SQL: How to count instances of Column A in Column B - sql

I have a table with 2 columns A, and B that represent a connection graph between the two.
A B
1 3
2 5
4 2
3 5
2 3
I need to find how many instances of column A occur in column B (including 0)
So for the example above I would need the result set
A OccurencesInB
1 0
2 1
3 2
4 0
The best I have so far is
SELECT B, COUNT(*) AS TABLE_COUNT
FROM TABLE
GROUP BY B
ORDER BY TABLE_COUNT DESC
This does not find the instances of A that do not occur in B, which is crucial.
Any assistance will be greatly appreciated!


Use a correlated sub-query:
SELECT A,
TABLE_COUNT = (SELECT COUNT(*)
FROM TableName t2
WHERE t2.B = t1.A)
FROM TableName t1
GROUP BY A
ORDER BY TABLE_COUNT DESC, A
Result:
A TABLE_COUNT
3 2
2 1
1 0
4 0

Related

SQL: Select rows with bigger than max in another table

Table "TBL1":
a
b
1
2
1
3
2
3
Table "TBL2":
a
b
1
2
1
3
I tried this:
SELECT a, b
FROM TBL1 Where a > MAX (tbl2.a);
Obviously it didn't work. Ideally the solution would work in sqlite.

You need a subquery for that:
SELECT a, b
FROM TBL1 Where a > (SELECT MAX(a) FROM tbl2);
Check the demo here.

Select field 1 Where Field 2 ='value1' And field 2 ='value2' when Field 1 is Same

I am currently having troubles with filtering my SQL records. I need something like what it results in the following concept: Table is
A B
1 1
1 3
2 1
2 2
2 3
2 4
3 1
3 2
I want to select value of A , where B=1 and B=2 And B=3 when same A .... result is
A
2
Please help

You can use aggregation:
select a
from mytable
where b in (1, 2, 3)
group by a
having count(*) = 3
This assumes no duplicates in the table - else, you need to change the having clause to:
having count(distinct b) = 3

SQL query to find the entries corresponding to the maximum count of each type

I have a table X in Postgres with the following entries
A B C
2 3 1
3 3 1
0 4 1
1 4 1
2 4 1
3 4 1
0 5 1
1 5 1
2 5 1
3 5 1
0 2 2
1 2 3
I would like to find out the entries having maximum of Column C for every kind of A and B i.e (group by B) with the most efficient query possible and return corresponding A and B.
Expected Output:
A B C
1 2 3
2 3 1
0 4 1
0 5 1
Please help me with this problem . Thank you

demo: db<>fiddle
Using DISTINCT ON:
SELECT DISTINCT ON (B)
A, B, C
FROM
my_table
ORDER BY B, C DESC, A
DISTINCT ON gives you exactly the first row for an ordered group. In this case B is grouped.
After ordering B (which is necessary): We first order the maximum C (with DESC) to the top of each group. Then (if there are tied MAX(C) values) we order the A to get the minimum A to the top.

Seems like it is a greatest n per group problem:
WITH cte AS (
SELECT *, RANK() OVER (PARTITION BY B ORDER BY C DESC, A ASC) AS rnk
FROM t
)
SELECT *
FROM cte
WHERE rnk = 1
You're not clear which A needs to be considered, the above returns the row with smallest A.

itseems to me you need max()
select A,B, max(c) from table_name
group by A,B

this will work:
select * from (SELECT t.*,
rank() OVER (PARTITION BY A,B order by C) rank
FROM tablename t)
where rank=1 ;

SQL combining of a COUNT with a WHERE in single query

Here is the data, call it table T
A B
-- --
1 14
2 15
3 16
4 1
4 3
4 6
4 9
4 12
4 15
I would like to get the value of A that has only one value and a B value of 15.
There are two rows where B=15 but there are 6 rows where A=4 and only one row where A=2.
So the correct SQL should return me the 2.
I have tried this but it returns both rows.
select A from T group by A,B having Count(A) = 1 and B = 15
This similarly fails:
select A from T where B = 15 group by A having count( A ) = 1

Try this:
select A
from T
group by A
having Count(A) = 1 and Max(B) = 15;
Your problem seems to be that you are grouping by both columns. You only want to group by A.
Admittedly, your query has group by A, T, but I think that is a typo, based on the described behavior.

You can check the count of B after grouping by A.
select A
from T
group by A
having Count(B) = 1 and max(B) = 15

Oracle: find duplicate rows in select query

My SQL query returns results with 4 columns "A", "B", "C", "D".
Suppose the results are:
A B C D
1 1 1 1
1 1 1 2
2 2 2 1
Is it possible to get the count of duplicate rows with columns "A", "B", "C" in each row.
e.g. the expected result is:
A B C D cnt
1 1 1 1 2
1 1 1 2 2
2 2 2 1 1
I tried using count(*) over. But it returns me the total number of rows returned by the query.
Another information is that in example I have mentioned only 3 columns based on which I need to check the count. But my actual query has such 8 columns. And number of rows in database are huge. So I think group by will not be a feasible option here.
Any hint is appreciable.
Thanks.

Maybe too late, but probably the count over as analytic function (aka window function) within oracle helps you. When I understand your request correctly, this should solve your problem :
create table sne_test(a number(1)
,b number(1)
,c number(1)
,d number(1)
,e number(1)
,f number(1));
insert into sne_test values(1,1,1,1,1,1);
insert into sne_test values(1,1,2,1,1,1);
insert into sne_test values(1,1,2,4,1,1);
insert into sne_test values(1,1,2,5,1,1);
insert into sne_test values(1,2,1,1,3,1);
insert into sne_test values(1,2,1,2,1,2);
insert into sne_test values(2,1,1,1,1,1);
commit;
SELECT a,b,c,d,e,f,
count(*) over (PARTITION BY a,b,c)
FROM sne_test;
A B C D E F AMOUNT
-- -- -- -- -- -- ------
1 1 1 1 1 1 1
1 1 2 4 1 1 3
1 1 2 1 1 1 3
1 1 2 5 1 1 3
1 2 1 1 3 1 2
1 2 1 2 1 2 2
2 1 1 1 1 1 1

To find duplicates you must group the data based on key column
select
count(*)
,empno
from
emp
group by
empno
having
count(*) > 1;
This allows you to aggregate by empno even when multiple records exist for each category (more than one).

You have to use a subquery where you get the count of rows, grouped by A, B and C. And then you join this subquery again with your table (or with your query), like this:
select your_table.A, your_table.B, your_table.C, your_table.D, cnt
from
your_table inner join
(SELECT A, B, C, count(*) as cnt
FROM your_table
GROUP BY A, B, C) t
on t.A = your_table.A
and t.B = your_table.B
and t.C = your_table.C