I want to find the count(distinct column name) wihtout using group by in hive.
my input is :
name id
a 2
a 3
a 4
b 1
c 4
c 4
d 7
d 9
my expected output is
name count
a 3
b 1
c 1
d 2
can some tell me how to achieve this without using group by. please help
A canonical solution with no explicit group by is select distinct with window functions:
select distinct name, count(distinct id) over (partition by name)
from t;
In your case, I strongly recommend the group by version:
select name, count(distinct id)
from t
group by name;
You can use subquery :
select distinct t.name,
(select count(distinct id) from table t1 where t1.name = t.name) as count
from table t;
However, GROUP BY is really appropriate way to do this.
just use count aggregate function with distinct keyword
select name,count(distinct id) as cnt from table
group by name
Related
I have table as:
ID PAYOR_NAME
---------- ------------
4 AETNAU
4 AETNA
2 UMR
3 CIGNA
1 METLIFE
Id needs to be one to one mapping with payor_name.But Id 4 is associated with multiple payor_name ,so it is considered as duplicate. So,I tried to find the duplicate by using:
select id, count(*) duplic_data
from (
select distinct id, payor_name
from offc.payor_collec
order by id) t1
group by id;
It is giving me the duplicate Id,But i am wondering is there any also way where we can find duplicates in one to one mappings?
There are multiple ways of finding it:
using exists
using group by and having
using analytical function, if you want all column data for that duplicate values (same as exists)
-
select id, payor_name, cnt as count_
from (
select id, payor_name,
count(1) over (partition by id) as cnt
from offc.payor_collec) t1
Where cnt > 1;
It will give you following result:
ID PAYOR_NAME COUNT_
---------- ------------ -------
4 AETNAU 2
4 AETNA 2
Cheers!!
How about simply using exists:
select pc.id, pc.payor_name, count(*)
from offc.payor_collec pc
where exists (select 1
from offc.payor_collec pc2
where pc2.id = pc.id and pc2.payor_name <> pc.payor_name
)
group by pc.id, pc.payor_name
order by pc.id, count(*) desc;
This also orders by the most frequent value, which might be helpful in figuring out the best name.
This could be a way:
select ID, count(*)
from offc.payor_collec
group by ID
having count(distinct PAYOR_NAME) > 1
Another option would be using a subquery containing having clause with having count(ID)>1
select *
from payor_collec
where ID in
( select ID
from payor_collec t
group by ID
having count(ID)>1 )
I'm looking for a way to find the top count value of a column by SQL.
If for example this is my data
id type
----------
1 A
1 B
1 A
2 C
2 D
2 D
I would like the result to be:
1 A
2 D
I'm looking for a way to do it without groping by the column I count (type in the example)
Thanks
Statistically, this is called the "mode". You can calculate it using window functions:
select id, type, cnt
from (select id, type, count(*) as cnt,
row_number() over (partition by id order by count(*) desc) as seqnum
from t
group by id, type
) t
where seqnum = 1;
If there are ties, then an arbitrary value is chosen from among the ties.
You are looking for the statistic mode (the most often ocurring value):
select id, stats_mode(type)
from mytable
group by id
order by id;
Not all DBMS support this however. Check your docs, wheher this function or a similar one is available in your DBMS.
Just GROUP BY id, type and keep the rows with the maximum counter:
select id, type
from tablename
group by id, type
having count(*) = (
select count(*) from tablename group by id, type order by count(*) desc limit 1
)
See the demo
Or
select id, type
from tablename
group by id, type
having count(*) = (
select max(t.counter) from (select count(*) counter from tablename group by id, type) t
)
See the demo
I have a table where storing details
ID NAME
1 A
2 A
1 A
I need the output like
ID Name Count
1,2 A 3
Please help to get the output like that in oracle select query
In Oracle, you can use listagg(), but it has no distinct option. So, use a subquery and two levels of aggregation:
select listagg(id, ',') within group (order by id) as id, name, sum(cnt)
from (select id, name, count(*) as cnt
from t
group by id, name
) x
group by name;
table a
column id : a a b b
column total : 1 2 1 3
how can i show? in one table without use compute
a 3 7
b 4 7
Do group by to sum each id's total. Do a sub-select to count total:
select id,
sum(total) as total,
(select sum(total) from a) as totalall
from a
group by id
Using window functions with a distinct, it can be simply expressed like this:
select distinct id,
sum(Total) over(partition by id) total,
Sum(Total) over () total_all
from mytable
SQL Fiddle
One way is to use OUTER APPLY. You could also set a variable to the sum of the table and call that variable.
select a.id, sum(a.total) as total, b.Grand as GrandTotal
from tablea a
outer apply
(select sum(total) as Grand from tablea) b
group by a.id
I have a table with 2 fields:
ID Name
-- -------
1 Alpha
2 Beta
3 Beta
4 Beta
5 Charlie
6 Charlie
I want to group them by name, with 'count', and a row 'SUM'
Name Count
------- -----
Alpha 1
Beta 3
Charlie 2
SUM 6
How would I write a query to add SUM row below the table?
