将运行总和分解为最大组大小/长度答案

【问题标题】：Break up running sum into maximum group size / length将运行总和分解为最大组大小/长度
【发布时间】：2021-06-04 09:22:37
【问题描述】：

我正在尝试将运行（有序）总和分解为一组最大值。当我实现以下示例逻辑时...

IF OBJECT_ID(N'tempdb..#t') IS NOT NULL DROP TABLE #t
SELECT TOP (ABS(CHECKSUM(NewId())) % 1000) ROW_NUMBER() OVER (ORDER BY name) AS ID,
    LEFT(CAST(NEWID() AS NVARCHAR(100)),ABS(CHECKSUM(NewId())) % 30) AS Description
    INTO #t
FROM sys.objects

DECLARE @maxGroupSize INT
SET @maxGroupSize = 100

;WITH t AS (
SELECT
    *,
    LEN(Description) AS DescriptionLength,
    SUM(LEN(Description)) OVER (/*PARTITION BY N/A */ ORDER BY ID) AS [RunningLength],
    SUM(LEN(Description)) OVER (/*PARTITION BY N/A */ ORDER BY ID)/@maxGroupSize AS GroupID
FROM #t
)
SELECT *, SUM(DescriptionLength) OVER (PARTITION BY GroupID) AS SumOfGroup
FROM t
ORDER BY GroupID, ID

我得到的组大于最大组大小（长度）100。

【问题讨论】：

标签： sql sql-server tsql sql-server-2012

【解决方案1】：

递归公用表表达式 (rcte) 将是解决此问题的一种方法。

样本数据

有限的一组固定样本数据。

create table data
(
  id int,
  description nvarchar(20)
);

insert into data (id, description) values
( 1, 'qmlsdkjfqmsldk'),
( 2, 'mldskjf'),
( 3, 'qmsdlfkqjsdm'),
( 4, 'fmqlsdkfq'),
( 5, 'qdsfqsdfqq'),
( 6, 'mds'),
( 7, 'qmsldfkqsjdmfqlkj'),
( 8, 'qdmsl'),
( 9, 'mqlskfjqmlkd'),
(10, 'qsdqfdddffd');

解决方案

对于每个递归步骤，评估 (r.group_running_length + len(d.description) <= @group_max_length) 是否必须扩展前一个组或必须在 case 表达式中启动一个新组。

将组目标大小设置为40，以更好地拟合样本数据。

declare @group_max_length int = 40;

with rcte as
(
  select d.id,
         d.description,
         len(d.description) as description_length,
         len(d.description) as running_length,
         1 as group_id,
         len(d.description) as group_running_length
  from data d
  where d.id = 1
union all
  select d.id,
         d.description,
         len(d.description),
         r.running_length + len(d.description),
         case
           when r.group_running_length + len(d.description) <= @group_max_length
           then r.group_id
           else r.group_id + 1
         end,
         case
           when r.group_running_length + len(d.description) <= @group_max_length
           then r.group_running_length + len(d.description)
           else len(d.description)
         end
  from rcte r
  join data d
    on d.id = r.id + 1
)
select r.id,
       r.description,
       r.description_length,
       r.running_length,
       r.group_id,
       r.group_running_length,
       gs.group_sum
from rcte r
cross apply ( select max(r2.group_running_length) as group_sum
              from rcte r2
              where r2.group_id = r.group_id ) gs -- group sum
order by r.id;

结果

包含运行组长度以及每行的组总和。

id  description       description_length  running_length  group_id  group_running_length  group_sum
--  ----------------  ------------------  --------------  --------  --------------------  ---------
1   qmlsdkjfqmsldk     14                   14            1         14                    33
2   mldskjf             7                   21            1         21                    33
3   qmsdlfkqjsdm       12                   33            1         33                    33
4   fmqlsdkfq           9                   42            2          9                    39
5   qdsfqsdfqq         10                   52            2         19                    39
6   mds                 3                   55            2         22                    39
7   qmsldfkqsjdmfqlkj  17                   72            2         39                    39
8   qdmsl               5                   77            3          5                    28
9   mqlskfjqmlkd       12                   89            3         17                    28
10  qsdqfdddffd        11                  100            3         28                    28

Fiddle 查看实际情况（包括随机数据版本）。

【讨论】：