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How to create lines from ordered groupbyed points in BigQuery?
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I have a BigQuery table:
create or replace table `project.table.mock` as (
select 1 as col0, 'a' as col1, 'x' as col2
union all
select 2 as col0, 'a' as col1, 'y' as col2
union all
select 4 as col0, 'b' as col1, 'z' as col2
union all
select 8 as col0, 'b' as col1, 'X' as col2
union all
select 7 as col0, 'b' as col1, 'Y' as col2
)
Visualization:
I would like to group by column col1, and array_agg the results from col2. I would like to have the elements appearing in each array to be sorted by col0.
I am now at:
select array_agg(col2) as col1arrays from `project.table.mock` group by col1;
which gives me:
The desired output in the second row would be [z, Y, X] (as the row where z appears in col2 has 4 in col0, the row where Y appears in col2 has 7 in col0 and the row where X appears in col2 has 8 in col0, and 4 < 7 < 8.
How can I achieve ordering within array_agg, as described above, in BigQuery?
You can add ORDER BY clause in ARRAY_AGG() function.
SELECT ARRAY_AGG(col2 ORDER BY col1 ASC) AS col1arrays
FROM `project.table.mock`
GROUP BY col1;
https://cloud.google.com/bigquery/docs/reference/standard-sql/aggregate_functions#array_agg
WITH mock as (
select 1 as col0, 'a' as col1, 'x' as col2
union all
select 2 as col0, 'a' as col1, 'y' as col2
union all
select 4 as col0, 'b' as col1, 'z' as col2
union all
select 8 as col0, 'b' as col1, 'X' as col2
union all
select 7 as col0, 'b' as col1, 'Y' as col2
)
select array_agg(col2 ORDER BY col0) as col1arrays from mock group by col1;
output:
+------------+
| col1arrays |
+------------+
| [x,y] |
| [z,Y,X] |
+------------+
Related
I am having below dummy table
select * from (
select 'A' as col1, 'B' as col2 from dual
union
select 'B' as col1, 'A' as col2 from dual
union
select 'A' as col1, 'C' as col2 from dual
union
select 'C' as col1, 'A' as col2 from dual
union
select 'A' as col1, 'D' as col2 from dual
)a
which will give output as below
col1 col2
A B
A C
A D
B A
C A
I wants to find the distinct values from that table like below
col1 col2
A B
A C
A D
first row can be A B or B A same as second can be A C or C A
Is it possible??
We got the solution for above problem which is below
select distinct least(col1, col2), greatest(col1, col2)
from the_table;
but if there is more than 2 column, then i wouldn't work
Let us assume the below scenario
Input
col1 col2 col3
A B E
A C E
A D E
B A F
C A E
Output
col1 col2 col3
A B E
A D E
B A F
C A E
then what would be the possible solution ?
Here is one method:
select col1, col2
from t
where col1 <= col2
union all
select col1, col2
from t
where col1 > col2 and
not exists (select 1 from t t2 where t2.col1 = t.col2 and t2.col2 = t.col1);
The following will work for Oracle and Postgres:
select distinct least(col1, col2), greatest(col1, col2)
from the_table;
Online example: http://rextester.com/BZXC69735
select DISTINCT * from (
select 'A' as col1, 'B' as col2 from dual
union
select 'B' as col1, 'A' as col2 from dual
union
select 'A' as col1, 'C' as col2 from dual
union
select 'C' as col1, 'A' as col2 from dual
union
select 'A' as col1, 'D' as col2 from dual
)a
select col1, col2 from t where col1 <= col2
union
select col2, col1 from t where col1 > col2
WE have below data in oracle database -
col1 col2
Z1 A
Z1 B
Z2 A
Z2 C
Z3 A
Z4 D
I want count on column two in such a way that -
Ouput -
col2 count
A 3 (Z1,Z2,Z3)
B 0 (Dont count if A is already present for record)
C 0
D 1 (Z4)
Best Regards
You can use window function rank() to achieve this.
