Sql 最少的桶安装在元件中

我有一张桌子，上面有桶和元素，如下所示。 如果某个元素可以放入桶中，则在“资格”列中为1 例如：如果你看下面的数据，元素x可以放在bucket-a、b、c中，而不是d和e中

我想找到最少的桶来分组我的元素。 在这种情况下，bucket c和d可以将所有元素分组到两个bucket中

我能在bigquery中动态高效地完成这项工作吗？原始数据并不像这样简单

with matrix as (
---element x 

 select "element-x" as element, "bucketa" bucket , 1 eligibilty 
 union all 
 select "element-x" as element, "bucketb" bucket , 1 eligibilty 
 union all 
 select "element-x" as element, "bucketc" bucket , 1 eligibilty 
 union all 
 select "element-x" as element, "bucketd" bucket , 0 eligibilty 
 union all 
 select "element-x" as element, "buckete" bucket , 0 eligibilty 

union all
---element y 

 select "element-y" as element, "bucketa" bucket , 0 eligibilty 
 union all 
 select "element-y" as element, "bucketb" bucket , 0 eligibilty 
 union all 
 select "element-y" as element, "bucketc" bucket , 1 eligibilty 
 union all 
 select "element-y" as element, "bucketd" bucket , 0 eligibilty 
 union all 
 select "element-y" as element, "buckete" bucket , 0 eligibilty 
union all
---element z

 select "element-z" as element, "bucketa" bucket , 1 eligibilty 
 union all 
 select "element-z" as element, "bucketb" bucket , 0 eligibilty 
 union all 
 select "element-z" as element, "bucketc" bucket , 1 eligibilty 
 union all 
 select "element-z" as element, "bucketd" bucket , 0 eligibilty 
 union all 
 select "element-z" as element, "buckete" bucket , 0 eligibilty 
union all
---element p

 select "element-p" as element, "bucketa" bucket , 0 eligibilty 
 union all 
 select "element-p" as element, "bucketb" bucket , 0 eligibilty 
 union all 
 select "element-p" as element, "bucketc" bucket , 1 eligibilty 
 union all 
 select "element-p" as element, "bucketd" bucket , 0 eligibilty 
 union all 
 select "element-p" as element, "buckete" bucket , 0 eligibilty 
union all
---element q

 select "element-q" as element, "bucketa" bucket , 1 eligibilty 
 union all 
 select "element-q" as element, "bucketb" bucket , 0 eligibilty 
 union all 
 select "element-q" as element, "bucketc" bucket , 0 eligibilty 
 union all 
 select "element-q" as element, "bucketd" bucket , 1 eligibilty 
 union all 
 select "element-q" as element, "buckete" bucket , 0 eligibilty 

union all
---element r

 select "element-r" as element, "bucketa" bucket , 0 eligibilty 
 union all 
 select "element-r" as element, "bucketb" bucket , 1 eligibilty 
 union all 
 select "element-r" as element, "bucketc" bucket , 0 eligibilty 
 union all 
 select "element-r" as element, "bucketd" bucket , 1 eligibilty 
 union all 
 select "element-r" as element, "buckete" bucket , 1 eligibilty 



) 

下面应该有用

with buckets_elements as ( 
  select array[struct(a), struct(b), struct(c), struct(d), struct(e)] buckets
  from (
    select 
      array_agg(if(bucket = 'bucketa' and eligibilty = 1, element, null) ignore nulls) a,
      array_agg(if(bucket = 'bucketb' and eligibilty = 1, element, null) ignore nulls) b,
      array_agg(if(bucket = 'bucketc' and eligibilty = 1, element, null) ignore nulls) c,
      array_agg(if(bucket = 'bucketd' and eligibilty = 1, element, null) ignore nulls) d,
      array_agg(if(bucket = 'buckete' and eligibilty = 1, element, null) ignore nulls) e
    from matrix
  )
), columns_names as (
  select array_agg(bucket order by bucket) cols
  from (select distinct bucket from matrix)
), columns_index as (
  select generate_array(0, array_length(cols) - 1) as arr  
  from columns_names
), buckets_combinations as (
  select  
    (select array_agg(
      case when n & (1<<pos) <> 0 then arr[offset(pos)] end 
      ignore nulls)
     from unnest(generate_array(0, array_length(arr) - 1)) pos
    ) as combo
  from columns_index cross join 
  unnest(generate_array(1, cast(power(2, array_length(arr)) - 1 as int64))) n
)
select 
  array(select cols[offset(i)] from columns_names, unnest(combo) i) winners
from (
  select combo, 
    rank() over(order by (select count(distinct  el) from unnest(val) v, unnest(v.a) el) desc, array_length(combo)) as rnk
  from (
    select any_value(c).combo, array_agg(buckets[offset(i)]) val
    from buckets_combinations c, unnest(combo) i, buckets_elements b
    group by format('%t', c)
  )
)
where rnk = 1           

如果应用于y中的样本数据，我们的问题-输出为

---

注意：我只是重复使用了上一个问题的答案，只是更改/调整了bucket\u元素和columns\u名称CTE以反映新的模式。其余的都是一样的：o

什么样的数据逻辑“bucket-c”和“bucket-d”将所有元素分组？