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;
Related
I'm stuck with how to write SQL statements, so I would appreciate it if you could teach me.
Current status
items table
id
session_id
item_id
competition_id
1
1
2
1
2
1
3
1
2
1
2
1
2
1
2
1
2
1
5
2
3
1
7
2
4
1
4
2
5
1
5
2
want to
grouping by competition_id,
Count the same numbers in item_id,Extract the most common numbers and their numbers.
For example
If competition_id is 1,item_id → 2 ,and the number is 3
If competition_id is 2,item_id → 5 ,and the number is 2
If competition_id is 3,・・・
If competition_id is 4,・・・
environment
macOS BigSur
ruby 2.7.0
Rails 6.1.1
sqlite
In statistics, what you are asking for is the mode, the most common value.
You can use aggregation and row_number():
select ct.*
from (select competition_id, item_id, count(*) as cnt,
row_number() over (partition by competition_id order by count(*) desc) as seqnum
from t
group by competition_id, item_id
) ci
where seqnum = 1;
In the event that there are ties, this returns only one of the values, arbitrarily. If you want all modes when there are ties use rank() instead of row_number().
i have a table like this
code Quantity
1 5
1 6
2 2
2 1-
3 4
.
.
how can made it like this
code Quantity remain
1 5 5
1 6 11
2 2 2
2 1- 1
3 4 4
.
.
Your query presumes an ordering of the rows. I will assume you have such a column.
Assuming the values are numbers (1- ???), then you can simply use a cumulative sum:
select t.*,
sum(quantity) over (partition by code order by ?) as remaining
from t;
The ? is for the column that specifies the ordering.
You can do a window sum, but you need a column to unambiguously order the records within groups sharing the same code. I assumed that this column is called id.
select t.*, sum(quantity) over(partition by code order by id) remain from mytable t
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
It's my first question here so I hope I can explain it well enough,
I want to order my data by amount of occurrences in the table.
My table is like this:
id Daynr
1 2
1 4
2 4
2 5
2 6
3 1
4 2
4 5
And I want it to sort it like this:
id Daynr
3 1
1 2
1 4
4 2
4 5
2 4
2 5
2 6
Player #3 has one day in the table, and Player #1 has 2.
My table is named "dayid"
Both id and Daynr are foreign keys, together making it a primary key
I hope this explains my problem enough, Please ask for more information it's my first time here.
Thanks in advance
You can do this by counting the number of times that things occur for each id. Most databases support window functions, so you can do this as:
select id, daynr
from (select t.*, count(*) over (partition by id) as cnt
from table t
) t
order by cnt, id;
You can also express this as a join:
select t.id, t.daynr
from table as t inner join
(select id, count(*) as cnt
from table
group by id
) as tg
on t.id = tg.id
order by tg.cnt, id;
Note that both of these include the id in the order by. That way, if two ids have the same count, all rows for the id will appear together.