select col2, count(case when rn = 1 then 1 end) cnt from (
select t.*,
rank() over (partition by col1 order by case when col2 = 'A' then 1 else 2 end) rn
from table t
) group by col2;
The most general solution to your propositions where each key COL1 is counted only in the first occurrence of the key COL2 (in alphabetical order)
WITH tab AS
(
SELECT 'Z1' col1, 'A' col2 FROM dual UNION ALL
SELECT 'Z1' col1, 'B' col2 FROM dual UNION ALL
SELECT 'Z2' col1, 'A' col2 FROM dual UNION ALL
SELECT 'Z2' col1, 'C' col2 FROM dual UNION ALL
SELECT 'Z3' col1, 'A' col2 FROM dual UNION ALL
SELECT 'Z4' col1, 'D' col2 FROM dual
), tab2 as (
select COL1, COL2,
row_number() over (partition by COL1 order by COL2) as rn
from tab)
select COL1, COL2,
case when rn = 1 then 1 else 0 end is_valid
from tab2
order by 1,2
;
COL1 COL2 IS_VALID
---- ---- ----------
Z1 A 1
Z1 B 0
Z2 A 1
Z2 C 0
Z3 A 1
Z4 D 1
The rest is simple group by with a SUM on IS_VALID
select COL2, sum(is_valid) cnt from tab3 -- TAB3 is the above row source
group by COL2
order by 1
COL2 CNT
---- ----------
A 3
B 0
C 0
D 1
Thanks Guys. But I could do this way -
select count(case
when (LISTAGG(col2,'-') WITHIN GROUP (ORDER BY col2)) like '%A%' then 1
else null
end) A,
count(case
when (LISTAGG(col2,'-') WITHIN GROUP (ORDER BY col2)) = 'B' then 1
else null
end) B,
count(case
when (LISTAGG(col2,'-') WITHIN GROUP (ORDER BY col2)) = 'C' then 1
else null
end) C,
count(case
when (LISTAGG(col2,'-') WITHIN GROUP (ORDER BY col2)) = 'D' then 1
else null
end) D
from T
GROUP BY col1
Thanks for your replies
Assume your table name is table_name, One way to do it is using this:
WITH table_a AS
(
SELECT DISTINCT col1
FROM table_name
WHERE col2 = 'A'
)
SELECT col2,
SUM(CASE WHEN col1 IN (SELECT col1 FROM table_a)
THEN DECODE(col2, 'A', 1, 0)
ELSE 1 END
) count
FROM table_name
GROUP BY col2
ORDER BY col2;
Tested ok:
WITH table_name AS
(
SELECT 'Z1' col1, 'A' col2 FROM dual UNION ALL
SELECT 'Z1' col1, 'B' col2 FROM dual UNION ALL
SELECT 'Z2' col1, 'A' col2 FROM dual UNION ALL
SELECT 'Z2' col1, 'C' col2 FROM dual UNION ALL
SELECT 'Z3' col1, 'A' col2 FROM dual UNION ALL
--SELECT 'Z4' col1, 'B' col2 FROM dual UNION ALL
SELECT 'Z4' col1, 'D' col2 FROM dual
)
, table_a AS
(
SELECT DISTINCT col1
FROM table_name
WHERE col2 = 'A'
)
SELECT col2,
SUM(CASE WHEN col1 IN (SELECT col1 FROM table_a)
THEN DECODE(col2, 'A', 1, 0)
ELSE 1 END
) count
FROM table_name
GROUP BY col2
ORDER BY col2;
You want to count each record where either col2 is 'A' or no 'A' record exists for col1.
select
col2,
count(
case
when col2 = 'A' or col1 not in (select col1 from table_name where col2 = 'A') then 1
end) as cnt
from table_name
group by col2;
select col2, count(case when col2 = col3 then 'x' end) as ct
from ( select col2, min(col2) over (partition by col1) as col3
from your_table
)
group by col2
order by col2 -- if needed
;
Explanation:
There is an inner query (a.k.a. "subquery") which returns one row for each row in the original table. It returns col2 as is, and an additional (new) column, labeled col3. col3 is calculated as the "first" or min() value of col2 (in alphabetical order) for all the rows in the original table that have the same value in col1 as the current row does. This is a typical example of an analytic function; partition by col1 is similar to group by col1 but it returns all the rows in the group (all the original rows from the original table) instead of one row per group, as would an aggregate function.
