In a column I have something like this:
Amount:
12
2x25
192
How is it possible to multiply in this example 2x25 to order it correctly ASC.
My starting point:
SELECT * FROM table
ORDER BY REPLACE(Amount,'x','*') ASC
TIA
frgtv10
try this
SELECT
CAST(if(Amount LIKE '%x%', SUBSTRING_INDEX(Amount, 'x', 1) *
SUBSTRING_INDEX(Amount, 'x', -1) , Amount) as unsigned ) as amount
FROM table1
ORDER BY Amount ASC
DEMO HERE
steps and explaining :
locate fields with x value
sbstring from left and right and multiply it.
then cast the multiplication as unsigned.
order it asc
As long as this is the only formula (multiplying 2 numbers), you should be able to hard-code it with INSTR, SUBSTRING, and CONVERT.
Related
I have a table which contains thousands of rows and I would like to calculate the 90th percentile for one of the fields, called 'round'.
For example, select the value of round which is at the 90th percentile.
I don't see a straightforward way to do this in MySQL.
Can somebody provide some suggestions as to how I may start this sort of calculation?
Thank you!
First, lets assume that you have a table with a value column. You want to get the row with 95th percentile value. In other words, you are looking for a value that is bigger than 95 percent of all values.
Here is a simple answer:
SELECT * FROM
(SELECT t.*, #row_num :=#row_num + 1 AS row_num FROM YOUR_TABLE t,
(SELECT #row_num:=0) counter ORDER BY YOUR_VALUE_COLUMN)
temp WHERE temp.row_num = ROUND (.95* #row_num);
Compare solutions:
Number of seconds it took on my server to get 99 percentile of 1.3 million rows:
LIMIT x,y with index and no where: 0.01 seconds
LIMIT x,y with no where: 0.7 seconds
LIMIT x,y with where: 2.3 seconds
Full scan with no where: 1.6 seconds
Full scan with where: 5.7 seconds
Fastest solution for large tables using LIMIT x,y ():
Get count of values: SELECT COUNT(*) AS cnt FROM t
Get nth value, where n = (cnt - 1) * (1 - 0.95) : SELECT k FROM t ORDER BY k DESC LIMIT n,1
This solution requires two queries, because mysql does not support specifying variables in LIMIT clause, except for stored procedures (can be optimized with stored procedure). Usually additional query overhead is very low
This solution can be further optimized if you add index to k column and do not use complex where clauses (like 0.01 second for table with 1 million rows, because sorting is not needed).
Implementation example in PHP (can calculate percentile not only of columns, but also of expressions):
function get_percentile($table, $where, $expr, $percentile) {
if ($where) $subq = "WHERE $where";
else $subq = "";
$r = query("SELECT COUNT(*) AS cnt FROM $table $subq");
$w = mysql_fetch_assoc($r);
$num = abs(round(($w['cnt'] - 1) * (100 - $percentile) / 100.0));
$q = "SELECT ($expr) AS prcres FROM $table $subq ORDER BY ($expr) DESC LIMIT $num,1";
$r = query($q);
if (!mysql_num_rows($r)) return null;
$w = mysql_fetch_assoc($r);
return $w['prcres'];
}
// Usage example
$time = get_percentile(
"state", // table
"service='Time' AND cnt>0 AND total>0", // some filter
"total/cnt", // expression to evaluate
80); // percentile
The SQL standard supports the PERCENTILE_DISC and PERCENTILE_CONT inverse distribution functions for precisely this job. Implementations are available in at least Oracle, PostgreSQL, SQL Server, Teradata. Unfortunately not in MySQL. But you can emulate PERCENTILE_DISC in MySQL 8 as follows:
SELECT DISTINCT first_value(my_column) OVER (
ORDER BY CASE WHEN p <= 0.9 THEN p END DESC /* NULLS LAST */
) x,
FROM (
SELECT
my_column,
percent_rank() OVER (ORDER BY my_column) p,
FROM my_table
) t;
This calculates the PERCENT_RANK for each row given your my_column ordering, and then finds the last row for which the percent rank is less or equal to the 0.9 percentile.
This only works on MySQL 8+, which has window function support.
