问题
I have two tables, apps and reviews (simplified for the sake of discussion):
apps table
id int
reviews table
id int
review_date date
app_id int (foreign key that points to apps)
2 questions:
1. How can I write a query / function to answer the following question?:
Given a series of dates from the earliest reviews.review_date to the latest reviews.review_date (incrementing by a day), for each date, D, which apps had the most reviews if the app's earliest review was on or later than D?
I think I know how to write a query if given an explicit date:
SELECT
apps.id,
count(reviews.*)
FROM
reviews
INNER JOIN apps ON apps.id = reviews.app_id
group by
1
having
min(reviews.review_date) >= '2020-01-01'
order by 2 desc
limit 10;
But I don't know how to query this dynamically given the desired date series and compile all this information in a single view.
2. What's the best way to model this data?
It would be nice to have the # of reviews at the time for each date as well as the app_id. As of now I'm thinking something that might look like:
... 2020-01-01_app_id | 2020-01-01_review_count | 2020-01-02_app_id | 2020-01-02_review_count ...
But I'm wondering if there's a better way to do this. Stitching the data together also seems like a challenge.
回答1:
I think this is what you are looking for:
Postgres 13 or newer
WITH cte AS ( -- MATERIALIZED
SELECT app_id, min(review_date) AS earliest_review, count(*)::int AS total_ct
FROM reviews
GROUP BY 1
)
SELECT *
FROM (
SELECT generate_series(min(review_date)
, max(review_date)
, '1 day')::date
FROM reviews
) d(review_window_start)
LEFT JOIN LATERAL (
SELECT total_ct, array_agg(app_id) AS apps
FROM (
SELECT app_id, total_ct
FROM cte c
WHERE c.earliest_review >= d.review_window_start
ORDER BY total_ct DESC
FETCH FIRST 1 ROWS WITH TIES -- new & hot
) sub
GROUP BY 1
) a ON true;
WITH TIES makes it a bit cheaper. Added in Postgres 13 (currently beta). See:
- Greater than or equal to ALL() and equal to MAX() speed
Postgres 12 or older
WITH cte AS ( -- MATERIALIZED
SELECT app_id, min(review_date) AS earliest_review, count(*)::int AS total_ct
FROM reviews
GROUP BY 1
)
SELECT *
FROM (
SELECT generate_series(min(review_date)
, max(review_date)
, '1 day')::date
FROM reviews
) d(review_window_start)
LEFT JOIN LATERAL (
SELECT total_ct, array_agg(app_id) AS apps
FROM (
SELECT total_ct, app_id
, rank() OVER (ORDER BY total_ct DESC) AS rnk
FROM cte c
WHERE c.earliest_review >= d.review_window_start
) sub
WHERE rnk = 1
GROUP BY 1
) a ON true;
db<>fiddle here
Same as above, but without WITH TIES.
We don't need to involve the table apps at all. The table reviews has all information we need.
The CTE cte computes earliest review & current total count per app. The CTE avoids repeated computation. Should help quite a bit.
It is always materialized before Postgres 12, and should be materialized automatically in Postgres 12 since it is used many times in the main query. Else you could add the keyword MATERIALIZED in Postgres 12 or later to force it. See:
- How to force evaluation of subquery before joining / pushing down to foreign server
The optimized generate_series() call produces the series of days from earliest to latest review. See:
- Generating time series between two dates in PostgreSQL
- Join a count query on a generate_series in postgres and also retrieve Null-values as "0"
Finally, the LEFT JOIN LATERAL you already discovered. But since multiple apps can tie for the most reviews, retrieve all winners, which can be 0 - n apps. The query aggregates all daily winners into an array, so we get a single result row per review_window_start. Alternatively, define tiebreaker(s) to get at most one winner. See:
- What is the difference between LATERAL and a subquery in PostgreSQL?
回答2:
If you are looking for hints, then here are a few:
- Are you aware of
generate_series()and how to use it to compose a table of dates given a start and end date? If not, then there are plenty of examples on this site. - To answer this question for any given date, you need to have only two measures for each app, and only one of these is used to compare an app against other apps. Your query in part 1 shows that you know what these two measures are.
- Hints 1 and 2 should be enough to get this done. The only thing I can add is for you not to worry about making the database do "too much work." That is what it is there to do. If it does not do it quickly enough, then you can think about optimizations, but before you get to that step, concentrate on getting the answer that you want.
Please comment if you need further clarification on this.
回答3:
The missing piece for me was lateral join. I can accomplish just about what I want using the following:
select
review_windows.review_window_start,
id,
review_total,
earliest_review
from
(
select
date_trunc('day', review_windows.review_windows) :: date as review_window_start
from
generate_series(
(
SELECT
min(reviews.review_date)
FROM
reviews
),
(
SELECT
max(reviews.review_date)
FROM
reviews
),
'1 year'
) review_windows
order by
1 desc
) review_windows
left join lateral (
SELECT
apps.id,
count(reviews.*) as review_total,
min(reviews.review_date) as earliest_review
FROM
reviews
INNER JOIN apps ON apps.id = reviews.app_id
where
reviews.review_date >= review_windows.review_window_start
group by
1
having
min(reviews.review_date) >= review_windows.review_window_start
order by
2 desc,
3 desc
limit
2
) apps_most_reviews on true;
来源：https://stackoverflow.com/questions/63310905/get-apps-with-the-highest-review-count-since-a-dynamic-series-of-days