如何将前几行的数组累积组合成一个聚合数组？（雪花/SQL）答案

【问题标题】：how to cumulatively combine arrays from previous rows into one aggregate array? (Snowflake/SQL)如何将前几行的数组累积组合成一个聚合数组？（雪花/SQL）
【发布时间】：2022-10-25 05:37:28
【问题描述】：

想象一下，我有一个包含两列的表，一个日期 DATE 和一个字符串 ITEMS 数组。

我想创建一个列 ITEMS_AGG ，其中包含前行中所有数组的聚合，即类似于：

DATE   ITEMS      ITEMS_AGG
1      a, b       a, b
2      a, c       a, b, c
3      b, c       a, b, c
4.     a, d       a, b, c, d
5.     a, b, e    a, b, c, d, e

等等

【问题讨论】：

标签： sql snowflake-cloud-data-platform

【解决方案1】：

具有不同定义的累积array_aggJavaScript UDTF.

样本数据：

CREATE OR REPLACE TABLE test(grp TEXT, date INT, ITEMS ARRAY)
AS
     SELECT  'X',1,  ARRAY_CONSTRUCT('a', 'b')       
UNION SELECT 'X',2,  ARRAY_CONSTRUCT('a', 'c')       
UNION SELECT 'X',3,  ARRAY_CONSTRUCT('b', 'c')       
UNION SELECT 'X',4,  ARRAY_CONSTRUCT('a', 'd')       
UNION SELECT 'X',5,  ARRAY_CONSTRUCT('a', 'b', 'e')
UNION SELECT 'Y',1,  ARRAY_CONSTRUCT('z')
UNION SELECT 'Y',2,  ARRAY_CONSTRUCT('y','x')
UNION SELECT 'Y',3,  ARRAY_CONSTRUCT('y');

功能：

CREATE OR REPLACE FUNCTION aggregate (TS ARRAY)
RETURNS table (output variant)
LANGUAGE JAVASCRIPT
STRICT
IMMUTABLE
AS '
{
  initialize: function(argumentInfo, context) {
        this.result = [];
    },
  processRow: function (row, rowWriter, context) {
       this.result = [...new Set(this.result.concat(row.TS))];          
       rowWriter.writeRow({OUTPUT: this.result.sort()});           
   }
 }
 ';

询问：

SELECT *
FROM test,  TABLE(aggregate(ITEMS) OVER(PARTITION BY grp ORDER BY date))
ORDER BY grp, date;

输出：

【讨论】：

【解决方案2】：

好吧，这并不完全是您想要的，您可以使用递归 cte 进行加倍聚合，因为此 SQL 没有 ARRAY_CAT(DISTINCT) ：

WITH data AS (
    SELECT column1 as date, split(column2, ',') as items FROM VALUES
    (1, 'a,b'),
    (2, 'a,c'),
    (3, 'b, c'),
    (4, 'a,d'),
    (5, 'a,b,e')
), rec AS (
    WITH RECURSIVE r_cte AS (
        SELECT date as date, items
        FROM data
        WHERE date = 1
        
        UNION ALL
        
        SELECT r.date+1 as r_date, array_cat(r.items, d.items) as items
        FROM r_cte r
        JOIN data d 
            ON r.date + 1 = d.date
    )
    SELECT * from r_cte
)
SELECT *
FROM rec;

DATE	ITEMS
1	[ "a", "b" ]
2	[ "a", "b", "a", "c" ]
3	[ "a", "b", "a", "c", "b", " c" ]
4	[ "a", "b", "a", "c", "b", " c", "a", "d" ]
5	[ "a", "b", "a", "c", "b", " c", "a", "d", "a", "b", "e" ]

但实际上你应该使用 Lukasz 解决方案。

【讨论】：

【解决方案3】：

派对迟到了，但如果您要过来，我们可以使用 lateral 和 array_union_agg 加入

with cte (grp, dt, items) as

(select 'x', 1, ['a', 'b'] union all   
 select 'x', 2, ['a', 'c'] union all        
 select 'x', 3, ['b', 'c'] union all        
 select 'x', 4, ['a', 'd'] union all        
 select 'x', 5, ['a', 'b', 'e'] union all 
 select 'y', 1, ['z'] union all 
 select 'y', 2, ['y','x'] union all 
 select 'y', 3, ['y'])


select *
from cte a, lateral(select array_union_agg(b.items) as items_agg 
                     from cte b 
                     where a.grp=b.grp and b.dt<=a.dt) t2
order by a.grp, a.dt

【讨论】：