I want a SQL query which should tell me that for each ID which value repeated most of time.
For example lets take the following table:
Id Value
1 10
1 20
1 10
1 10
2 1
1 3
Desired Output
Id Value Count
1 10 3
2 1 1
From above example, it shows that for Id 1, Value 10 was repeated most of times and for Id 2, value 1 was repeated most of times
Any suggestion would be really appreciated.
Use rank to number the id's based on their value counts in descending order and pick up the 1st ranked rows.
select id, value, cnt
from (select id, value, count(*) as cnt,
rank() over (partition by id order by count(*) desc) as rnk
from t
group by id, value) x
where rnk = 1
Based on Gordon's comment, if you need only one value per id in case of ties, use row_number instead of rank, as rank returns all the ties in value counts.
Related
my goal is to put the value of the 1 row in every row of the new column.
First value in this example is the number 10.
The New Table is showing my goal.
Table
Product ID Name Value
1 ABC 10
2 XYZ 22
3 LMM 8
New Table
Product ID Name Value New Column
1 ABC 10 10
2 XYZ 22 10
3 LMM 8 10
I would fetch the value with the row_rumber function, but how i get that value in every row?
You can use the first_value() window function:
select product_id, name, value,
first_value(value) over (order by product_id) as new_column
from the_table
order by product_id;
Rows in a table have no implied sort order. So the "first row" can only be defined when an order by is present.
Assuming you want to pick the first one according to the product ID, you can do:
select *,
( select value
from (select *, row_number() over(order by product_id) as rn from t) x
where rn = 1
) as new_column
from t
I've got this table ratings:
id
user_id
type
value
0
0
Rest
4
1
0
Bar
3
2
0
Cine
2
3
0
Cafe
1
4
1
Rest
4
5
1
Bar
3
6
1
Cine
2
7
1
Cafe
5
8
2
Rest
4
9
2
Bar
3
10
3
Cine
2
11
3
Cafe
5
I want to have a table with a row for every pair (user_id, type) for the top 3 rated types through all users (ranked by sum(value) across the whole table).
Desired result:
user_id
type
value
0
Rest
4
0
Cafe
1
0
Bar
3
1
Rest
4
1
Cafe
5
1
Bar
3
2
Rest
4
3
Cafe
5
2
Bar
3
I was able to do this with two queries, one to get the top 3 and then another to get the rows where the type matches the top 3 types.
Does someone know how to fit this into a single query?
Get rows per user for the 3 highest ranking types, where types are ranked by the total sum of their value across the whole table.
So it's not exactly about the top 3 types per user, but about the top 3 types overall. Not all users will have rows for the top 3 types, even if there would be 3 or more types for the user.
Strategy:
Aggregate to get summed values per type (type_rnk).
Take only the top 3. (Break ties ...)
Join back to main table, eliminating any other types.
Order result by user_id, type_rnk DESC
SELECT r.user_id, r.type, r.value
FROM ratings r
JOIN (
SELECT type, sum(value) AS type_rnk
FROM ratings
GROUP BY 1
ORDER BY type_rnk DESC, type -- tiebreaker
LIMIT 3 -- strictly the top 3
) v USING (type)
ORDER BY user_id, type_rnk DESC;
db<>fiddle here
Since multiple types can have the same ranking, I added type to the sort order to break ties alphabetically by their name (as you did not specify otherwise).
Turns out, we don't need window functions - the ones with OVER and, optionally, PARTITION for this. (Since you asked in a comment).
I think you just want row_number(). Based on your results, you seem to want three rows per type, with the highest value:
select t.*
from (select t.*,
row_number() over (partition by type order by value desc) as seqnum
from t
) t
where seqnum <= 3;
Your description suggests that you might just want this per user, which is a slight tweak:
select t.*
from (select t.*,
row_number() over (partition by user order by value desc) as seqnum
from t
) t
where seqnum <= 3;
Consider the following table:
ID GroupId Rank
1 1 1
2 1 2
3 1 1
4 2 10
5 2 1
6 3 1
7 4 5
I need an sql (for MS-SQL) select query selecting a single Id for each group with the lowest rank. Each group needs to only return a single ID, even if there are two with the same rank (as 1 and 2 do in the above table). I've tried to select the min value, but the requirement that only one be returned, and the value to be returned is the ID column, is throwing me.
Does anyone know how to do this?
Use row_number():
select t.*
from (select t.*,
row_number() over (partition by groupid order by rank) as seqnum
from t
) t
where seqnum = 1;
I have a table of consecutive ids (integers, 1 ... n), and values (integers), like this:
Input Table:
id value
-- -----
1 1
2 1
3 2
4 3
5 1
6 1
7 1
Going down the table i.e. in order of increasing id, I want to count how many times in a row the same value has been seen consecutively, i.e. the position in a run:
Output Table:
id value position in run
-- ----- ---------------
1 1 1
2 1 2
3 2 1
4 3 1
5 1 1
6 1 2
7 1 3
Any ideas? I've searched for a combination of windowing functions including lead and lag, but can't come up with it. Note that the same value can appear in the value column as part of different runs, so partitioning by value may not help solve this. I'm on Hive 1.2.
One way is to use a difference of row numbers approach to classify consecutive same values into one group. Then a row number function to get the desired positions in each group.
Query to assign groups (Running this will help you understand how the groups are assigned.)
select t.*
,row_number() over(order by id) - row_number() over(partition by value order by id) as rnum_diff
from tbl t
Final Query using row_number to get positions in each group assigned with the above query.
select id,value,row_number() over(partition by value,rnum_diff order by id) as pos_in_grp
from (select t.*
,row_number() over(order by id) - row_number() over(partition by value order by id) as rnum_diff
from tbl t
) t
I have a single table, where I want to return a list of the MAX(id) GROUPed by another identifier. However I have a third column that, when it meets a certain criteria, "trumps" rows that don't meet that criteria.
Probably easier to explain with an example. Sample table has:
UniqueId (int)
GroupId (int)
IsPriority (bit)
Raw data:
UniqueId GroupId IsPriority
-----------------------------------
1 1 F
2 1 F
3 1 F
4 1 F
5 1 F
6 2 T
7 2 T
8 2 F
9 2 F
10 2 F
So, because no row in groupId 1 has IsPriority set, we return the highest UniqueId (5). Since groupId 2 has rows with IsPriority set, we return the highest UniqueId with that value (7).
So output would be:
5
7
I can think of ways to brute force this, but I am looking to see if I can do this in a single query.
SQL Fiddle Demo
WITH T
AS (SELECT *,
ROW_NUMBER() OVER (PARTITION BY GroupId
ORDER BY IsPriority DESC, UniqueId DESC ) AS RN
FROM YourTable)
SELECT UniqueId,
GroupId,
IsPriority
FROM T
WHERE RN = 1