如何通过在 bigquery sql 中进行分组字符串比较来返回同一列中字符串值的差异？

Question

I have a table of products with a lot of products with an example like this:

product	brand
colgate smile 250gr	colgate
colgate fresh breath 250gr	colgate
colgate mint 250gr	colgate
relx pod pro mango - 1pod	relx
relx pod pro lychee - 1pod	relx
soju jinro chamisul green grape 360ml	jinro
soju jinro chamisul strawberry 360ml	jinro
soju jinro chamisul apple grape 360ml	jinro

into

product	brand	word
colgate smile 250gr	colgate	smile
colgate fresh breath 250gr	colgate	fresh breath
colgate mint 250gr	colgate	mint
relx pod pro mango - 1pod	relx	mango
relx pod pro lychee - 1pod	relx	lychee
soju jinro chamisul green grape 360ml	jinro	green grape
soju jinro chamisul strawberry 360ml	jinro	strawberry
soju jinro chamisul apple 360ml	jinro	apple

I want to group by brand and get difference in string and return that as new column. How do I do a transformation? and check for regexp_contains(str_1, str_2_split)=false and return the value?

Answer 1

Consider below naïve approach

• split product to distinct word
• identify words that are repeated in all rows of the same brand
• join back to original table and remove (replace with empty string) all such words
• whatever left - trim it and [optionally] replace occurrence of multiple space with just one space

So, query would look like below

with common_words as (
  select brand, 
    r'' || array_to_string(array(
      select word
      from t.words word
      group by word
      having count(*) = cnt
    ), '|') words
  from (
    select brand, count(*) cnt, array_concat_agg(words) words
    from (
      select brand, array(
          select distinct word
          from unnest(split(product, ' ')) word
        ) words
      from your_table
    )
    group by brand
  ) t
)
select product, brand, 
  regexp_replace(trim(regexp_replace(product, words, '')), r'\s+', ' ') as diff
from your_table
join common_words
using (brand)    

if applied to sample data in your question - output is