_keys: [friend_a_id: :id, friend_b_id: :id]
many_to_many :reverse_friends, User,
join_through: "friendships",
join_keys: [friend_b_id: :id, friend_a_id: :id]
many_to_many :topics, Topic,
join_through: "follows"
has_many :tweets, Tweet
timestamps()
end
One note is that friendship can be tricky. To get all the friends I am with, I need to query for both direct and reverse friends. Maybe someone can show the way to make this friends association more elegant.
So You can query for direct tweets
q = from t in Tweet, where: t.user_id == ^the_user_id_you_want
Here it is not complete, but You can see the idea
q = from t in Tweet, where: t.user_id in [^friend_1_id, ^friend_2_id…]
I saw here https://hexdocs.pm/ecto/Ecto.Query.html#select/3 that You can use or_where, but I did not yet made the one query you are looking for 
So, in theory (not tested), You shoud be able to do
q = from t in Tweet, where: t.user_id == ^the_user_id_you_want,
or_where: t.user_id in [^friend_1_id, ^friend_2_id…],
or_where: t.topic_id in [^topic_that_user_follows_id, …]
Then, use a distinct: true, an order by date… and whatever You want to do
Please show your schema, so we can compare our way of doing this.
