1

I have a PostgreSQL table with the following schema and data:

CREATE TABLE IF NOT EXISTS T(
     id uuid PRIMARY KEY,
     username varchar(15),
     person varchar(10),
     tweets int,
     followers int,
     following int,
     likes int,
     created_at date) 
;





  id
  username
  person
  tweets
  followers
  following
  likes
  created_at






  3fa34100-d688-4051-a687-ec49d05e7212
  renok
  null
  110
  6
  0
  0
  2020-10-10




  bab9ceb9-2770-49ea-8489-77e5d763a223
  Lydia_C
  test user2
  515
  1301
  1852
  1677
  2020-10-10




  4649077a-9188-4821-a1ec-3b38608ea44a
  Kingston_Sav
  null
  2730
  1087
  1082
  1339
  2020-10-10




  eef80836-e140-4adc-9598-8b612ab1825b
  TP_s
  null
  1835
  998
  956
  1832
  2020-10-10




  fd3ff8c7-0994-40b6-abe0-915368ab9ae5
  DKSnr4
  null
  580
  268
  705
  703
  2020-10-10




  3fa34100-d688-4051-a687-ec49d05e7312
  renok
  null
  119
  6
  0
  0
  2020-10-12




  bab9ceb9-2770-49ea-8489-77e5d763a224
  Lydia_C
  test user2
  516
  1301
  1852
  1687
  2020-10-12




  4649077a-9188-4821-a1ec-3b38608ea44B
  Kingston_Sav
  null
  2737
  1090
  1084
  1342
  2020-10-12




  eef80836-e140-4adc-9598-8b612ae1835c
  TP_s
  null
  1833
  998
  957
  1837
  2020-10-12




  fd3ff8c7-0994-40b6-abe0-915368ab7ab5
  DKSnr4
  null
  570
  268
  700
  703
  2020-10-12

I intend to get the biggest difference between the most recent date and the next most recent date for each unique username and the find the username with the largest margin (difference) for example..In the above table the most recent date is 2020-10-12 and the next most recent date is 2020-10-10.

So I want to get something like this

 id                                   | username     | person | tweets | followers | following | likes | created_at | prev_followers | gain
 :----------------------------------- | :----------- | :----- | -----: | --------: | --------: | ----: | :--------- | -------------: | ---:
 4649077a-9188-4821-a1ec-3b38608ea44a | Kingston_Sav | null   |   2737 |      1090 |      1084 |  1342 | 2020-10-12 |           1087 |    3
basebandit
  • 13
  • 4

2 Answers2

0
WITH 
cte1 AS ( SELECT DISTINCT created_at
          FROM t
          ORDER BY 1 DESC LIMIT 2 ),
cte2 AS ( SELECT src.*, 
                 LEAD(src.followers) OVER (PARTITION BY src.id 
                                           ORDER BY src.created_at DESC) prev_followers,
                 src.followers - LEAD(src.followers) OVER (PARTITION BY src.id 
                                                           ORDER BY src.created_at DESC) gain
          FROM t src
          JOIN cte1 ON src.created_at = cte1.created_at )
SELECT *
FROM cte2
WHERE gain IS NOT NULL
ORDER BY gain DESC LIMIT 1;

fiddle

whenever there are two records with the same difference margin..It always shows one instead of two.

WITH 
cte1 AS ( SELECT DISTINCT created_at
          FROM t
          ORDER BY 1 DESC LIMIT 2 ),
cte2 AS ( SELECT src.*, 
                 LEAD(src.followers) OVER (PARTITION BY src.id 
                                           ORDER BY src.created_at DESC) prev_followers,
                 src.followers - LEAD(src.followers) OVER (PARTITION BY src.id 
                                                           ORDER BY src.created_at DESC) gain
          FROM t src
          JOIN cte1 ON src.created_at = cte1.created_at ),
cte3 AS ( SELECT *, RANK() OVER (ORDER BY gain DESC) rnk
          FROM cte2
          WHERE gain IS NOT NULL )
SELECT *
FROM cte3
WHERE rnk = 1;

fiddle

Akina
  • 20,750
  • 2
  • 20
  • 22
0

Many ways lead to Rome. The below should be a good one (fast and flexible) to "find the username with the largest margin"

Assuming all involved columns are defined NOT NULL. And each username can only have one entry per day. Else you have to do more.

WITH cte AS (
   SELECT *, dense_rank() OVER (ORDER BY created_at DESC) AS rnk
   FROM   tbl
   )
SELECT d1.*
     , d2.followers AS prev_followers
     , d1.followers - d2.followers AS gain
FROM  (SELECT * FROM cte WHERE rnk = 1) d1
JOIN  (SELECT * FROM cte WHERE rnk = 2) d2 USING (username)
ORDER  BY gain DESC
        , d1.followers, username  -- added tiebreaker
LIMIT  1;

The CTE named cte attaches rank numbers with dense_rank() (not rank(), not row_number()). Then join the latest day (rnk = 1) with the one before (rnk = 2) and calculate the gain. Obviously, users must have entries for both days to qualify. Finally order by the gain and take the first row.

Note the added ORDER BY expressions to try and break possible ties: there can be multiple users with the same gain, so you have to define how to deal with that. One way is to add tiebreakes. In my example, a user with a smaller absolute number of followers is preferred (higher relative gain), and the alphabetically first wins if that's still ambiguous.

Or you return all "winners":

Again, many ways ... Postgres 13 added the standard SQL clause WITH TIES for that purpose exactly:

WITH cte AS (
   SELECT *, dense_rank() OVER (ORDER BY created_at DESC) AS rnk
   FROM   tbl
   )
SELECT d1.*
     , d2.followers AS prev_followers
     , d1.followers - d2.followers AS gain
FROM  (SELECT * FROM cte WHERE rnk = 1) d1
JOIN  (SELECT * FROM cte WHERE rnk = 2) d2 USING (username)
ORDER  BY gain DESC
FETCH  FIRST 1 ROWS WITH TIES;

db<>fiddle here

Detailed explanation for WITH TIES:

Erwin Brandstetter
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  • 633