SELECT name, COUNT(name) AS count
FROM table
GROUP BY name
UNION ALL
SELECT 'SUM' name, COUNT(name)
FROM table
OUTPUT:
name count
-------------------------------------------------- -----------
alpha 1
beta 3
Charlie 2
SUM 6
SELECT name, COUNT(name) AS count, SUM(COUNT(name)) OVER() AS total_count
FROM Table GROUP BY name
Without specifying which rdbms you are using
Have a look at this demo
SQL Fiddle DEMO
SELECT Name, COUNT(1) as Cnt
FROM Table1
GROUP BY Name
UNION ALL
SELECT 'SUM' Name, COUNT(1)
FROM Table1
That said, I would recomend that the total be added by your presentation layer, and not by the database.
This is a bit more of a SQL SERVER Version using Summarizing Data Using ROLLUP
SQL Fiddle DEMO
SELECT CASE WHEN (GROUPING(NAME) = 1) THEN 'SUM'
ELSE ISNULL(NAME, 'UNKNOWN')
END Name,
COUNT(1) as Cnt
FROM Table1
GROUP BY NAME
WITH ROLLUP
Try this:
SELECT ISNULL(Name,'SUM'), count(*) as Count
FROM table_name
Group By Name
WITH ROLLUP
all of the solution here are great but not necessarily can be implemented for old mysql servers (at least at my case). so you can use sub-queries (i think it is less complicated).
select sum(t1.cnt) from
(SELECT column, COUNT(column) as cnt
FROM
table
GROUP BY
column
HAVING
COUNT(column) > 1) as t1 ;
Please run as below :
Select sum(count)
from (select Name,
count(Name) as Count
from YourTable
group by Name); -- 6
The way I interpreted this question is needing the subtotal value of each group of answers. Subtotaling turns out to be very easy, using PARTITION:
SUM(COUNT(0)) OVER (PARTITION BY [Grouping]) AS [MY_TOTAL]
This is what my full SQL call looks like:
SELECT MAX(GroupName) [name], MAX(AUX2)[type],
COUNT(0) [count], SUM(COUNT(0)) OVER(PARTITION BY GroupId) AS [total]
FROM [MyView]
WHERE Active=1 AND Type='APP' AND Completed=1
AND [Date] BETWEEN '01/01/2014' AND GETDATE()
AND Id = '5b9xxxxx-xxxx-xxxx-xxxx-xxxxxxxxxxxx' AND GroupId IS NOT NULL
GROUP BY AUX2, GroupId
The data returned from this looks like:
name type count total
Training Group 2 Cancelation 1 52
Training Group 2 Completed 41 52
Training Group 2 No Show 6 52
Training Group 2 Rescheduled 4 52
Training Group 3 NULL 4 10535
Training Group 3 Cancelation 857 10535
Training Group 3 Completed 7923 10535
Training Group 3 No Show 292 10535
Training Group 3 Rescheduled 1459 10535
Training Group 4 Cancelation 2 27
Training Group 4 Completed 24 27
Training Group 4 Rescheduled 1 27
You can use union to joining rows.
select Name, count(*) as Count from yourTable group by Name
union all
select "SUM" as Name, count(*) as Count from yourTable
For Sql server you can try this one.
SELECT ISNULL([NAME],'SUM'),Count([NAME]) AS COUNT
FROM TABLENAME
GROUP BY [NAME] WITH CUBE
with cttmp
as
(
select Col_Name, count(*) as ctn from tab_name group by Col_Name having count(Col_Name)>1
)
select sum(ctn) from c
You can use ROLLUP
select nvl(name, 'SUM'), count(*)
from table
group by rollup(name)
Use it as
select Name, count(Name) as Count from YourTable
group by Name
union
Select 'SUM' , COUNT(Name) from YourTable
I am using SQL server and the following should work for you:
select cast(name as varchar(16)) as 'Name', count(name) as 'Count'
from Table1
group by Name
union all
select 'Sum:', count(name)
from Table1
I required having count(*) > 1 also. So, I wrote my own query after referring some the above queries
SYNTAX:
select sum(count) from (select count(`table_name`.`id`) as `count` from `table_name` where {some condition} group by {some_column} having count(`table_name`.`id`) > 1) as `tmp`;
Example:
select sum(count) from (select count(`table_name`.`id`) as `count` from `table_name` where `table_name`.`name` IS NOT NULL and `table_name`.`name` != '' group by `table_name`.`name` having count(`table_name`.`id`) > 1) as `tmp`;
You can try group by on name and count the ids in that group.
SELECT name, count(id) as COUNT FROM table group by name
After the query, run below to get the total row count
select ##ROWCOUNT
select sum(s) from
(select count(Col_name) as s from Tab_name group by Col_name having count(*)>1)c