I was trying to solve this for quite some time and then I found the following answer. Honestly brilliant. Also quite fast even for big tables (the table where I used it contained approx 5 mil records and needed a couple of seconds).
SELECT
CAST(SUBSTRING_INDEX(SUBSTRING_INDEX( GROUP_CONCAT(field_name ORDER BY
field_name SEPARATOR ','), ',', 95/100 * COUNT(*) + 1), ',', -1) AS DECIMAL)
AS 95th Per
FROM table_name;
As you can imagine just replace table_name and field_name with your table's and column's names.
For further information check Roland Bouman's original post
In MySQL 8 there is the ntile window function you can use:
SELECT SomeTable.ID, SomeTable.Round
FROM SomeTable
JOIN (
SELECT SomeTable, (NTILE(100) OVER w) AS Percentile
FROM SomeTable
WINDOW w AS (ORDER BY Round)
) AS SomeTablePercentile ON SomeTable.ID = SomeTablePercentile.ID
WHERE Percentile = 90
LIMIT 1
https://dev.mysql.com/doc/refman/8.0/en/window-function-descriptions.html#function_ntile
http://www.artfulsoftware.com/infotree/queries.php#68
SELECT
a.film_id ,
ROUND( 100.0 * ( SELECT COUNT(*) FROM film AS b WHERE b.length <= a.length ) / total.cnt, 1 )
AS percentile
FROM film a
CROSS JOIN (
SELECT COUNT(*) AS cnt
FROM film
) AS total
ORDER BY percentile DESC;
This can be slow for very large tables
As pert Tony_Pets answer, but as I noted on a similar question: I had to change the calculation slightly, for example the 90th percentile - "90/100 * COUNT(*) + 0.5" instead of "90/100 * COUNT(*) + 1". Sometimes it was skipping two values past the percentile point in the ordered list, instead of picking the next higher value for the percentile. Maybe the way integer rounding works in mysql.
ie:
.... SUBSTRING_INDEX(SUBSTRING_INDEX( GROUP_CONCAT(fieldValue ORDER BY fieldValue SEPARATOR ','), ',', 90/100 * COUNT(*) + 0.5), ',', -1) as 90thPercentile ....
The most common definition of a percentile is a number where a certain percentage of scores fall below that number. You might know that you scored 67 out of 90 on a test. But that figure has no real meaning unless you know what percentile you fall into. If you know that your score is in the 95th percentile, that means you scored better than 95% of people who took the test.
This solution works also with the older MySQL 5.7.
SELECT *, #row_num as numRows, 100 - (row_num * 100/(#row_num + 1)) as percentile
FROM (
select *, #row_num := #row_num + 1 AS row_num
from (
SELECT t.subject, pt.score, p.name
FROM test t, person_test pt, person p, (
SELECT #row_num := 0
) counter
where t.id=pt.test_id
and p.id=pt.person_id
ORDER BY score desc
) temp
) temp2
-- optional: filter on a minimal percentile (uncomment below)
-- having percentile >= 80
An alternative solution that works in MySQL 8: generate a histogram of your data:
ANALYZE TABLE my_table UPDATE HISTOGRAM ON my_column WITH 100 BUCKETS;
And then just select the 95th record from information_schema.column_statistics:
SELECT v,c FROM information_schema.column_statistics, JSON_TABLE(histogram->'$.buckets',
'$[*]' COLUMNS(v VARCHAR(60) PATH '$[0]', c double PATH '$[1]')) hist
WHERE column_name='my_column' LIMIT 95,1
And voila! You will still need to decide whether you take the lower or upper limit of the percentile, or perhaps take an average - but that is a small task now. Most importantly - this is very quick, once the histogram object is built.
Credit for this solution: lefred's blog.
I Have a Table which has a Column of Type 'varchar' of size 255 char. The sample values look like this
Test/16-17/1
Test/16-17/2
test/16-17/10
Test/16-17/11
Test/16-17/20
Test/16-17/22
However on Using ORDER BY clause in the Select the Result is in following order
Test/16-17/1
Test/16-17/10
Test/16-17/11
Test/16-17/2
Test/16-17/20
Test/16-17/22
How Do I get the Result in the Chronolgical Order of Numerical Values in the End of the values?
