I need to write a stored procedure or table function to return a new data table as a new data source.
I wish to loop through the original table for every 5 rows base on the invoice ID column (it's possible not start from 1), the first 5 rows add to the left of the new table and the second 5 rows add to the right of the new table, the third 5 rows to the left and so on.
For example, Here is the original table:
Here is the expect table:
Thanks in advance!
declare #rowCount int = 5;
with cte as (
select *,( (IN_InvoiceID-1) / #rowCount ) % 2 group1
,( (IN_InvoiceID-1) / #rowCount ) group2
,IN_InvoiceID % #rowCount group3
from T
)
select * from cte
select T1.INID,T1.IN_InvoiceID,T1.IN_InvoiceAmount,T2.INID,T2.IN_InvoiceID,T2.IN_InvoiceAmount
from CTE T1
left join CTE T2 on T2.group1 = 1 and T1.group2 = T2.group2-1 and T1.group3 = T2.group3
where T1.group1 = 0
Test DDL
CREATE TABLE T
([INID] varchar(38), [IN_InvoiceID] int, [IN_InvoiceAmount] int)
;
INSERT INTO T
([INID], [IN_InvoiceID], [IN_InvoiceAmount])
VALUES
('DB3E17E6-35C5-41:121-93B1-F809BF6B2972', 1, 2999),
('3212F048-8213-4FCC-AB64-121485B77D4E43', 2, 3737),
('E3526373-A204-40F5-801C-7F8302A4E5E2', 3, 3175),
('76CC9C19-BF79-4E8A-8034-A33805AD3390', 4, 391),
('EC7A2FBC-B62D-4865-88DE-A8097975F125', 5, 1206),
('52AD3046-21331-4F0A-BD1D-67F232C54244', 6, 402),
('CA48F132-A9F5-4516-9E58-CDEE6644AAD1', 7, 1996),
('02E10C31-CAB2-4220-B66A-CEE5E67A9378', 8, 3906),
('98F1EEFF-B07A-4B65-87F4-E165264284DD', 9, 2575),
('91EBDD8B-B73C-470C-8900-DD66078483DB', 10, 2965),
('6E2490E5-C4DE-4833-877F-1590F7BDC1B8', 11, 1603),
('00985921-AC3C-4E3E-BAE1-7F58302F831A', 12, 1302)
;
Result:
Could you please check article Display Data in Multiple Columns using SQL showing with example case how a database developer can show the list of data rows in a columnar mode using Row_Number() function and mode arithmetic expression
You need to add additional columns from the same row that is different in the sample
Seems as if you want to split the table into 2 tables with alternating 5 rows. An easy way to do this would be:
Take data into a temp table having an extra column (lets say
grouping_id)
Update the grouping id so that each 5 rows have the same id. You can
use in_invoiceId % 5 (the nod function). After this step the first 5
rows will have grouping_id 0, next 5 will have 1, next will have 2
(assuming your invoice id is incremented +1 for all rows).
You can just do a normal select with where clause for odd and even grouping_id
Ideally, you can manage with the 2 tables Master and detail table.
