创建总和为 12 的数字组

Question

Bigquery 中的表在从 1 到 100 的队列中具有 100 个或更多从 0.5 到 6 的数字，增量为 0.5。 您需要依次创建数字组。每组的数字之和为12。 图中是一个表格示例，第 1 组应由数字 1、2、3、6 组成，加起来为 12。缺少的数字 4、5 应进入下一组 我将不胜感激任何提示和想法。

我尝试先将它们分组以获得小于或等于12的总和，然后在第二步中添加缺少的数字以在每组中得到12

订单号	数字
1	2.0
2	4.5
3	5.0
4	3.5
5	3.5
6	0.5
7	1.5
8	4.5
9	4.5
10	1.5

   WITH cte AS (SELECT
    order_num,
    num,
    SUM(num) OVER (ORDER BY order_num) AS cumulative_sum
    FROM
    `ms_jzs6s5.test_numbers`)
, groupHours as (
    SELECT

  order_num,
  num,
  cumulative_sum,
  CASE
    WHEN cumulative_sum <= CAST(12 AS FLOAT64) THEN num
    ELSE num - GREATEST(cumulative_sum - CAST(12 AS FLOAT64), 0)
  END AS plan,
  IF(MOD(CAST(cumulative_sum AS NUMERIC), 12) <> 0, CEIL(cumulative_sum / 12), CEIL(cumulative_sum / 12) + 1) AS group_num AS group_num,
FROM
  cte 
)

, Lastgroup as (
SELECT
groupHours.*
FROM
  groupHours
JOIN (
  SELECT MAX(group_num) maxgroup
  FROM groupHours
  UNION ALL
  SELECT MAX(group_num) - 1 maxgroup
  FROM groupHours) ON groupHours.group_num = maxgroup
)

, groupLastHours as (
SELECT
group_num,
12 - SUM(num) AS lastHour

FROM
  groupHours
GROUP BY group_num
)

(
SELECT 1 as iteration, * 
FROM groupHours

UNION ALL

SELECT 2 iteration, 
Lastgroup.*
FROM Lastgroup
LEFT JOIN (
    SELECT 
    groupLastHours.group_num,
    Lastgroup.num,
    MIN(order_num) order_num,
      FROM groupLastHours
      LEFT JOIN Lastgroup ON lastHour = Lastgroup.num
      GROUP BY 
      groupLastHours.group_num,
    groupLastHours.lastHour,
    Lastgroup.num
) lsh ON Lastgroup.num = lsh.num
)
ORDER BY group_num, iteration

Answer 1

这不是一个智能分组，但它会累积创建组，每组的总和不超过 12 ...

with t as (
  select * from unnest([
    struct(1 as order_num, 2.0 as num),
    (2,4.5), (3,5.0), (4,3.5), (5,3.5), (6,0.5), (7,1.5), (8,4.5), (9,4.5), (10,1.5)
  ])
)

select * 
  ,sum(num) over (order by order_num) as cumusum
  ,cast(floor(sum(num) over (order by order_num)/12) as int64) as g
  ,sum(num) over (order by order_num) - cast(floor(sum(num) over (order by order_num)/12) as int64)*12 as g_fill_sum
from t

也许有人可以将其变成更好的解决方案......