To see what the inner query does by itself, select it and run it in your favorite front-end. You may add col1 to the select in the inner query - that will make what's going on in this query even clearer. You'll get the initial table, with one more column, col3, that shows the "min" col2 for each value of col1. I didn't include col1 in the subquery because I don't need it, but add it back to see what the subquery really does.
Then in the outer query I take the results from the inner query and I group by col2. For each col2 I count just how many times it is equal to the "min" value of col2 for the corresponding col1 value. That's what the case expression does in the count() function; when col2 is not equal to col3, then case returns null (by default) so the expression - and therefore the row - is not counted.
I should add that the query written this way assumes there are no duplicate (col1, col2) rows in the original table. If there are, then the inner subquery should select from a sub-subquery; line 3 of my code should be
from (select distinct col1, col2 from your_table)
Use the below script:
SELECT A.COL2, NVL(B.CNT, 0) AS CNT
FROM (SELECT DISTINCT COL2 FROM TET) A
LEFT JOIN (SELECT COL2, COUNT(COL2) AS CNT
FROM (SELECT SUBSTR(F, 1, INSTR(F, ',') - 1) AS COL2,
ROW_NUMBER() OVER(PARTITION BY SUBSTR(F, 1, INSTR(F, ',') - 1) ORDER BY SUBSTR(F, 1, INSTR(F, ',') - 1)) AS U
FROM (SELECT COL1,
LISTAGG(COL2, ',') WITHIN GROUP(ORDER BY COL2) || ',' AS F
FROM TET
GROUP BY COL1)) A
GROUP BY COL2) B
ON A.COL2 = B.COL2
ORDER BY A.COL2;
I know the basics in SQL programming and I know how to apply some tricks in SQL Server in order to get the result set, but I don't know all tricks in Oracle.
I have these columns:
col1 col2 col3
And I wrote this query
SELECT
col1, MAX(col3) AS mx3
FROM
myTable
GROUP BY
col1
And I need to get the value of col2 in the same row where I found the max value of col3, do you know some trick to solve this problem?
The easiest way to do this, IMHO, is not to use max, but the window function rank:
SELECT col1 , col2, col3
FROM (SELECT col1, col2, col3,
RANK() OVER (PARTITION BY col1 ORDER BY col3 DESC) rk
FROM myTable) t
WHERE rk = 1
BTW, the same syntax should also work for MS SQL-Server and most other modern databases, with MySQL being the notable exception.
A couple of different ways to do this:
In both cases I'm treating your initial query as either a common table expression or as an inline view and joining it back to the base table to get your added column. The trick here is that the INNER JOIN eliminates all the records not in your max query.
