I am looking to explode a row into multiple rows based on a column[integer] value, I am trying to do this using presto
Below is an example
id
count
1
5
2
2
expected output
id
count
1
5
1
5
1
5
1
5
1
5
2
2
2
2
in the above example, id 1 has to be repeated 5 times and id 2 has to be repeated 2 times based on the count.
Based on my experience, presto doesnt support recursive CTE.
Any help would be appreciated.
Thanks
You could make the count into array with REPEAT and then CROSS JOIN.
Your input:
CREATE TABLE test AS
SELECT id, count
FROM (
VALUES
(1, 5),
(2, 2)
) AS x (id, count)
Then:
SELECT id, t.count
FROM test
CROSS JOIN UNNEST(repeat(count, count)) AS t (count)
You don't need recursive CTE's here, you can use sequence or repeat to generate an array for corresponding length and then flatten it with unnest:
-- sample data
WITH dataset (id, count) AS (
VALUES (1, 5),
(2, 2)
)
-- query
select id, count
from dataset,
unnest (repeat(id, count)) as t (ignore)
-- unnest (sequence(0, count - 1)) as t (ignore)
Output:
id
count
1
5
1
5
1
5
1
5
1
5
2
2
2
2
Related
I am trying to process data within the same table.
Input:
Table
id sort value
1 1 1
2 1 8
3 2 0
4 1 2
What I want to achieve is obtain for each id, the first encountered value for all value equal to its sort, and this ordered by id.
Output
Table
id sort value new
1 1 1 1
2 1 8 1
3 2 0 0
4 1 2 1
I tried to self join the table, but I constantly get relation not found. I tried with a case statement but I don't see how can I connect to the same table, I get the same error, relation not found.
The beauty of SQL is that many requirements (yours included) can be verbosely described in very similar way they are finally coded:
with t(id, sort, value ) as (values
(1, 1, 1),
(2, 1, 8),
(3, 2, 0),
(4, 1, 2)
)
select t.*
, first_value(value) over (partition by sort order by id) as "new"
from t
order by id
id
sort
value
new
1
1
1
1
2
1
8
1
3
2
0
0
4
1
2
1
fiddle
For example, lets say I have
id
values
1
10
1
12
1
10
2
2
2
5
2
4
then i would want sql to return
id
values
1
32
2
11
This is very basic sql.
select id, sum(values) as values
from foo
group by id
Select ID,sum(values)
From table
Group by ID;
As an example, I have the following table:
T | S
------
1 | 5
1 | 6
1 | 7
2 | 6
2 | 7
3 | 6
Query: array [1,2]
I want to select all values in S that have the value 1 AND 2 in the T Column.
So in the above example I should get as a result (6,7) because only 6 and 7 have for column T the values 1 and 2.
But i do not want to have 5 in my results as 5 does not have 2 in the T column.
How would I do this in sequelize?
how do i make (1,2) to be used as an array?
Either you insert the array joined as comma-separated literal into the query text (variant 1) or you join the array into one string literal and transfer it iinto the query as a parameter (variant 2).
Variant 1
SELECT s
FROM sourcetable
WHERE t IN (1,2) -- separate filter values
GROUP BY s
HAVING COUNT(DISTINCT t) = 2 -- unique values count
Variant 2
SELECT s
FROM sourcetable
WHERE FIND_IN_SET(t, '1,2') -- separate filter values
GROUP BY s
HAVING COUNT(DISTINCT t) = 2 -- unique values count
If (s,t) is unique then DISTINCT keyword may be removed.
Using SQL here. Trying to select all rows where the column value is unique within that specific partition.
Have tried:
select *
from dataTable
where value in ( select value
from dataTable
group by tv_id, value
having count(*) > 1)
but it returns the full table-- i think the issue is that the values for many of tv_ids are identical and overlap.
What I have:
tv_id value
1 1
1 2
1 2
1 3
2 1
2 1
2 2
2 3
2 4
3 1
3 1
3 2
What I want:
tv_id value
1 2
1 2
2 1
2 1
3 1
3 1
I have a bunch of tv_ids and essentially, I only want the rows where the value is not unique within each tv_id.
Ex: I don't want tv_id, value: 3, 2 because it is the only combination in the data.
Thanks in advance!
Maybe something like this does the trick
Oracle Option
I include this oracle version because it enables you to understand better what are you querying.
select tv_id, value
from dataTable
where (tv_id, value) in (
select tv_id, value
from dataTable
group by tv_id, value
having count(1) > 1
)
SQL
But this is a standard sql version that will work with almost any database engine
select tv_id, value
from dataTable d1
join (
select tv_id, value
from dataTable
group by tv_id, value
having count(1) > 1
) d2
on d1.tv_id=d2.tv_id
and d1.value=d2.value
You need to query the same table twice because the group by makes a distinct in your data, so you won't retrieve duplicated rows as you show in your expected output.
I have an input:
id
1
2
3
4
5
6
7
8
9
10
I want get even and odd columns separately by columns in specified output like this
id col
1 2
3 4
5 6
7 8
9 10
here id and col are separate columns id contains the odd number and col contains the even number for specified input
SELECT MIN(id) as id, MAX(id) as col
FROM YourTable
GROUP BY FLOOR((id+1)/2)
For IDs 1 and 2, (id+1)/2 are 2/2 = 1 and 3/2 = 1.5, respectively, and FLOOR then returns 1 for both of them. Similarly, for 3 and 4, this is 2, and so on. So it groups all the input rows into pairs based on this formula. Then it uses MIN and MAX within each group to get the lower and higher IDs of the pairs.
Joined the table on itself
select *
from yourTable tA
left join yourTable tb on tA.id = (tB.id - 1)
where tA.id % 2 <> 0
If you use SQL you can try:
SELECT CASE WHEN column % 2 = 1
THEN column
ELSE null
END AS odds,
CASE WHEN column % 2 = 2
THEN column
ELSE null
END AS even
FROM yourtable
but not exactl as you ask
To show odd:
Select * from MEN where (RowID % 2) = 1
To show even:
Select * from MEN where (RowID % 2) = 0
Now, just join those two result sets and that's it.
Source