汇总一次取k的行的所有组合 [英] Aggregate all combinations of rows taken k at a time

问题描述

我正在尝试为表中行的子集的字段计算聚合函数.问题是我想一次找到k个行的每个组合的均值-因此，对于所有行，我想找到(说)10行的每个组合的均值.所以:

I am trying to calculate an aggregate function for a field for a subset of rows in a table. The problem is that I'd like to find the mean of every combination of rows taken k at a time --- so for all the rows, I'd like to find (say) the mean of every combination of 10 rows. So:

 id | count
----|------
  1 |  5
  2 |  3
  3 |  6
...
 30 | 16

应该给我

ids 1..10的平均值；ID 1，3..11;ID 1、4..12，以此类推.我知道这会产生很多行.

mean of ids 1..10; ids 1, 3..11; ids 1, 4..12, and so so. I know this will yield a lot of rows.

对于从数组中找到组合，有一些答案.我可以通过编程来做到这一点，一次获取30个ID 10，然后 SELECT 对其进行编码.是否可以使用 PARTITION BY ， TABLESAMPLE 或其他功能(例如python的

There are SO answers for finding combinations from arrays. I could do this programmatically by taking 30 ids 10 at a time and then SELECTing them. Is there a way to do this with PARTITION BY, TABLESAMPLE, or another function (something like python's itertools.combinations())? (TABLESAMPLE by itself won't guarantee which subset of rows I am selecting as far as I can tell.)

推荐答案

引用的答案中描述的方法是静态的.一个更方便的解决方案可能是使用递归.

The method described in the cited answer is static. A more convenient solution may be to use recursion.

示例数据:

drop table if exists my_table;
create table my_table(id int primary key, number int);
insert into my_table values
(1, 5), 
(2, 3), 
(3, 6), 
(4, 9), 
(5, 2);

查询可在5个元素集中找到2个元素子集(k组合，k = 2):

Query which finds 2 element subsets in 5 element set (k-combination with k = 2):

with recursive recur as (
    select 
        id, 
        array[id] as combination, 
        array[number] as numbers, 
        number as sum
    from my_table
union all
    select 
        t.id, 
        combination || t.id, 
        numbers || t.number, 
        sum+ number
    from my_table t
    join recur r on r.id < t.id 
    and cardinality(combination) < 2            -- param k
)
select combination, numbers, sum/2.0 as average -- param k
from recur
where cardinality(combination) = 2              -- param k

 combination | numbers |      average       
-------------+---------+--------------------
 {1,2}       | {5,3}   | 4.0000000000000000
 {1,3}       | {5,6}   | 5.5000000000000000
 {1,4}       | {5,9}   | 7.0000000000000000
 {1,5}       | {5,2}   | 3.5000000000000000
 {2,3}       | {3,6}   | 4.5000000000000000
 {2,4}       | {3,9}   | 6.0000000000000000
 {2,5}       | {3,2}   | 2.5000000000000000
 {3,4}       | {6,9}   | 7.5000000000000000
 {3,5}       | {6,2}   | 4.0000000000000000
 {4,5}       | {9,2}   | 5.5000000000000000
(10 rows)   

对于k = 3的相同查询给出:

The same query for k = 3 gives:

 combination | numbers |      average       
-------------+---------+--------------------
 {1,2,3}     | {5,3,6} | 4.6666666666666667
 {1,2,4}     | {5,3,9} | 5.6666666666666667
 {1,2,5}     | {5,3,2} | 3.3333333333333333
 {1,3,4}     | {5,6,9} | 6.6666666666666667
 {1,3,5}     | {5,6,2} | 4.3333333333333333
 {1,4,5}     | {5,9,2} | 5.3333333333333333
 {2,3,4}     | {3,6,9} | 6.0000000000000000
 {2,3,5}     | {3,6,2} | 3.6666666666666667
 {2,4,5}     | {3,9,2} | 4.6666666666666667
 {3,4,5}     | {6,9,2} | 5.6666666666666667
(10 rows)

当然，如果不需要，您可以从查询中删除数字.

Of course, you can remove numbers from the query if you do not need them.

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