SELECT A.*,
FROM myTable A
INNER JOIN (SELECT col1 , MAX( col3 ) AS mx3 FROM myTable GROUP BY col1) B
on A.Col1=B.Col1
and B.mx3 = A.Col3
or
with CTE AS (SELECT col1 , MAX( col3 ) AS mx3 FROM myTable GROUP BY col1)
SELECT A.*
FROM MyTable A
INNER JOIN CTE
on A.col1 = B.Col1
and A.col3= cte.mx3
Here's an alternative that's just a slight extension of your existing group by query (ie. doesn't require querying the same table more than once):
with mytable as (select 1 col1, 1 col2, 1 col3 from dual union all
select 1 col1, 2 col2, 2 col3 from dual union all
select 1 col1, 1 col2, 3 col3 from dual union all
select 1 col1, 3 col2, 3 col3 from dual union all
select 2 col1, 10 col2, 1 col3 from dual union all
select 2 col1, 23 col2, 2 col3 from dual union all
select 2 col1, 12 col2, 2 col3 from dual)
SELECT
col1,
MAX(col2) keep (dense_rank first order by col3 desc) mx2,
MAX(col3) AS mx3
FROM
myTable
GROUP BY
col1;
COL1 MX2 MX3
---------- ---------- ----------
1 3 3
2 23 2
I have the following table;
column 1 column 2 column 3
1 2 X
1 2 X
1 2 Y
1 3 Z
1 3 X
I need to write an SQL query to get the output as;
1 2 X (because X is the maximum occurrence)
1 3 Z or X(because number of occurrence of Z or X is same)
How do i do this ?
I think i have a solution for you, try this script using the functions RANK(), ROW_NUMBER() & DENSE_RANK(), you choose the function that fits with your needs :
with temp as (
select 1 as col1, 2 AS col2, 'X' as col3 union all
select 1 as col1, 2 AS col2, 'Y' as col3 union all
select 1 as col1, 2 AS col2, 'X' as col3 union all
select 1 as col1, 3 AS col2, 'Z' as col3 union all
select 1 as col1, 3 AS col2, 'T' as col3 union all
select 1 as col1, 3 AS col2, 'Y' as col3 union all
select 1 as col1, 3 AS col2, 'Y' as col3 union all
select 1 as col1, 4 AS col2, 'Y' as col3 union all
select 1 as col1, 4 AS col2, 'W' as col3)
,temp2 AS (
select
col1
,col2
,col3
,COUNT(1) nb_occurence
,RANK() OVER(PARTITION BY col1,col2 ORDER BY COUNT(1) DESC) Ordre_RANK
,ROW_NUMBER() OVER(PARTITION BY col1,col2 ORDER BY COUNT(1) DESC) Ordre_ROW_NUMBER
,DENSE_RANK() OVER(PARTITION BY col1,col2 ORDER BY COUNT(1) DESC) Ordre_DENSE_RANK
from temp
GROUP BY
col1
,col2
,col3 )
SELECT *
FROM temp2
--WHERE Ordre_RANK = 1
--WHERE Ordre_ROW_NUMBER = 1
--WHERE Ordre_DENSE_RANK = 1
I hope this will help you.
Here is my sample table
Col1 Col2
A 1
B 1
A 1
B 2
C 3
I want to be able to select distinct records where all rows have the same value in Col1 and Col2. So my answer should be
Col1 Col2
A 1
C 3
I tried
SELECT Col1, Col2 FROM Table GROUP BY Col1, Col2
This gives me
Col1 Col2
A 1
B 1
B 2
C 3
which is not the result I am looking for. Any tips would be appreciated.
Try this out:
SELECT col1, MAX(col2) aCol2 FROM t
GROUP BY col1
HAVING COUNT(DISTINCT col2) = 1
Output:
| COL1 | ACOL2 |
|------|-------|
| A | 1 |
| C | 3 |
Fiddle here.
Basically, this makes sure that amount the different values for col2 are unique for a given col1.
Try this:
SELECT * FROM MYTABLE
GROUP BY Col1, Col2
HAVING COUNT(*)>1
For example SQLFiddle here
you can try either of the below -
select col1, col2 from
(
select 'A' Col1 , 1 Col2
from dual
union all
select 'B' , 1
from dual
union all
select 'A' ,1
from dual
union all
select 'B' ,2
from dual
)
group by col1, col2
having count(*) >1;
OR
select col1, col2
from
(
select col1, col2, row_number() over (partition by col1, col2 order by col1, col2) cnt
from
(
select 'A' Col1 , 1 Col2
from dual
union all
select 'B' , 1
from dual
union all
select 'A' ,1
from dual
union all
select 'B' ,2
from dual
)
)
where cnt>1;