But due to my curiosity, I am able to solve and give the answer as
Declare #table table(id int identity, invoice_id int)
; WITH Numbers AS
(
SELECT n = 1
UNION ALL
SELECT n + 1
FROM Numbers
WHERE n+1 <= 50
)
insert into #table SELECT n
FROM Numbers
Select (a.id )%5 ,* from #table a join #table b on a.id+5 = b.id and a.id != b.id
;WITH Numbers AS
(
SELECT n = 1, o = 5
UNION ALL
SELECT n + 10, o = o+10
FROM Numbers
WHERE n+1 <= 50
)
select a.id ParentId,a.invoice_id ParentInvoiceId, --b.n, b.o,
c.invoice_id childInvoiceID from #table a
join Numbers b on a.id between b.n and b.o
left join #table c on a.id + 5 = c.id
Here is my solution
First i create grps based on whether the in_invoiceid is divisible by 5 or not.(Ignore the remainders)
After that i create a category to indicate between alternative groups(ie by checking if the remainder is 0 or otherise)
Then its a matter of dense_ranking the records on the basis of the category field ordered by in_invoiceid
Lastly a join with category=1 rows with same dense_rank as those records in category=0
create table Invoicetable(IN_ID varchar(100), IN_InvoiceID int)
INSERT INTO Invoicetable (IN_ID, IN_InvoiceID)
VALUES
('2345-BCDE-6645-1DDF', 1),
('2345-BCDE-6645-3DDF', 2),
('2345-BCDE-6645-4DDF', 3),
('2345-BCDE-6645-5DDF', 4),
('2345-BCDE-6645-6DDF', 5),
('2345-BCDE-6645-7DDF', 6),
('2345-BCDE-6645-aDDF', 7),
('2345-BCDE-6645-sDDF', 8),
('2345-BCDE-6645-dDDF', 9),
('2345-BCDE-6645-dDDF', 10),
('2345-BCDE-6645-dDDF', 11),
('2345-BCDE-6645-dDDF', 12);
with data
as (
select *
,(in_invoiceid-1)/5 as grp
,case when ((in_invoiceid-1)/5)%2=0 then '1' else '0' end as category
,dense_rank() over(partition by case when ((in_invoiceid-1)/5)%2=0 then '1' else '0' end
order by in_invoiceid) as rnk
from invoicetable a
)
select *
from data a
left join data b
on a.rnk=b.rnk
and b.category=0
where a.category=1
Here is db fiddle link.
https://dbfiddle.uk/?rdbms=sqlserver_2017&fiddle=287f101737c580ca271940764b2536ae
You may try with the following approach. Dividing the table is done with (((ROW_NUMBER() OVER (ORDER BY IN_InvoiceID) - 1) / 5) % 2 = 0) which groups records in left and right groups.
CREATE TABLE #InvoiceTable(
IN_ID varchar(24),
IN_InvoiceID int
)
INSERT INTO #InvoiceTable (IN_ID, IN_InvoiceID)
VALUES
('2345-BCDE-6645-1DDF', 1),
('2345-BCDE-6645-3DDF', 2),
('2345-BCDE-6645-4DDF', 3),
('2345-BCDE-6645-5DDF', 4),
('2345-BCDE-6645-6DDF', 5),
('2345-BCDE-6645-7DDF', 6),
('2345-BCDE-6645-aDDF', 7),
('2345-BCDE-6645-sDDF', 8),
('2345-BCDE-6645-dDDF', 9),
('2345-BCDE-6645-dDDF', 10),
('2345-BCDE-6645-dDDF', 11),
('2345-BCDE-6645-dDDF', 12);
WITH cte AS (
SELECT
IN_ID,
IN_InvoiceID,
CASE
WHEN (((ROW_NUMBER() OVER (ORDER BY IN_InvoiceID) - 1) / 5) % 2 = 0) THEN 'L'
ELSE 'R'
END AS IN_Position
FROM #InvoiceTable
),
cteL AS (
SELECT IN_ID, IN_InvoiceID, ROW_NUMBER() OVER (ORDER BY IN_InvoiceID) AS IN_RowNumber
FROM cte
WHERE IN_Position = 'L'
),
cteR AS (
SELECT IN_ID, IN_InvoiceID, ROW_NUMBER() OVER (ORDER BY IN_InvoiceID) AS IN_RowNumber
FROM cte
WHERE IN_Position = 'R'
)
SELECT cteL.IN_ID, cteL.IN_InvoiceID, cteR.IN_ID, cteR.IN_InvoiceID
FROM cteL
LEFT JOIN cteR ON (cteL.IN_RowNumber = cteR.IN_RowNumber)
Output:
IN_ID IN_InvoiceID IN_ID IN_InvoiceID
2345-BCDE-6645-1DDF 1 2345-BCDE-6645-7DDF 6
2345-BCDE-6645-3DDF 2 2345-BCDE-6645-aDDF 7
2345-BCDE-6645-4DDF 3 2345-BCDE-6645-sDDF 8
2345-BCDE-6645-5DDF 4 2345-BCDE-6645-dDDF 9
2345-BCDE-6645-6DDF 5 2345-BCDE-6645-dDDF 10
2345-BCDE-6645-dDDF 11 NULL NULL
2345-BCDE-6645-dDDF 12 NULL NULL
What's the simplest (and hopefully not too slow) way to calculate the median with MySQL? I've used AVG(x) for finding the mean, but I'm having a hard time finding a simple way of calculating the median. For now, I'm returning all the rows to PHP, doing a sort, and then picking the middle row, but surely there must be some simple way of doing it in a single MySQL query.