If it follows the same / and have 2 digits in the column, you can use
order by right(vtest,2)
ORDER BY CONVERT( RIGHT( test_column_name,
LENGTH(test_column_name) - LOCATE('/', REVERSE(test_column_name))
), UNSIGNED INTEGER);
OR
ORDER BY CONVERT(substring_index(id, '/', -1), UNSIGNED INTEGER);
For your particular examples, you can order by length and then the value:
order by length(vtest), vtest
I don't know if this will work for other rows not in the question.
This will work no matter what the first 10 characters are:
ORDER BY CONCAT(LEFT(vtest, 10), CASE WHEN LENGTH(vtest) = 12
THEN CONCAT(0,RIGHT(vtest,1))
ELSE RIGHT(vtest,2) END)
Select the end part after the last "/", and then multiply it by 1 to convert it to a number:
ORDER BY RIGHT(test_column, LENGTH(test_column) - LOCATE('/', REVERSE(test_column))) * 1;
OR
ORDER BY substring_index(id, '/', -1) * 1;
This is basically the same as the answer by Keyur Panchal, but it uses implicit conversion which has several advantages: 1) Easier to read/write, 2) You don't need to worry about the type to convert to, 3) It ignores characters if they happen to exist in some values.
I have a variance report query here I need the 'Variance' to not have 10 decimal points in the Variance Column. What is the most convenient way to round Variance results to the 100th?
WITH A AS
(
select
A.FACTORY,
A.JOB_NUMBER,
A.PROCESS_STAGE,
A.PART_CODE,
B.PART_DESC_1,
A.INPUT_QTY_STD,
A.QUANTITY_INPUT,
A.QUANTITY_OUTSTANDING,
A.INPUT_QTY_ACTUAL,
(A.QUANTITY_OUTSTANDING*100/NULLIF(A.INPUT_QTY_STD,0)) as variance,
A.ACTUAL_CLOSE_DATE
from
(select * from [man_prod].[dbo].[JOB_STAGE_LINES]
where JOB_NUMBER in (select JOB_NUMBER from JOB_OUTPUTS where
BF_QTY_ACTUAL<>0
and ABS(DATEDIFF(HOUR,ACTUAL_CLOSE_DATE,GETDATE())) < 12 and STATUS_FLAG='C'
)) A
join fin_prod.dbo.PRODUCT_MASTER B
ON A.PART_CODE=B.PART_CODE
WHERE
A.INPUT_QTY_STD<>0 and
A.QUANTITY_OUTSTANDING <>0
)
SELECT * FROM A WHERE A.variance >10.000000 OR A.variance <-10
order by PROCESS_STAGE asc ,PART_CODE asc, variance desc ;
The Variance column comes out at 00.0000000000 i need it to display 00.000 or 00.000000
Help is greatly appreciated
Use the MySQL ROUND() function, the second argument is the number of decimal places if it is positive.
ROUND((A.QUANTITY_OUTSTANDING*100/NULLIF(A.INPUT_QTY_STD,0)), 3) as variance,
In this example if the value is 0.0000000000 it would be rounded to 3 decimal places, or 0.000.
You can use the TRUNCATE option:
TRUNCATE((A.QUANTITY_OUTSTANDING*100/NULLIF(A.INPUT_QTY_STD,0)), 3) as variance,
or use the ROUND if you are looking for rounding(as suggested by doublesharp)
ROUND((A.QUANTITY_OUTSTANDING*100/NULLIF(A.INPUT_QTY_STD,0)), 3) as variance,
Using Convert to convert it to a decimal of the desired length is what i prefer when i am not actually rounding the value, just formatting.
CONVERT(DECIMAL(10,3),10000)
I have this sql query
SELECT `price` FROM `used_cars` ORDER BY `price` DESC
So I obviously want to order by price from high to low. However, it seems to be taking the first digit and sorting by that. My theory is that it is treating it like a string, and as the column is a varchar it makes sense. However, this is third party data, so I am stuck with it. How can I order so that the larger numbers come first?
