Get amount of distinct values across multiple columns - google-bigquery

Question

I am using #standardSQL BigQuery and want to be able to count unique values across columns.
I want to count the number of distinct values across my user funnel per visit_id which are stored in the following columns: first_pgroup,second,third,forth,fifth,sixth,seventh,eighth,ninth,tenth.
I have the following 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
My goal is to be able to create a new column which states the unique values in each row.
The desired output is:
Here is my query:
> 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

Following approached worked based on Mikahils answer:
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

Answer 2

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

Answer 3

If you want to use just sql to solve this, below approach is much simple and works across multiple databases
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