Example data:
id | val
--------
1 4
2 7
3 2
4 2
5 9
6 8
7 3
Sorting on val gives 2 2 3 4 7 8 9, so the median should be 4, versus SELECT AVG(val) which == 5.
In MariaDB / MySQL:
SELECT AVG(dd.val) as median_val
FROM (
SELECT d.val, #rownum:=#rownum+1 as `row_number`, #total_rows:=#rownum
FROM data d, (SELECT #rownum:=0) r
WHERE d.val is NOT NULL
-- put some where clause here
ORDER BY d.val
) as dd
WHERE dd.row_number IN ( FLOOR((#total_rows+1)/2), FLOOR((#total_rows+2)/2) );
Steve Cohen points out, that after the first pass, #rownum will contain the total number of rows. This can be used to determine the median, so no second pass or join is needed.
Also AVG(dd.val) and dd.row_number IN(...) is used to correctly produce a median when there are an even number of records. Reasoning:
SELECT FLOOR((3+1)/2),FLOOR((3+2)/2); -- when total_rows is 3, avg rows 2 and 2
SELECT FLOOR((4+1)/2),FLOOR((4+2)/2); -- when total_rows is 4, avg rows 2 and 3
Finally, MariaDB 10.3.3+ contains a MEDIAN function
I just found another answer online in the comments:
For medians in almost any SQL:
SELECT x.val from data x, data y
GROUP BY x.val
HAVING SUM(SIGN(1-SIGN(y.val-x.val))) = (COUNT(*)+1)/2
Make sure your columns are well indexed and the index is used for filtering and sorting. Verify with the explain plans.
select count(*) from table --find the number of rows
Calculate the "median" row number. Maybe use: median_row = floor(count / 2).
Then pick it out of the list:
select val from table order by val asc limit median_row,1
This should return you one row with just the value you want.
I found the accepted solution didn't work on my MySQL install, returning an empty set, but this query worked for me in all situations that I tested it on:
SELECT x.val from data x, data y
GROUP BY x.val
HAVING SUM(SIGN(1-SIGN(y.val-x.val)))/COUNT(*) > .5
LIMIT 1
Unfortunately, neither TheJacobTaylor's nor velcrow's answers return accurate results for current versions of MySQL.
Velcro's answer from above is close, but it does not calculate correctly for result sets with an even number of rows. Medians are defined as either 1) the middle number on odd numbered sets, or 2) the average of the two middle numbers on even number sets.
So, here's velcro's solution patched to handle both odd and even number sets:
SELECT AVG(middle_values) AS 'median' FROM (
SELECT t1.median_column AS 'middle_values' FROM
(
SELECT #row:=#row+1 as `row`, x.median_column
FROM median_table AS x, (SELECT #row:=0) AS r
WHERE 1
-- put some where clause here
ORDER BY x.median_column
) AS t1,
(
SELECT COUNT(*) as 'count'
FROM median_table x
WHERE 1
-- put same where clause here
) AS t2
-- the following condition will return 1 record for odd number sets, or 2 records for even number sets.
WHERE t1.row >= t2.count/2 and t1.row <= ((t2.count/2) +1)) AS t3;
To use this, follow these 3 easy steps:
Replace "median_table" (2 occurrences) in the above code with the name of your table
Replace "median_column" (3 occurrences) with the column name you'd like to find a median for
If you have a WHERE condition, replace "WHERE 1" (2 occurrences) with your where condition
I propose a faster way.
Get the row count:
SELECT CEIL(COUNT(*)/2) FROM data;
Then take the middle value in a sorted subquery:
SELECT max(val) FROM (SELECT val FROM data ORDER BY val limit #middlevalue) x;
I tested this with a 5x10e6 dataset of random numbers and it will find the median in under 10 seconds.
