获取跨多个列的不同值的数量

Question

我正在使用#standardSQL BigQuery 并希望能够跨列计算唯一值。

我想计算每个 visit_id 我的用户漏斗中不同值的数量，这些值存储在以下列中：first_pgroup、second、third、forth、fifth、sixth、seventh、第八，第九，第十。

我有以下table:

visit_id	first_pgroup	second	third	forth	fifth	sixth	seventh	eighth	ninth	tenth
1	cloth	cloth	cloth	cloth	cloth	cloth	cloth	cloth	cloth	cloth
2	shoes	cloth	beauty
3	beauty
4	home&living	cloth	home&living	shoes	accessories
5	shoes	shoes	shoes	shoes	shoes	shoes

我的目标是能够创建一个新列来说明每行中的唯一值。

期望的输出是：

这是我的查询：

>        WITH
>             config AS (
>                 SELECT
>                 --SET time frame 
>                 DATE "2022-01-01" AS start_date,
>                 DATE "2022-01-01" AS end_date   ),
>         
>         
>         PDP_table AS (
>             SELECT
>             DISTINCT
>                 request_id AS request_id_PDP,
>                 eventType,
>                 SAFE_CAST(pid AS STRING) AS pid,
>                 SPLIT(pgroup , '/')[SAFE_OFFSET(0)] as product_group
>         
>             FROM
>                 `project.table`,
>                 config
>             WHERE
>                 DATE(PARTITIONTIME) BETWEEN config.start_date
>                 AND config.end_date
>                 AND pid IS NOT NULL
>                 and eventType = "view" ),
>         
>        table AS (   SELECT
>             timestamp,
>             visit_id,
>             request_id,   FROM
>             `project.table.2`,
>             config   WHERE
>             DATE(PARTITIONTIME) BETWEEN config.start_date
>               AND config.end_date
>               AND site IN ('Live2',
>                 'App'
>                 )AND country = 'de'),
>               raw as (   SELECT
>             cast((b.product_group) as string)  AS first_pgroup,
>             cast(LEAD( b.product_group, 1) OVER (PARTITION BY visit_id ORDER BY request_id) as string) AS second,    cast( LEAD(
> b.product_group, 2) OVER (PARTITION BY visit_id ORDER BY request_id)as
> string) AS third,
>             cast(LEAD( b.product_group, 3) OVER (PARTITION BY visit_id ORDER BY request_id)as string) AS forth,
>             cast(LEAD( b.product_group, 4) OVER (PARTITION BY visit_id ORDER BY request_id) as string) AS fifth,    cast( LEAD(
> b.product_group, 5) OVER (PARTITION BY visit_id ORDER BY request_id)
> as string) AS sixth,  cast( LEAD( b.product_group, 6) OVER (PARTITION
> BY visit_id ORDER BY request_id) as string) AS seventh,    cast( LEAD(
> b.product_group, 7) OVER (PARTITION BY visit_id ORDER BY request_id
> )as string) AS eighth, cast( LEAD( b.product_group, 8) OVER (PARTITION
> BY visit_id ORDER BY request_id) as string) AS ninth,
>             cast(LEAD( b.product_group, 9) OVER (PARTITION BY visit_id ORDER BY request_id) as string) AS tenth   FROM
>             table a   JOIN
>             PDP_table b   ON
>             b.request_id_PDP = a.request_id
>             )  
>              select *, (
>             select count(distinct category) - 1
>             from unnest(values(replace(to_json_string(t), 'null', '"null"'))) category
>             where category != 'null'   ) as category_count from raw

Answer 1

考虑以下方法

create temp function values(input string) returns array<string> language js as """
  return Object.values(JSON.parse(input));
""";
select *, (
    select count(distinct category) - 1
    from unnest(values(replace(to_json_string(t), 'null', '"null"'))) category
    where category != 'null'
  ) as category_count
from your_table t

如果应用于我们问题中的样本数据 - 输出是

Answer 2

如果您只想使用 sql 来解决这个问题，下面的方法非常简单并且适用于多个数据库

with main_table
as 
(select visit_id,first_group group_value from table
  union all
select visit_id,second_group from table
  union all
select visit_id,third_group from table
  union all
select visit_id,fourth_group from table
  union all
select visit_id,fifth_group from table
  union all
select visit_id,sixth_group from table
  union all
select visit_id,seventh_group from table)
select s.*,count(distinct m.group_value) distint_count
from main_table m
left join table s
   on m.visit_id=s.visit_id

Answer 3

以下方法基于 Mikahils 的回答：

select *,  (select count(distinct category) from
   unnest([first_pgroup,second,third,forth,fifth,sixth,seventh,eighth,ninth,tenth])
     as category  where category != 'null') as cat_count from raw

获取跨多个列的不同值的数量

Get amount of distinct values across multiple columns

google-bigquery