So this is an example of how they are ordered
9698
8999
8988
8900
5983
4988
4984
42441
40949
3995
3995
38995
37685
36999
35983
34990
34785
32999
30594
29999
29999
2862
28000
27995
You should convert the column to a numeric data type. You can do that in the table definition, or in the query itself, for example with:
... ORDER BY `price`+0 DESC
CAST should work:
SELECT CAST(price AS UNSIGNED) AS NumPrice
FROM used_cars
ORDER BY NumPrice DESC
This should work:
(...)
ORDER BY CAST (price AS INT)
This will work but is bad performance
SELECT price FROM used_cars ORDER BY CAST(price AS int) DESC
I'm having trouble with calculating the median of a list of values, not the average.
I found this article
Simple way to calculate median with MySQL
It has a reference to the following query which I don't understand properly.
SELECT x.val from data x, data y
GROUP BY x.val
HAVING SUM(SIGN(1-SIGN(y.val-x.val))) = (COUNT(*)+1)/2
If I have a time column and I want to calculate the median value, what do the x and y columns refer to?
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.
This will find an arbitrary percentile by replacing the COUNT(*)/2 with COUNT(*)*n where n is the percentile (.5 for median, .75 for 75th percentile, etc).
val is your time column, x and y are two references to the data table (you can write data AS x, data AS y).
EDIT:
To avoid computing your sums twice, you can store the intermediate results.
CREATE TEMPORARY TABLE average_user_total_time
(SELECT SUM(time) AS time_taken
FROM scores
WHERE created_at >= '2010-10-10'
and created_at <= '2010-11-11'
GROUP BY user_id);
Then you can compute median over these values which are in a named table.
EDIT: Temporary table won't work here. You could try using a regular table with "MEMORY" table type. Or just have your subquery that computes the values for the median twice in your query. Apart from this, I don't see another solution. This doesn't mean there isn't a better way, maybe somebody else will come with an idea.
First try to understand what the median is: it is the middle value in the sorted list of values.
Once you understand that, the approach is two steps:
sort the values in either order
pick the middle value (if not an odd number of values, pick the average of the two middle values)
Example:
Median of 0 1 3 7 9 10: 5 (because (7+3)/2=5)
Median of 0 1 3 7 9 10 11: 7 (because 7 is the middle value)
So, to sort dates you need a numerical value; you can get their time stamp (as seconds elapsed from epoch) and use the definition of median.
Finding median in mysql using group_concat
Query:
SELECT
IF(count%2=1,
SUBSTRING_INDEX(substring_index(data_str,",",pos),",",-1),
(SUBSTRING_INDEX(substring_index(data_str,",",pos),",",-1)
+ SUBSTRING_INDEX(substring_index(data_str,",",pos+1),",",-1))/2)
as median
FROM (SELECT group_concat(val order by val) data_str,
CEILING(count(*)/2) pos,
count(*) as count from data)temp;
Explanation:
Sorting is done using order by inside group_concat function
Position(pos) and Total number of elements (count) is identified. CEILING to identify position helps us to use substring_index function in the below steps.
Based on count, even or odd number of values is decided.
Odd values: Directly choose the element belonging to the pos using substring_index.
Even values: Find the element belonging to the pos and pos+1, then add them and divide by 2 to get the median.
Finally the median is calculated.
If you have a table R with a column named A, and you want the median of A, you can do as follows:
SELECT A FROM R R1
WHERE ( SELECT COUNT(A) FROM R R2 WHERE R2.A < R1.A ) = ( SELECT COUNT(A) FROM R R3 WHERE R3.A > R1.A )
Note: This will only work if there are no duplicated values in A. Also, null values are not allowed.
Simplest ways me and my friend have found out... ENJOY!!
SELECT count(*) INTO #c from station;
select ROUND((#c+1)/2) into #final;
SELECT round(lat_n,4) from station a where #final-1=(select count(lat_n) from station b where b.lat_n > a.lat_n);
Here is a solution that is easy to understand. Just replace Your_Column and Your_Table as per your requirement.
SET #r = 0;
SELECT AVG(Your_Column)
FROM (SELECT (#r := #r + 1) AS r, Your_Column FROM Your_Table ORDER BY Your_Column) Temp
WHERE
r = (SELECT CEIL(COUNT(*) / 2) FROM Your_Table) OR
r = (SELECT FLOOR((COUNT(*) / 2) + 1) FROM Your_Table)
Originally adopted from this thread.