获取SQL中每X行的平均值 [英] Get Average value for each X rows in SQL

问题描述

我们说我有下表

+----+-------+
| Id | Value |
+----+-------+
|  1 |   2.0 |
|  2 |   8.0 |
|  3 |   3.0 |
|  4 |   9.0 |
|  5 |   1.0 |
|  6 |   4.0 |
|  7 |   2.5 |
|  8 |   6.5 |
+----+-------+

我想绘制这些值，但是由于我的真实表有成千上万个值，因此我考虑了获取和平均每个X行的值.我有什么办法可以做到这一点，即每2或4行，如下所示:

I want to plot these values, but since my real table has thousands of values, I thought about getting and average for each X rows. Is there any way for me to do so for, ie, each 2 or 4 rows, like below:

2
+-----+------+
| 1-2 |  5.0 |
| 3-4 |  6.0 |
| 5-6 |  2.5 |
| 7-8 |  4.5 |
+-----+------+

4
+-----+------+
| 1-4 |  5.5 |
| 5-8 |  3.5 |
+-----+------+

还可以根据表中的总行数来使此X值动态化吗?例如，如果我有1000行，则将基于每200行(1000/5)计算平均值，但是如果我有20行，则应基于每4行(20/5)计算平均值.

Also, is there any way to make this X value dynamic, based on the total number of rows in my table? Something like, if I have 1000 rows, the average will be calculated based on each 200 rows (1000/5), but if I have 20, calculate it based on each 4 rows (20/5).

我知道如何以编程方式执行此操作，但是有什么方法可以使用纯SQL做到这一点?

I know how to do that programmatically, but is there any way to do so using pure SQL?

我需要它在mysql上工作.

I need it to work on mysql.

推荐答案

根据您的DBMS，类似的方法将起作用:

Depending on your DBMS, something like this will work:

SELECT
   ChunkStart = Min(Id),
   ChunkEnd = Max(Id),
   Value = Avg(Value)
FROM
   (
      SELECT
         Chunk = NTILE(5) OVER (ORDER BY Id),
         *
      FROM
         YourTable
   ) AS T
GROUP BY
   Chunk
ORDER BY 
   ChunkStart;

根据您的请求，无论有多少行，这都会创建5个组或块.

This creates 5 groups or chunks no matter how many rows there are, as you requested.

如果您没有窗口功能，则可以伪造它:

If you have no windowing functions you can fake it:

SELECT
   ChunkStart = Min(Id),
   ChunkEnd = Max(Id),
   Value = Avg(Value)
FROM
   YourTable
GROUP BY
   (Id - 1) / (((SELECT Count(*) FROM YourTable) + 4) / 5)
;

我在这里做了一些假设，例如Id以1开头并且没有空格，并且如果事情没有平均分配，那么您希望最后一组太小而不是太大.我还假设整数除法将导致出现在SQL Server女士中.

I made some assumptions here such as Id beginning with 1 and there being no gaps, and that you would want the last group too small instead of too big if things didn't divide evenly. I also assumed integer division would result as in Ms SQL Server.

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