在 Redshift 中查找下一个最旧的行

Question

I have a table called user_activity in Redshift that has department, user_id, activity_type, activity_id, activity_date.

I'd like to query a daily report of how many days since the last event (of any type). Using CROSS APPLY (SQL Server) or LATERAL JOIN (Postgres 9+), I'd do something like...

SELECT d.date, a.last_activity_date
FROM date_table d
CROSS JOIN (
            SELECT DISTINCT user_id FROM activity_table
        ) u
CROSS APPLY (
                SELECT TOP 1 activity_date as last_activity_date
                FROM activity_table
                WHERE user_id = u.user_id AND activity_date <= d.date
                ORDER BY activity_date DESC
            ) a

For now, I write it similar to the below, but it is a bit slow and I am afraid it'll only get slower.

with user_activity as (
    select distinct activity_date, user_id from activity_table
)
select
    d.date, u.user_id,
    max(u.activity_date) as last_activity_date
from date_table d
inner join user_activity u on u.activity_date <= d.date
where d.date between '2020-01-01' and current_date
group by 1, 2

Can someone suggest a good alternative for my needs or for CROSS APPLY / LATERAL JOIN.

Answer 1

As you are seeing cross joining and inequality joining will slow down as you data grows and are generally not the approach you want in Redshift. This is due to the data size increase that comes with this type of action when applied to large data.tables that are typical in Redshift.

You want to use window functions to perform this type of analysis. But you will need to step back and rethink how you will structure the SQL. A MAX(activity_date) window function, partitioned by user_id and ordered by date and with a frame clause of all preceding rows, will find the most recent activity to any activity.

Now this will produce only rows for user_ids and dates that exist in the data.table and it looks like you want 1 row for each date for each user_id, right? To do this you need to UNION in a frame of data that has 1 row for each date for each user_id ahead of the window function. You will need NULLs in for the other columns so that the data widths match. You will also want the dates in a separate column from activity_date. Now all dates for all user ids will be in the source and the window function will give you the result you want.

You also ask 'how is this better than the joins?' Well in the joins you are replicating all the data records by the number of dates which can get really big. In this approach you just have the original data records plus one row per user_id per date (which is the size of your output) and as the number of records per user_id grows this approach doesn't.

——— Request to modify asker's code per comments made to their approach ———

Your code is definitely on the right track as you have removed the massive inequality join of your original. I made 2 comments about it. The first is that I believe you need GROUP BY user_id, date to prevent multiple rows per user_id per date that would result if there are records for the same user_id on a single date with differing activity_types. This is a simple oversight.

The second is to state that I intended for you to use UNION ALL, not LEFT JOIN, in combining the actual data and the user_id/date framework. Your approach works fine but I have found that unioning with very large amounts of data is generally faster than joining but you do need to make sure the columns match up. Either way we end up with a data segment with 3 columns - 2 date columns, one with NULLs for framework rows, and 1 user_id. Your approach is fine and the difference in performance is likely very small unless you have huge tables.

Since you asked for a rewrite, here it is with both changes. (NOTE: my laptop is in the shop so I don't have ready access to Redshift at the moment and this SQL is untested. If the intent is not clear from this and you need me to debug it will be delayed by a few days. I'm keeping your setup methods and SQL structure.)

with date_table as (
    select '2000-01-01'::date as date
    union all
    select '2000-01-02'::date
    union all
    select '2000-01-03'::date
    union all
    select '2000-01-04'::date
    union all
    select '2000-01-05'::date
    union all
    select '2000-01-06'::date
),
users as (
    select 1 as user_id
    union all
    select 2
    union all
    select 3
),
user_activity as (
    select 1 as user_id, '2000-01-01'::date as activity_date
    union all
    select 1 as user_id, '2000-01-04'::date as activity_date
    union all
    select 3 as user_id, '2000-01-03'::date as activity_date
    union all
    select 1 as user_id, '2000-01-05'::date as activity_date
    union all
    select 1 as user_id, '2000-01-06'::date as activity_date
),
user_dates as (
    select d.date, u.user_id
    from date_table d
    cross join users u
),
user_date_activity as (
    select cal_date, user_id,
        lag(max(activity_date), 1) ignore nulls over (partition by user_id order by date) as last_activity_date
    from (
        Select user_id, date as cal_date, NULL as activity_date from user_dates
        Union all
        Select user_id, activity_date as cal_date, activity_date from user_activity 
    )
    Group by user_id, cal_date
)
select * from user_date_activity
order by user_id, cal_date```

Answer 2

This was my query based on Bill's answer.

with date_table as (
    select '2000-01-01'::date as date
    union all
    select '2000-01-02'::date
    union all
    select '2000-01-03'::date
    union all
    select '2000-01-04'::date
    union all
    select '2000-01-05'::date
    union all
    select '2000-01-06'::date
),
users as (
    select 1 as user_id
    union all
    select 2
    union all
    select 3
),
user_activity as (
    select 1 as user_id, '2000-01-01'::date as activity_date
    union all
    select 1 as user_id, '2000-01-04'::date as activity_date
    union all
    select 3 as user_id, '2000-01-03'::date as activity_date
    union all
    select 1 as user_id, '2000-01-05'::date as activity_date
    union all
    select 1 as user_id, '2000-01-06'::date as activity_date
),
user_dates as (
    select d.date, u.user_id
    from date_table d
    cross join users u
),
user_date_activity as (
    select ud.date, ud.user_id,
        lag(ua.activity_date, 1) ignore nulls over (partition by ud.user_id order by ud.date) as last_activity_date
    from user_dates ud
    left join user_activity ua on ud.date = ua.activity_date and ud.user_id = ua.user_id
)
select * from user_date_activity
order by user_id, date