I am working with SQL Server 2008
If I have a Table as such:
Code Value
-----------------------
4 240
4 299
4 210
2 NULL
2 3
6 30
6 80
6 10
4 240
2 30
How can I find the median AND group by the Code column please?
To get a resultset like this:
Code Median
-----------------------
4 240
2 16.5
6 30
I really like this solution for median, but unfortunately it doesn't include Group By:
https://stackoverflow.com/a/2026609/106227
The solution using rank works nicely when you have an odd number of members in each group, i.e. the median exists within the sample, where you have an even number of members the rank method will fall down, e.g.
1
2
3
4
The median here is 2.5 (i.e. half the group is smaller, and half the group is larger) but the rank method will return 3. To get around this you essentially need to take the top value from the bottom half of the group, and the bottom value of the top half of the group, and take an average of the two values.
WITH CTE AS
( SELECT Code,
Value,
[half1] = NTILE(2) OVER(PARTITION BY Code ORDER BY Value),
[half2] = NTILE(2) OVER(PARTITION BY Code ORDER BY Value DESC)
FROM T
WHERE Value IS NOT NULL
)
SELECT Code,
(MAX(CASE WHEN Half1 = 1 THEN Value END) +
MIN(CASE WHEN Half2 = 1 THEN Value END)) / 2.0
FROM CTE
GROUP BY Code;
Example on SQL Fiddle
In SQL Server 2012 you can use PERCENTILE_CONT
SELECT DISTINCT
Code,
Median = PERCENTILE_CONT(0.5) WITHIN GROUP (ORDER BY Value) OVER(PARTITION BY Code)
FROM T;
Example on SQL Fiddle
SQL Server does not have a function to calculate medians, but you could use the ROW_NUMBER function like this:
WITH RankedTable AS (
SELECT Code, Value,
ROW_NUMBER() OVER (PARTITION BY Code ORDER BY VALUE) AS Rnk,
COUNT(*) OVER (PARTITION BY Code) AS Cnt
FROM MyTable
)
SELECT Code, Value
FROM RankedTable
WHERE Rnk = Cnt / 2 + 1
To elaborate a bit on this solution, consider the output of the RankedTable CTE:
Code Value Rnk Cnt
---------------------------
4 240 2 3 -- Median
4 299 3 3
4 210 1 3
2 NULL 1 2
2 3 2 2 -- Median
6 30 2 3 -- Median
6 80 3 3
6 10 1 3
Now from this result set, if you only return those rows where Rnk equals Cnt / 2 + 1 (integer division), you get only the rows with the median value for each group.
Related
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 table such as this:
Id, ReportId, UserId
1 1 1
2 2 1
3 3 1
4 4 1
5 1 2
6 2 2
7 3 2
8 1 3
9 2 3
10 1 4
My table has thousands of records, above is just an example of the table structure simplified for purpose of understanding the problem.
I'm trying to figure out what at what percentile a user sits based on how many reports he has read.
I've been looking into PERCENTILE_CONT and PERCENTILE_DISC functions, but I fail to understand them properly. https://learn.microsoft.com/en-us/sql/t-sql/functions/percentile-cont-transact-sql
What confuses me most is that what it appears to me is that these functions are trying to find the 50th percentile, not percentile for a specific record.
Maybe I'm just not understanding this correctly. Is there a better way?
EDIT:
To clarify. I want to know at what percentile a specific user (in this case user with id 1) sits based on how many reports they have read. If they read the most reports they would be at a higher percentile, what is that percentile? Lets say there are 100 users exactly, then the person with most reports read would be 1st percentile.
Update #2
One of these should do it:
select
a.UserId,
a.reports_read,
PERCENTILE_DISC(0.5) WITHIN GROUP (ORDER BY a.reports_read) OVER (partition by UserId) AS percentile_d,
PERCENTILE_CONT(0.5) WITHIN GROUP (ORDER BY a.reports_read) OVER (partition by UserId) AS percentile_c,
PERCENT_RANK() OVER(ORDER BY a.reports_read ) percent_rank,
CUME_DIST() OVER(ORDER BY a.reports_read ) AS cumulative_distance
from
(select UserId, count(distinct(ReportId)) as reports_read
from #tmp
group by UserId
) a
It gives the following results:
UserId reports_read percentile_d percentile_c percent_rank cumulative_distance
4 1 1 1 0 0.25
3 2 2 2 0.33333 0.5
2 3 3 3 0.66667 0.75
1 6 6 6 1 1
I hope this helps.
