最小的水桶，以适应元素 [英] Fewest buckets to fit in the elements

问题描述

我有一张桌子，上面有水桶和下面的元素.如果某个元素可以放入存储桶中，则在资格"列中为1例如:如果您查看下面的数据，则元素x可以容纳在存储区a，b，c中，而不能容纳在d和e中.

I have a table with buckets and elements like below. If an element can fit in a bucket it is 1 in the eligibility column For example: If you look at the data below, element x can fit in bucket-a,b,c and not in d and e

我想找到最少的存储桶来对元素进行分组.在这种情况下，存储桶c和d可以将所有元素归为两个存储桶.

I want to find the fewest buckets to group my elements. In this case, buckets c and d could group all the elements in just two buckets.

我能在bigquery中动态有效地做到这一点吗?原始数据并非如此简单.

Any idea if i can do this in bigquery dynamically and efficiently ? original data is not as simple as this.

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 



) 

推荐答案

下面应该可以工作

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           

如果应用于您的问题中的样本数据，则输出为

if applied to sample data in y our question - output is

注意:我只是重复使用上一个问题的答案，只是更改/调整了 buckets_elements 和 columns_names CTE以反映新的架构.其余的都完全相同:o)

Note: I simply reused answer for previous question and just changed / adjusted buckets_elements and columns_names CTEs to reflect new schema. All the rest is exactly the same :o)

这篇关于最小的水桶，以适应元素的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！