Install and use this mysql statistical functions: http://www.xarg.org/2012/07/statistical-functions-in-mysql/
After that, calculate median is easy:
SELECT median(val) FROM data;
A comment on this page in the MySQL documentation has the following suggestion:
-- (mostly) High Performance scaling MEDIAN function per group
-- Median defined in http://en.wikipedia.org/wiki/Median
--
-- by Peter Hlavac
-- 06.11.2008
--
-- Example Table:
DROP table if exists table_median;
CREATE TABLE table_median (id INTEGER(11),val INTEGER(11));
COMMIT;
INSERT INTO table_median (id, val) VALUES
(1, 7), (1, 4), (1, 5), (1, 1), (1, 8), (1, 3), (1, 6),
(2, 4),
(3, 5), (3, 2),
(4, 5), (4, 12), (4, 1), (4, 7);
-- Calculating the MEDIAN
SELECT #a := 0;
SELECT
id,
AVG(val) AS MEDIAN
FROM (
SELECT
id,
val
FROM (
SELECT
-- Create an index n for every id
#a := (#a + 1) mod o.c AS shifted_n,
IF(#a mod o.c=0, o.c, #a) AS n,
o.id,
o.val,
-- the number of elements for every id
o.c
FROM (
SELECT
t_o.id,
val,
c
FROM
table_median t_o INNER JOIN
(SELECT
id,
COUNT(1) AS c
FROM
table_median
GROUP BY
id
) t2
ON (t2.id = t_o.id)
ORDER BY
t_o.id,val
) o
) a
WHERE
IF(
-- if there is an even number of elements
-- take the lower and the upper median
-- and use AVG(lower,upper)
c MOD 2 = 0,
n = c DIV 2 OR n = (c DIV 2)+1,
-- if its an odd number of elements
-- take the first if its only one element
-- or take the one in the middle
IF(
c = 1,
n = 1,
n = c DIV 2 + 1
)
)
) a
GROUP BY
id;
-- Explanation:
-- The Statement creates a helper table like
--
-- n id val count
-- ----------------
-- 1, 1, 1, 7
-- 2, 1, 3, 7
-- 3, 1, 4, 7
-- 4, 1, 5, 7
-- 5, 1, 6, 7
-- 6, 1, 7, 7
-- 7, 1, 8, 7
--
-- 1, 2, 4, 1
-- 1, 3, 2, 2
-- 2, 3, 5, 2
--
-- 1, 4, 1, 4
-- 2, 4, 5, 4
-- 3, 4, 7, 4
-- 4, 4, 12, 4
-- from there we can select the n-th element on the position: count div 2 + 1
If MySQL has ROW_NUMBER, then the MEDIAN is (be inspired by this SQL Server query):
WITH Numbered AS
(
SELECT *, COUNT(*) OVER () AS Cnt,
ROW_NUMBER() OVER (ORDER BY val) AS RowNum
FROM yourtable
)
SELECT id, val
FROM Numbered
WHERE RowNum IN ((Cnt+1)/2, (Cnt+2)/2)
;
The IN is used in case you have an even number of entries.
If you want to find the median per group, then just PARTITION BY group in your OVER clauses.
Rob
Most of the solutions above work only for one field of the table, you might need to get the median (50th percentile) for many fields on the query.
I use this:
SELECT CAST(SUBSTRING_INDEX(SUBSTRING_INDEX(
GROUP_CONCAT(field_name ORDER BY field_name SEPARATOR ','),
',', 50/100 * COUNT(*) + 1), ',', -1) AS DECIMAL) AS `Median`
FROM table_name;
You can replace the "50" in example above to any percentile, is very efficient.
Just make sure you have enough memory for the GROUP_CONCAT, you can change it with:
SET group_concat_max_len = 10485760; #10MB max length
More details: http://web.performancerasta.com/metrics-tips-calculating-95th-99th-or-any-percentile-with-single-mysql-query/
I have this below code which I found on HackerRank and it is pretty simple and works in each and every case.
SELECT M.MEDIAN_COL FROM MEDIAN_TABLE M WHERE
(SELECT COUNT(MEDIAN_COL) FROM MEDIAN_TABLE WHERE MEDIAN_COL < M.MEDIAN_COL ) =
(SELECT COUNT(MEDIAN_COL) FROM MEDIAN_TABLE WHERE MEDIAN_COL > M.MEDIAN_COL );
You could use the user-defined function that's found here.
