SQLite - Create dummy variable vector/string from multiple columns - sqlite

Question

I have some data that looks like this:
UserID Category
------ --------
1 a
1 b
2 c
3 b
3 a
3 c
A I'd like to binary-encode this grouped by UserID: three different values exist in Category, so a binary encoding would be something like:
UserID encoding
------ --------
1 "1, 1, 0"
2 "0, 0, 1"
3 "1, 1, 1"
i.e., all three values are present for UserID = 3, so the corresponding vector is "1, 1, 1".
Is there a way to do this without doing a bunch of CASE WHEN statements? There may be dozens of possible values in Category

Answer 1

Cross join the distinct users to distinct categories and left join to the table.
Then use GROUP_CONCAT() window function which supports an ORDER BY clause, to collect the 0s and 1s:
WITH
users AS (SELECT DISTINCT UserID FROM tablename),
categories AS (
SELECT DISTINCT Category, DENSE_RANK() OVER (ORDER BY Category) rn
FROM tablename
),
cte AS (
SELECT u.UserID, c.rn,
'"' || GROUP_CONCAT(t.UserID IS NOT NULL)
OVER (PARTITION BY u.UserID ORDER BY c.rn) || '"' encoding
FROM users u CROSS JOIN categories c
LEFT JOIN tablename t
ON t.UserID = u.UserID AND t.Category = c.Category
)
SELECT DISTINCT userID,
FIRST_VALUE(encoding) OVER (PARTITION BY UserID ORDER BY rn DESC) encoding
FROM cte
ORDER BY userID
This will work for any number of categories.
See the demo.
Results:
UserID
encoding
1
"1,1,0"
2
"0,0,1"
3
"1,1,1"

Answer 2

First create an encoding table to explicit establish order of categories in the bitmap:
create table e (Category int, Encoding int);
insert into e values ('a', 1), ('b', 2), ('c', 4);
First generate a list of users u (cross) joined with the encoding table e to get a fully populated (UserId, Category, Encoding) table. Then left join the fully populated table with the user supplied data t. The right hand side t can now be used to drive if we need to set a bit or not:
select
u.UserId,
'"' ||
group_concat(case when t.UserId is null then 0 else 1 end, ', ')
|| '"' 'encoding'
from
(select distinct UserID from t) u
join e
left natural join t
group by 1
order by e.Encoding
and it gives the expected result:
1|"1, 1, 0"
2|"0, 0, 1"
3|"1, 1, 1"