BigQuery SQL - 根据多列的最大值创建新列

Question

I have a table contains info about customers and their purchases amount of each type of food. I want to create new columns that is the most freq type of food they have purchased. Is there an efficient way to do this?

I tried using case when and do one-to-one comparison, but it got very tedious.

Sample data:

Cust_ID	apple_type1	apple_type2	apple_type3	apple_type4	apple_type5	apple_type6
1	2	0	0	3	6	1
2	0	0	0	1	0	1
3	4	2	1	1	0	1
4	5	5	5	0	0	0
5	0	0	0	0	0	0

--WANT

Cust_ID	freq_apple_type_buy
1	type5
2	type4 and type6
3	type1
4	type1 and type2 and type3
5	unknown

Answer 1

This uses UNPIVOT to turn your columns in to rows. Then uses RANK() to assign each row a rank, which means if multiple rows are matched in quantity, they share the same rank.

It then selects only the products with rank=1 (possibly multiple rows, if multiple products are tied for first place)

WITH
  normalised_and_ranked AS
(
  SELECT
    cust_id,
    product,
    qty,
    RANK() OVER (PARTITION BY cust_id ORDER BY qty DESC) AS product_rank,
    ROW_NUMBER() OVER (PARTITION BY cust_id ORDER BY qty DESC) AS product_row
  FROM
     yourData
  UNPIVOT(
    qty FOR product IN (apple_type1, apple_type2, apple_type3, apple_type4, apple_type5, apple_type6)
  )
)
SELECT
  cust_id,
  CASE WHEN qty = 0 THEN NULL ELSE product END   AS product,
  CASE WHEN qty = 0 THEN NULL ELSE qty END   AS qty
FROM
  normalised_and_ranked
WHERE
  (product_rank = 1 AND qty > 0)
  OR
  (product_row = 1)

Edit: fudge added to ensure row of nulls returned if all qty are 0.

(Normally I'd just not return a row for such customers.)

Answer 2

Consider below approach

select Cust_ID, if(count(1) = any_value(all_count), 'unknown', string_agg(type, ' and ')) freq_apple_type_buy
from (
  select *, count(1) over(partition by Cust_ID) all_count
  from (
    select Cust_ID, replace(arr[offset(0)], 'apple_', '') type,cast(arr[offset(1)] as int64) value
    from data t,
    unnest(split(translate(to_json_string((select as struct * except(Cust_ID) from unnest([t]))), '{}"', ''))) kv,
    unnest([struct(split(kv, ':') as arr)])
  )
  where true qualify 1 = rank() over(partition by Cust_ID order by value desc)
)
group by Cust_ID    

if applied to sample data in your question - output is