Building off of velcro's answer, for those of you having to do a median off of something that is grouped by another parameter:
SELECT grp_field, t1.val FROM (
SELECT grp_field, #rownum:=IF(#s = grp_field, #rownum + 1, 0) AS row_number,
#s:=IF(#s = grp_field, #s, grp_field) AS sec, d.val
FROM data d, (SELECT #rownum:=0, #s:=0) r
ORDER BY grp_field, d.val
) as t1 JOIN (
SELECT grp_field, count(*) as total_rows
FROM data d
GROUP BY grp_field
) as t2
ON t1.grp_field = t2.grp_field
WHERE t1.row_number=floor(total_rows/2)+1;
Takes care about an odd value count - gives the avg of the two values in the middle in that case.
SELECT AVG(val) FROM
( SELECT x.id, x.val from data x, data y
GROUP BY x.id, x.val
HAVING SUM(SIGN(1-SIGN(IF(y.val-x.val=0 AND x.id != y.id, SIGN(x.id-y.id), y.val-x.val)))) IN (ROUND((COUNT(*))/2), ROUND((COUNT(*)+1)/2))
) sq
My code, efficient without tables or additional variables:
SELECT
((SUBSTRING_INDEX(SUBSTRING_INDEX(group_concat(val order by val), ',', floor(1+((count(val)-1) / 2))), ',', -1))
+
(SUBSTRING_INDEX(SUBSTRING_INDEX(group_concat(val order by val), ',', ceiling(1+((count(val)-1) / 2))), ',', -1)))/2
as median
FROM table;
Single query to archive the perfect median:
SELECT
COUNT(*) as total_rows,
IF(count(*)%2 = 1, CAST(SUBSTRING_INDEX(SUBSTRING_INDEX( GROUP_CONCAT(val ORDER BY val SEPARATOR ','), ',', 50/100 * COUNT(*)), ',', -1) AS DECIMAL), ROUND((CAST(SUBSTRING_INDEX(SUBSTRING_INDEX( GROUP_CONCAT(val ORDER BY val SEPARATOR ','), ',', 50/100 * COUNT(*) + 1), ',', -1) AS DECIMAL) + CAST(SUBSTRING_INDEX(SUBSTRING_INDEX( GROUP_CONCAT(val ORDER BY val SEPARATOR ','), ',', 50/100 * COUNT(*)), ',', -1) AS DECIMAL)) / 2)) as median,
AVG(val) as average
FROM
data
Optionally, you could also do this in a stored procedure:
DROP PROCEDURE IF EXISTS median;
DELIMITER //
CREATE PROCEDURE median (table_name VARCHAR(255), column_name VARCHAR(255), where_clause VARCHAR(255))
BEGIN
-- Set default parameters
IF where_clause IS NULL OR where_clause = '' THEN
SET where_clause = 1;
END IF;
-- Prepare statement
SET #sql = CONCAT(
"SELECT AVG(middle_values) AS 'median' FROM (
SELECT t1.", column_name, " AS 'middle_values' FROM
(
SELECT #row:=#row+1 as `row`, x.", column_name, "
FROM ", table_name," AS x, (SELECT #row:=0) AS r
WHERE ", where_clause, " ORDER BY x.", column_name, "
) AS t1,
(
SELECT COUNT(*) as 'count'
FROM ", table_name, " x
WHERE ", where_clause, "
) AS t2
-- the following condition will return 1 record for odd number sets, or 2 records for even number sets.
WHERE t1.row >= t2.count/2
AND t1.row <= ((t2.count/2)+1)) AS t3
");
-- Execute statement
PREPARE stmt FROM #sql;
EXECUTE stmt;
END//
DELIMITER ;
-- Sample usage:
-- median(table_name, column_name, where_condition);
CALL median('products', 'price', NULL);
My solution presented below works in just one query without creation of table, variable or even sub-query.
Plus, it allows you to get median for each group in group-by queries (this is what i needed !):
SELECT `columnA`,
SUBSTRING_INDEX(SUBSTRING_INDEX(GROUP_CONCAT(`columnB` ORDER BY `columnB`), ',', CEILING((COUNT(`columnB`)/2))), ',', -1) medianOfColumnB
FROM `tableC`
-- some where clause if you want
GROUP BY `columnA`;
It works because of a smart use of group_concat and substring_index.
But, to allow big group_concat, you have to set group_concat_max_len to a higher value (1024 char by default).
