Question

我有一个家庭关系数据库：

with example_data as(
  SELECT 'abc' as relative_1, 'def' as relative_2
  union all
  SELECT 'abc' as relative_1, '123' as relative_2
  union all
  SELECT 'def' as relative_1, '334' as relative_2
  union all
  SELECT 'fdc' as relative_1, '123' as relative_2
  union all
  SELECT 'fgl' as relative_1, '342' as relative_2
)

如何基于这些数据创建完整的家族，以便获得以下结果：

我编写的用于创建所需输出的代码似乎根本不实际，实际上，对于一个有10万行输入的表，我的查询在第5个自联接之后达到了6小时的限制。我不担心最终可能会将整个表连接成一个长串-我知道一个家族中只有这么多家庭成员。

此外，如果可以将结果返回为嵌套表，其中list_all_relatives作为重复字段，并且在relative_1和相应的远距离亲戚之间步数最少，则效果会更好。

我的低效率代码会在图像中返回结果：

  SELECT 'abc' as relative_1, 'def' as relative_2, 'abc' as list_0
  union all
  SELECT 'abc' as relative_1, '123' as relative_2, 'abc' as list_0
  union all
  SELECT 'def' as relative_1, '334' as relative_2, 'def' as list_0
  union all
  SELECT 'fdc' as relative_1, '123' as relative_2, 'fdc' as list_0
  union all
  SELECT 'fgl' as relative_1, '342' as relative_2, 'fgl' as list_0
)
,
step_0 as (
 SELECT relative_1, relative_2,
        ARRAY_TO_STRING(ARRAY(SELECT DISTINCT x FROM UNNEST(SPLIT(concat(relative_1,',',relative_2,',',list_0), ',')) AS x ORDER BY x), ',') AS combined_list
  from raw_data
)
,
step_1 as (
SELECT relative_1, relative_2, list_1,
       ARRAY_TO_STRING(ARRAY(SELECT DISTINCT x FROM UNNEST(SPLIT(concat(relative_1,',',relative_2,',',list_1), ',')) AS x ORDER BY x), ',') AS combined_list
from 
    step_0
  left join
    (select relative_2,combined_list as list_1 from step_0)
  using(relative_2)
)
,
step_2 as (
SELECT distinct * except (combined_list,list_1), 
       ARRAY_TO_STRING(ARRAY(SELECT DISTINCT x FROM UNNEST(SPLIT(concat(combined_list,',',list_2), ',')) AS x ORDER BY x), ',') AS combined_list,
from 
    step_1
left join
    (select relative_1,combined_list as list_2 from step_1)
  using(relative_1) 
)
,
step_3 as (
SELECT distinct * except (combined_list,list_2), 
       ARRAY_TO_STRING(ARRAY(SELECT DISTINCT x FROM UNNEST(SPLIT(concat(combined_list,',',list_3), ',')) AS x ORDER BY x), ',') AS combined_list,
from 
    step_2
left join
    (select relative_2,combined_list as list_3 from step_2)
  using(relative_2) 
)
,
step_4 as (
SELECT distinct * except (combined_list,list_3), 
       ARRAY_TO_STRING(ARRAY(SELECT DISTINCT x FROM UNNEST(SPLIT(concat(combined_list,',',list_4), ',')) AS x ORDER BY x), ',') AS combined_list,
from 
    step_3
left join
    (select relative_1,combined_list as list_4 from step_3)
  using(relative_1) 
),

step_N as (
SELECT *
from step_4 
)
,
step_prefinal as (
SELECT distinct
        relative_1,list_4, combined_list, 
        1+length(REGEXP_REPLACE(REGEXP_REPLACE(LOWER(combined_list), '[a-z]', ''),'[0-9]','')) as n_elements_in_list,
        max(1+length(REGEXP_REPLACE(REGEXP_REPLACE(LOWER(combined_list), '[a-z]', ''),'[0-9]',''))) over (partition by relative_1) as longest_list_relatives
from step_N
)
,
step_final as (
SELECT relative_1, combined_list, n_elements_in_list,
       count(*) over (partition by relative_1) as cnt_lists_per_relative,
       max(n_elements_in_list) over (partition by relative_1) as max_elements
from 
  step_prefinal 
 where true
 and longest_list_relatives = n_elements_in_list
 group by 1,2,3
)
,
stats as (
SELECT cnt_lists_per_relative, count(distinct relative_1) as cnt,   
       max(max_elements) as max_elements
from 
step_final
group by 1
order by 1
)

SELECT relative_1, combined_list as list_all_relatives 
from step_final
where true

Answer 1

我能够创建代码以更简单的方式重现您的输出。

我已使用提供的example_data重现它。我已经使用UDF和JavaScript实现了所需的输出。下面是代码：

#Custom UDF to return a list of strings from an array
CREATE TEMP FUNCTION
  rel (relatives ARRAY<String>)
  RETURNS string
  LANGUAGE js AS '''
  return relatives;
''';

#provided data
WITH example_data AS(
  SELECT 'abc' AS relative_1, 'def' AS relative_2
  UNION ALL
  SELECT 'abc' AS relative_1, '123' AS relative_2
  UNION ALL
  SELECT 'def' AS relative_1, '334' AS relative_2
  UNION ALL
  SELECT 'fdc' AS relative_1, '123' AS relative_2
  UNION ALL
  SELECT 'fgl' AS relative_1, '342' AS relative_2
),

#Manipulating the data
data AS 
(
  SELECT
    t2.key,
    ARRAY_AGG(CAST(relative_2 AS string) ) AS relatives 
  FROM example_data t1
  LEFT JOIN ( SELECT DISTINCT relative_1 AS key FROM example_data
    GROUP BY relative_1) t2
  ON key=relative_1
  GROUP BY 1
  ORDER BY key 
)

#selecting the desired fields and using the UDF
SELECT key, rel(relatives) AS list_of_relatives FROM data
ORDER BY key

如您所见，第一步是创建一个UDF，它接收Strings的嵌套字段，并为每个“ Key ”简单地在字符串列表中返回它们。声明 example_data 后，在步骤2 中，必须执行一些数据操作。为了实现如下表：

如您所见，该表在嵌套列中具有 relative_1 （ 命名为键 ）和亲戚。

然后，选择所需的字段。这意味着键和 UDF 为 list_of_relatives ，它们是在第一步中编写的，输出为下方：

最后，请注意， list_of_relatives 不再是嵌套字段。而是一个 String ，每个值都用逗号分隔。如下所示：

Bigquery：根据亲属表格查找家族氏族

1 个答案: