SQL - Return count of consecutive days where value was unchanged - sql

Question

I have a table like
date
ticker
Action
'2022-03-01'
AAPL
BUY
'2022-03-02'
AAPL
SELL.
'2022-03-03'
AAPL
BUY.
'2022-03-01'
CMG
SELL.
'2022-03-02'
CMG
HOLD.
'2022-03-03'
CMG
HOLD.
'2022-03-01'
GPS
SELL.
'2022-03-02'
GPS
SELL.
'2022-03-03'
GPS
SELL.
I want to do a group by ticker then count all the times that Actions have sequentially been the value that they are as of the last date, here it's 2022-03-03. ie for this example table it'd be like;
ticker
NumSequentialDaysAction
AAPL
0
CMG
1
GPS
2
Fine to pass in 2022-03-03 as a value, don't need to figure that out on the fly.
Tried something like this
---Table Creation---
CREATE TABLE UserTable
([Date] DATETIME2, [Ticker] varchar(5), [Action] varchar(5))
;
INSERT INTO UserTable
([Date], [Ticker], [Action])
VALUES
('2022-03-01' , 'AAPL' , 'BUY'),
('2022-03-02' , 'AAPL' , 'SELL'),
('2022-03-03' , 'AAPL' , 'BUY'),
('2022-03-01' , 'CMG' , 'SELL'),
('2022-03-02' , 'CMG' , 'HOLD'),
('2022-03-03' , 'CMG' , 'HOLD'),
('2022-03-01' , 'GPS' , 'SELL'),
('2022-03-02' , 'GPS' , 'SELL'),
('2022-03-03' , 'GPS' , 'SELL')
;
---Attempted Solution---
I'm thinking that I need to do a sub query to get the last value and join on itself to get the matching values. Then apply a window function, ordered by date to see that the proceeding value is sequential.
WITH CTE AS (SELECT Date, Ticker, Action,
ROW_NUMBER() OVER (PARTITION BY Ticker, Action ORDER BY Date) as row_num
FROM UserTable)
SELECT Ticker, COUNT(DISTINCT Date) as count_of_days
FROM CTE
WHERE row_num = 1
GROUP BY Ticker;
WITH CTE AS (SELECT Date, Ticker, Action,
DENSE_RANK() OVER (PARTITION BY Ticker ORDER BY Action,Date) as rank
FROM table)
SELECT Ticker, COUNT(DISTINCT Date) as count_of_days
FROM CTE
WHERE rank = 1
GROUP BY Ticker;

Answer 1

You can do this with the help of the LEAD function like so. You didn't specify which RDBMS you're using. This solution works in PostgreSQL:
WITH "withSequential" AS (
SELECT
ticker,
(LEAD("Action") OVER (PARTITION BY ticker ORDER BY date ASC) = "Action") AS "nextDayIsSameAction"
FROM UserTable
)
SELECT
ticker,
SUM(
CASE
WHEN "nextDayIsSameAction" IS TRUE THEN 1
ELSE 0
END
) AS "NumSequentialDaysAction"
FROM "withSequential"
GROUP BY ticker

Answer 2

Here is a way to do this using gaps and islands solution.
Thanks for sharing the create and insert scripts, which helps to build the solution quickly.
dbfiddle link.
https://dbfiddle.uk/rZLDTrNR
with data
as (
select date
,ticker
,action
,case when lag(action) over(partition by ticker order by date) <> action then
1
else 0
end as marker
from usertable
)
,interim_data
as (
select *
,sum(marker) over(partition by ticker order by date) as grp_val
from data
)
,interim_data2
as (
select *
,count(*) over(partition by ticker,grp_val) as NumSequentialDaysAction
from interim_data
)
select ticker,NumSequentialDaysAction
from interim_data2
where date='2022-03-03'

Answer 3

Another option, you could use the difference between two row_numbers approach as the following:
select [Ticker], count(*)-1 NumSequentialDaysAction -- you could use (distinct) to remove duplicate rows
from
(
select *,
row_number() over (partition by [Ticker] order by [Date]) -
row_number() over (partition by [Ticker], [Action] order by [Date]) grp
from UserTable
where [date] <= '2022-03-03'
) RN_Groups
/* get only rows where [Action] = last date [Action] */
where [Action] = (select top 1 [Action] from UserTable T
where T.[Ticker] = RN_Groups.[Ticker] and [date] <= '2022-03-03'
order by [Date] desc)
group by [Ticker], [Action], grp
See demo