You can set it like that (for current sql session) :
SET SESSION group_concat_max_len = 10000;
-- up to 4294967295 in 32-bits platform.
More infos for group_concat_max_len: https://dev.mysql.com/doc/refman/5.1/en/server-system-variables.html#sysvar_group_concat_max_len
Another riff on Velcrow's answer, but uses a single intermediate table and takes advantage of the variable used for row numbering to get the count, rather than performing an extra query to calculate it. Also starts the count so that the first row is row 0 to allow simply using Floor and Ceil to select the median row(s).
SELECT Avg(tmp.val) as median_val
FROM (SELECT inTab.val, #rows := #rows + 1 as rowNum
FROM data as inTab, (SELECT #rows := -1) as init
-- Replace with better where clause or delete
WHERE 2 > 1
ORDER BY inTab.val) as tmp
WHERE tmp.rowNum in (Floor(#rows / 2), Ceil(#rows / 2));
Knowing exact row count you can use this query:
SELECT <value> AS VAL FROM <table> ORDER BY VAL LIMIT 1 OFFSET <half>
Where <half> = ceiling(<size> / 2.0) - 1
SELECT
SUBSTRING_INDEX(
SUBSTRING_INDEX(
GROUP_CONCAT(field ORDER BY field),
',',
((
ROUND(
LENGTH(GROUP_CONCAT(field)) -
LENGTH(
REPLACE(
GROUP_CONCAT(field),
',',
''
)
)
) / 2) + 1
)),
',',
-1
)
FROM
table
The above seems to work for me.
I used a two query approach:
first one to get count, min, max and avg
second one (prepared statement) with a "LIMIT #count/2, 1" and "ORDER BY .." clauses to get the median value
These are wrapped in a function defn, so all values can be returned from one call.
If your ranges are static and your data does not change often, it might be more efficient to precompute/store these values and use the stored values instead of querying from scratch every time.
as i just needed a median AND percentile solution, I made a simple and quite flexible function based on the findings in this thread. I know that I am happy myself if I find "readymade" functions that are easy to include in my projects, so I decided to quickly share:
function mysql_percentile($table, $column, $where, $percentile = 0.5) {
$sql = "
SELECT `t1`.`".$column."` as `percentile` FROM (
SELECT #rownum:=#rownum+1 as `row_number`, `d`.`".$column."`
FROM `".$table."` `d`, (SELECT #rownum:=0) `r`
".$where."
ORDER BY `d`.`".$column."`
) as `t1`,
(
SELECT count(*) as `total_rows`
FROM `".$table."` `d`
".$where."
) as `t2`
WHERE 1
AND `t1`.`row_number`=floor(`total_rows` * ".$percentile.")+1;
";
$result = sql($sql, 1);
if (!empty($result)) {
return $result['percentile'];
} else {
return 0;
}
}
Usage is very easy, example from my current project:
...
$table = DBPRE."zip_".$slug;
$column = 'seconds';
$where = "WHERE `reached` = '1' AND `time` >= '".$start_time."'";
$reaching['median'] = mysql_percentile($table, $column, $where, 0.5);
$reaching['percentile25'] = mysql_percentile($table, $column, $where, 0.25);
$reaching['percentile75'] = mysql_percentile($table, $column, $where, 0.75);
...
Here is my way . Of course, you could put it into a procedure :-)
SET #median_counter = (SELECT FLOOR(COUNT(*)/2) - 1 AS `median_counter` FROM `data`);
SET #median = CONCAT('SELECT `val` FROM `data` ORDER BY `val` LIMIT ', #median_counter, ', 1');
PREPARE median FROM #median;
EXECUTE median;
You could avoid the variable #median_counter, if you substitude it:
SET #median = CONCAT( 'SELECT `val` FROM `data` ORDER BY `val` LIMIT ',
(SELECT FLOOR(COUNT(*)/2) - 1 AS `median_counter` FROM `data`),
', 1'
);
PREPARE median FROM #median;
EXECUTE median;
After reading all previous ones they didn't match with my actual requirement so I implemented my own one which doesn't need any procedure or complicate statements, just I GROUP_CONCAT all values from the column I wanted to obtain the MEDIAN and applying a COUNT DIV BY 2 I extract the value in from the middle of the list like the following query does :
(POS is the name of the column I want to get its median)
(query) SELECT
SUBSTRING_INDEX (
SUBSTRING_INDEX (
GROUP_CONCAT(pos ORDER BY CAST(pos AS SIGNED INTEGER) desc SEPARATOR ';')
, ';', COUNT(*)/2 )
, ';', -1 ) AS `pos_med`
FROM table_name
GROUP BY any_criterial
I hope this could be useful for someone in the way many of other comments were for me from this website.
