mysql speed of query - selecting max value of every 3 rows

Question

I have a table that holds price information. I need to select the max value of every three rows. EXAMPLE:

Table `daily_high`
 ____ _______
| ID | HIGH  |
| 1  | 24.65 |
| 2  | 24.93 |
| 3  | 26.02 |
| 4  | 25.33 |
| 5  | 25.16 |
| 6  | 25.91 |
| 7  | 26.05 |
| 8  | 28.13 |
| 9  | 27.07 |
|____|_______|


Desired output to new table (ID will be auto-increment so don't assume an association exists between this ID 1 and the daily_high ID 1:
 ____ ___________
| ID | 3MaxHIGH  |
|____|___________|
| 1  | 26.02     |
| 2  | 25.91     |
| 3  | 28.13     |
|____|___________|

I want to compare IDs 1,2, and 3 to determine the high value among them. Then once I have compared 1-3, I want to move on to 4 through 6, then 7 through 9, etc until I've done this for all values contained in the table (currently about 400,000 values). I have written code that uses

SELECT max(HIGH) FROM daily_high as dh1 JOIN (SELECT max(HIGH) FROM daily_high WHERE id >= dh1 AND id < (dh1.id + 3))

This works but is horribly slow. I've tried using the SELECT statement where I identify the column values to be pull for display, meaning between the SELECT and FROM parts of the query.

I've tried to use JOIN to join all 3 rows onto the same table for comparison but it too is horribly slow. By slow I mean just under 10 seconds to gather information for 20 rows. This means that the query has analyzed 60 rows (20 groups of 3) in 9.65879893303 seconds (I didn't make this up, I used microtime() to calculate it.

Anyone have any suggestions for faster code than what I've got?

Keep in mind that my actual table is not the same as what I've posted above, but it the concept is the same.

Thanks for any help.

Answer 1

如果你持续识别，你可以做到这一点

SELECT floor(id/3) as range, max(HIGH) FROM daily_high GROUP BY range;

Answer 2

Why not to use DIV operator for grouping your aggregation:

SELECT (id-1) DIV 3 + 1 AS ID, MAX(high) AS 3MaxHIGH 
FROM daily_high 
GROUP BY (id-1) DIV 3

This query gives the same result.

ID  3MaxHIGH
1   26.02
2   25.91
3   28.13

I was unable to run your query, and I believe that this one is faster.

UPD : To ensure that you have valid groups for your ranges, use this query:

select id, high, (id-1) div 3 + 1 from daily_high

result:

id  high    (id-1) div 3 + 1
1   24.65   1
2   24.93   1
3   26.02   1
4   25.33   2
5   25.16   2
6   25.91   2
7   26.05   3
8   28.13   3
9   27.07   3

Answer 3

Fuller answer with an example. The following code will do what I think you want.

SELECT FLOOR((row - 1) / 3), MAX(Sub1.high)
FROM (SELECT @row := @row + 1 as row, daily_high.*
FROM daily_high, (SELECT @row := 0) r) Sub1
GROUP BY FLOOR((row - 1) / 3) 
ORDER BY Sub1.ID

Answer 4

The below query worked for me on a test table. perhaps not the best, but the other solutions failed on my test table.

This does require the ID's to be sequential. Also be sure to put an index on High aswell for speed.

SELECT FLOOR(T1.Id/3)+1 AS Id, ROUND(GREATEST(T1.High, T2.High, T3.High),2) AS High FROM `daily_high` T1, `daily_high` T2, `daily_high` T3
WHERE T2.Id=T1.Id+1 
AND T3.Id=T2.Id+1
AND MOD(T1.Id, 3)=1

Answer 5

logic: if(id is divisible by 3, id/3-1, id/3)

select if(mod(id,3) = 0,floor(id/3)-1,floor(id/3)) as group_by_col , max(HIGH)
FROM daily_high GROUP BY group_by_col;