Improve SQLAlchemy update efficiency

Question

I have two tables, users (~200.000) and tweet (~2.000.000) I need to update all users, with the number of tweets, favourites(of their tweets), replies, retweets. This is inside a script:

@classmethod
def get_user_tweet_counts(cls, user_id):
    return (db_session
        .query(
            func.sum(Tweet.favorite_count).label('favorite_count'),
            func.sum(Tweet.retweet_count).label('retweet_count'),
            func.sum(Tweet.reply_count).label('reply_count'),
            func.count(Tweet.id).label('tweet_count'))
        .filter(Tweet.user_id == user_id)
        .group_by(Tweet.user_id).first())  # This will always be one result, should I query differently?

db_session:

engine = create_engine('postgresql://tweetsql:tweetsql@127.0.0.1/tweetsql')
db_session = scoped_session(sessionmaker(autocommit=False,
                                         autoflush=True,
                                         bind=engine))
Base = declarative_base()
Base.query = db_session.query_property()

The 10 minute loop:

for user in all_users:
    update_count += 1
    aggregation_result = Tweet.get_user_tweet_counts(user.id)
    user.total_tweet_favourites = aggregation_result[0] or 0
    user.total_tweet_retweets = aggregation_result[1] or 0
    user.total_tweet_replies = aggregation_result[2] or 0
    user.tweet_count = aggregation_result[3] or 0
User.save()  # this just calls db_session.commit()
# We only commit the session once to speed things up

User and Tweet are declared like:

User(Base), Tweet(Base) (from the db_session snippet).

While this is running, python hits 80% cpu, and ~600mb memory. How can I make this better? Tweet has index on user_id and its own id.

score 2 · Accepted Answer · edited May 23 '17 at 12:01

Here is a great answer by the author of SQLAlchemy. Basically, you'll want to bypass the ORM if you need to scale to large number of rows.

In your particular situation, you can write a single query to achieve the same result using SQL aggregation:

UPDATE users SET
  total_tweet_favourites = aggregated.total_tweet_favourites,
  total_tweet_retweets = aggregated.total_tweet_retweets,
  total_tweet_replies = aggregated.total_tweet_replies,
  tweet_count = aggregated.tweet_count
FROM (
  SELECT
    users.id AS id,
    SUM(tweets.favorite_count) AS total_tweet_favourites,
    SUM(tweets.retweet_count) AS total_tweet_retweets,
    SUM(tweets.reply_count) AS total_tweet_replies,
    COUNT(tweets.id) AS tweet_count
  FROM users JOIN tweets ON tweets.user_id = users.id
  GROUP BY users.id
) aggregated
WHERE users.id = aggregated.id;

To translate this to SQLAlchemy:

aggregated = session \
    .query(
        User.id.label("id"),
        func.sum(Tweet.favorite_count).label("total_tweet_favourites"),
        func.sum(Tweet.retweet_count).label("total_tweet_retweets"),
        func.sum(Tweet.reply_count).label("total_tweet_replies"),
        func.count(Tweet.id).label("tweet_count")) \
    .select_from(User) \
    .join(Tweet) \
    .group_by(User.id) \
    .subquery() \
    .alias("aggregated")
query = User.__table__ \
    .update() \
    .values(
        total_tweet_favourites=aggregated.c.total_tweet_favourites,
        total_tweet_retweets=aggregated.c.total_tweet_retweets,
        total_tweet_replies=aggregated.c.total_tweet_replies,
        tweet_count=aggregated.c.tweet_count) \
    .where(User.__table__.c.id == aggregated.c.id)
session.execute(query)

That looks more complicated that plain SQL :O. Makes sense though. Thanks I will look into it. — Giannis, Feb 01 '17 at 10:28
There currently is no FK between User and Tweet, tweet only has user_id. Is there a way to make this work without a join ? — Giannis, Feb 01 '17 at 18:17

Improve SQLAlchemy update efficiency

1 Answers1