Based on #bob's answer, this generalizes the query to have the ability to return multiple medians, grouped by some criteria.
Think, e.g., median sale price for used cars in a car lot, grouped by year-month.
SELECT
period,
AVG(middle_values) AS 'median'
FROM (
SELECT t1.sale_price AS 'middle_values', t1.row_num, t1.period, t2.count
FROM (
SELECT
#last_period:=#period AS 'last_period',
#period:=DATE_FORMAT(sale_date, '%Y-%m') AS 'period',
IF (#period<>#last_period, #row:=1, #row:=#row+1) as `row_num`,
x.sale_price
FROM listings AS x, (SELECT #row:=0) AS r
WHERE 1
-- where criteria goes here
ORDER BY DATE_FORMAT(sale_date, '%Y%m'), x.sale_price
) AS t1
LEFT JOIN (
SELECT COUNT(*) as 'count', DATE_FORMAT(sale_date, '%Y-%m') AS 'period'
FROM listings x
WHERE 1
-- same where criteria goes here
GROUP BY DATE_FORMAT(sale_date, '%Y%m')
) AS t2
ON t1.period = t2.period
) AS t3
WHERE
row_num >= (count/2)
AND row_num <= ((count/2) + 1)
GROUP BY t3.period
ORDER BY t3.period;
create table med(id integer);
insert into med(id) values(1);
insert into med(id) values(2);
insert into med(id) values(3);
insert into med(id) values(4);
insert into med(id) values(5);
insert into med(id) values(6);
select (MIN(count)+MAX(count))/2 from
(select case when (select count(*) from
med A where A.id<B.id)=(select count(*)/2 from med) OR
(select count(*) from med A where A.id>B.id)=(select count(*)/2
from med) then cast(B.id as float)end as count from med B) C;
?column?
----------
3.5
(1 row)
OR
select cast(avg(id) as float) from
(select t1.id from med t1 JOIN med t2 on t1.id!= t2.id
group by t1.id having ABS(SUM(SIGN(t1.id-t2.id)))=1) A;
Often, we may need to calculate Median not just for the whole table, but for aggregates with respect to our ID. In other words, calculate median for each ID in our table, where each ID has many records. (good performance and works in many SQL + fixes problem of even and odds, more about performance of different Median-methods https://sqlperformance.com/2012/08/t-sql-queries/median )
SELECT our_id, AVG(1.0 * our_val) as Median
FROM
( SELECT our_id, our_val,
COUNT(*) OVER (PARTITION BY our_id) AS cnt,
ROW_NUMBER() OVER (PARTITION BY our_id ORDER BY our_val) AS rn
FROM our_table
) AS x
WHERE rn IN ((cnt + 1)/2, (cnt + 2)/2) GROUP BY our_id;
Hope it helps
MySQL has supported window functions since version 8.0, you can use ROW_NUMBER or DENSE_RANK (DO NOT use RANK as it assigns the same rank to same values, like in sports ranking):
SELECT AVG(t1.val) AS median_val
FROM (SELECT val,
ROW_NUMBER() OVER(ORDER BY val) AS rownum
FROM data) t1,
(SELECT COUNT(*) AS num_records FROM data) t2
WHERE t1.row_num IN
(FLOOR((t2.num_records + 1) / 2),
FLOOR((t2.num_records + 2) / 2));
A simple way to calculate Median in MySQL
set #ct := (select count(1) from station);
set #row := 0;
select avg(a.val) as median from
(select * from table order by val) a
where (select #row := #row + 1)
between #ct/2.0 and #ct/2.0 +1;
The most simple and fast way to calculate median in mysql.
select x.col
from (select lat_n,
count(1) over (partition by 'A') as total_rows,
row_number() over (order by col asc) as rank_Order
from station ft) x
where x.rank_Order = round(x.total_rows / 2.0, 0)