I have a table called DATA on Microsoft SQL Server 2008 R2 with three non-nullable integer fields: ID, Sequence, and Value. Sequence values with the same ID will be consecutive, but can start with any value. I need a query that will return a count of consecutive rows with the same ID and Value.
For example, let's say I have the following data:
ID Sequence Value
-- -------- -----
1 1 1
5 1 100
5 2 200
5 3 200
5 4 100
10 10 10
I want the following result:
ID Start Value Count
-- ----- ----- -----
1 1 1 1
5 1 100 1
5 2 200 2
5 4 100 1
10 10 10 1
I tried
SELECT ID, MIN([Sequence]) AS Start, Value, COUNT(*) AS [Count]
FROM DATA
GROUP BY ID, Value
ORDER BY ID, Start
but that gives
ID Start Value Count
-- ----- ----- -----
1 1 1 1
5 1 100 2
5 2 200 2
10 10 10 1
which groups all rows with the same values, not just consecutive rows.
Any ideas? From what I've seen, I believe I have to left join the table with itself on consecutive rows using ROW_NUMBER(), but I am not sure exactly how to get counts from that.
Thanks in advance.
You can use Sequence - ROW_NUMBER() OVER (ORDER BY ID, Val, Sequence) AS g to create a group:
SELECT
ID,
MIN(Sequence) AS Sequence,
Val,
COUNT(*) AS cnt
FROM
(
SELECT
ID,
Sequence,
Sequence - ROW_NUMBER() OVER (ORDER BY ID, Val, Sequence) AS g,
Val
FROM
yourtable
) AS s
GROUP BY
ID, Val, g
Please see a fiddle here.
I have a table like this:
[challenge_log]
User_id | challenge | Try | Points
==============================================
1 1 1 5
1 1 2 8
1 1 3 10
1 2 1 5
1 2 2 8
2 1 1 5
2 2 1 8
2 2 2 10
I want the overall average points. To do so, i believe i need 3 steps:
Step 1 - Get the MAX value (of points) of each user in each challenge:
User_id | challenge | Points
===================================
1 1 10
1 2 8
2 1 5
2 2 10
Step 2 - SUM all the MAX values of one user
User_id | Points
===================
1 18
2 15
Step 3 - The average
AVG = SUM (Points from step 2) / number of users = 16.5
Can you help me find a query for this?
You can get the overall average by dividing the total number of points by the number of distinct users. However, you need the maximum per challenge, so the sum is a bit more complicated. One way is with a subquery:
select sum(Points) / count(distinct userid)
from (select userid, challenge, max(Points) as Points
from challenge_log
group by userid, challenge
) cl;
You can also do this with one level of aggregation, by finding the maximum in the where clause:
select sum(Points) / count(distinct userid)
from challenge_log cl
where not exists (select 1
from challenge_log cl2
where cl2.userid = cl.userid and
cl2.challenge = cl.challenge and
cl2.points > cl.points
);
Try these on for size.
Overall Mean
select avg( Points ) as mean_score
from challenge_log
Per-Challenge Mean
select challenge ,
avg( Points ) as mean_score
from challenge_log
group by challenge
If you want to compute the mean of each users highest score per challenge, you're not exactly raising the level of complexity very much:
Overall Mean
select avg( high_score )
from ( select user_id ,
challenge ,
max( Points ) as high_score
from challenge_log
) t
Per-Challenge Mean
select challenge ,
avg( high_score )
from ( select user_id ,
challenge ,
max( Points ) as high_score
from challenge_log
) t
group by challenge
After step 1 do
SELECT USER_ID, AVG(POINTS)
FROM STEP1
GROUP BY USER_ID
You can combine step 1 and 2 into a single query/subquery as follows:
Select BestShot.[User_ID], AVG(cast (BestShot.MostPoints as money))
from (select tLog.Challenge, tLog.[User_ID], MostPoints = max(tLog.points)
from dbo.tmp_Challenge_Log tLog
Group by tLog.User_ID, tLog.Challenge
) BestShot
Group by BestShot.User_ID
The subquery determines the most points for each user/challenge combo, and the outer query takes these max values and uses the AVG function to return the average value of them. The last Group By tells SQL to average all the values across each User_ID.